6.1.2: Time Dependence - The Integrated Rate Law
- To apply rate laws to zeroth, first and second order reactions.
Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth , first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is t o know how to determine the reaction order from experimental data.
Zeroth-Order Reactions
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is
\[\text{rate} = k. \nonumber \]
We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1} \]
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of \(−k\). The value of \(k\) is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of \(k\), a positive value.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
\[[A] = [A]_0 − kt \label{14.4.2} \]
where \([A]_0\) is the initial concentration of reactant \(A\). Equation \ref{14.4.2} has the form of the algebraic equation for a straight line,
\[y = mx + b, \nonumber \]
with \(y = [A]\), \(mx = −kt\), and \(b = [A]_0\).)
In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N 2 O on a platinum (Pt) surface to produce N 2 and O 2 , which occurs at temperatures ranging from 200°C to 400°C:
\[\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3} \]
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N 2 O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N 2 O to react with the entire Pt surface, doubling or quadrupling the N 2 O concentration will have no effect on the reaction rate. At very low concentrations of N 2 O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N 2 O concentration. The reaction rate is as follows:
\[\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4} \]
Thus the rate at which N 2 O is consumed and the rates at which N 2 and O 2 are produced are independent of concentration. As shown in Figure \(\PageIndex{2}\) , the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N 2 O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where \ce{NAD^{ +}}\) (nicotinamide adenine dinucleotide) and \(\ce{NADH}\) (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero ( Figure \(\PageIndex{3a}\)). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure \(\PageIndex{3}\)).
These examples illustrate two important points:
- In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
- A linear change in concentration with time is a clear indication of a zeroth-order reaction.
First-Order Reactions
In a first-order reaction , the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5} \]
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s −1 ).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
\[[A] = [A]_0e^{−kt} \label{14.4.6} \]
where \([A]_0\) is the initial concentration of reactant \(A\) at \(t = 0\); \(k\) is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation \(\ref{14.4.6}\) predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation \(\ref{14.4.6}\) and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of \(A\) and \(t\):
\[\ln[A] = \ln[A]_0 − kt \label{14.4.7} \]
Because Equation \(\ref{14.4.7}\) has the form of the algebraic equation for a straight line,
\[y = mx + b, \nonumber \]
with \(y = \ln[A]\) and \(b = \ln[A]_0\), a plot of \(\ln[A]\) versus \(t\) for a first-order reaction should give a straight line with a slope of \(−k\) and an intercept of \(\ln[A]_0\). Either the differential rate law ( Equation \(\ref{14.4.5}\) ) or the integrated rate law ( Equation \(\ref{14.4.7}\) ) can be used to determine whether a particular reaction is first order.
First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure \(\PageIndex{5}\) is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure \(\PageIndex{4}\) have been studied extensively to find ways of maximizing the concentration of the active species.
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Ta ble \(\PageIndex{1}\). The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin.
| Experiment | [Cisplatin] 0 (M) | Initial Rate (M/min) |
|---|---|---|
| 1 | 0.0060 | 9.0 × 10 −6 |
| 2 | 0.012 | 1.8 × 10 −5 |
| 3 | 0.024 | 3.6 × 10 −5 |
| 4 | 0.030 | 4.5 × 10 −5 |
Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table \(\PageIndex{1}\) shows that the reaction rate doubles [(1.8 × 10 −5 M/min) ÷ (9.0 × 10 −6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10 −5 M/min) ÷ (9.0 × 10 −6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k [cisplatin] 1 . Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table \(\PageIndex{1}\). For example, substituting the values for Experiment 3 into Equation \(\ref{14.4.5}\) ,
3.6 × 10 −5 M/min = k (0.024 M)
1.5 × 10 −3 min −1 = k
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
\[\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber \]
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
| Experiment | [CH 3 CH 2 Cl] 0 (M) | Initial Rate (M/s) |
|---|---|---|
| 1 | 0.010 | 1.6 × 10 −8 |
| 2 | 0.015 | 2.4 × 10 −8 |
| 3 | 0.030 | 4.8 × 10 −8 |
| 4 | 0.040 | 6.4 × 10 −8 |
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
- Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
- Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH 3 CH 2 Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH 3 CH 2 Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k [CH 3 CH 2 Cl].
C We can calculate the rate constant ( k ) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10 −8 M/s = k (0.010 M)
1.6 × 10 −6 s −1 = k
Sulfuryl chloride (SO 2 Cl 2 ) decomposes to SO 2 and Cl 2 by the following reaction:
\[SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber \]
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
| Experiment | [SO 2 Cl 2 ] 0 (M) | Initial Rate (M/s) |
|---|---|---|
| 1 | 0.0050 | 1.10 × 10 −7 |
| 2 | 0.0075 | 1.65 × 10 −7 |
| 3 | 0.0100 | 2.20 × 10 −7 |
| 4 | 0.0125 | 2.75 × 10 −7 |
- Answer
- first order; k = 2.2 × 10 −5 s −1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure \(\PageIndex{6a}\) shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C.
The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure \(\PageIndex{6}\) . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure \(\PageIndex{6}\) for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
\[\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}}
\\[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}}
\\[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber \]
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min −1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure \(\PageIndex{6}\) are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Video Example Using the First-Order Integrated Rate Law Equation:
Example Using the First-Order Integrated Rate Law Equation(opens in new window) [youtu.be]
If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M ( k = 1.6 × 10 −6 s −1 ) ?
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
- Substitute values for the initial concentration ([A] 0 ) and the calculated rate constant for the reaction ( k ) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t .
- Given a concentration [A], solve the integrated rate law for time t .
Solution
The exponential form of the integrated rate law for a first-order reaction ( Equation \(\ref{14.4.6}\) ) is [A] = [A] 0 e − kt .
A Having been given the initial concentration of ethyl chloride ([A] 0 ) and having the rate constant of k = 1.6 × 10 −6 s −1 , we can use the rate law to calculate the concentration of the reactant at a given time t . Substituting the known values into the integrated rate law,
\[\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt}
\\[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}})
\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \]
We could also have used the logarithmic form of the integrated rate law ( Equation \(\ref{14.4.7}\) ):
\[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt
\\[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})]
\\[4pt] &=-3.912-0.0576=-3.970 \nonumber
\\[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber
\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \]
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for \(t\). Eq uation \(\ref{14.4.7}\) gives the following:
\[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt
\\[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t}
\\[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right)
\\[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber \]
In the exercise in Example \(\PageIndex{1}\), you found that the decomposition of sulfuryl chloride (\(\ce{SO2Cl2}\)) is first order, and you calculated the rate constant at 320°C.
- Use the form(s) of the integrated rate law to find the amount of \(\ce{SO2Cl2}\) that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C.
- How long would it take for 90% of the SO 2 Cl 2 to decompose?
- Answer a
- 0.0252 M
- Answer b
- 29 h
Second-Order Reactions
The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form
\[\ce{2A → products.}\nonumber \]
A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8} \]
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M −1 ·s −1 ). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
\[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9} \]
Because Equation \(\ref{14.4.9}\) has the form of an algebraic equation for a straight line, y = mx + b , with y = 1/[A] and b = 1/[A] 0 , a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A] 0 .
Second-order reactions generally have the form 2A → products or A + B → products.
Video Discussing the Second-Order Integrated Rate Law Equation: Second-Order Integrated Rate Law Equation(opens in new window) [youtu.be]
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO 2 to NO and O 2 and the decomposition of HI to I 2 and H 2 . Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
Figure \(\PageIndex{7}\)
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law ( Equation \(\ref{14.4.8}\) ) or the integrated rate law ( Equation \(\ref{14.4.9}\) ).
| Time (min) | [Monomer] (M) | Instantaneous Rate (M/min) |
|---|---|---|
| 10 | 0.0044 | 8.0 × 10 −5 |
| 26 | 0.0034 | 5.0 × 10 −5 |
| 44 | 0.0027 | 3.1 × 10 −5 |
| 70 | 0.0020 | 1.8 × 10 −5 |
| 120 | 0.0014 | 8.0 × 10 −6 |
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table \(\PageIndex{2}\) . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
\[\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber \]
Because (1.7) 2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
rate ∝ [monomer] 2
This means that the reaction is second order in the monomer. Using Equation \(\ref{14.4.8}\) and the data from any row in Table \(\PageIndex{2}\) , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
\[\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber \]
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in Figure \(\PageIndex{8a}\) . The measurements show that the concentration of the monomer (initially 5.4 × 10 −3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in Figure \(\PageIndex{8b}\) . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M −1 ·min −1 , which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
\[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber \]
Experimental data for the reaction at 300°C and four initial concentrations of NO 2 are listed in the following table:
| Experiment | [NO 2 ] 0 (M) | Initial Rate (M/s) |
|---|---|---|
| 1 | 0.015 | 1.22 × 10 −4 |
| 2 | 0.010 | 5.40 × 10 −5 |
| 3 | 0.0080 | 3.46 × 10 −5 |
| 4 | 0.0050 | 1.35 × 10 −5 |
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
- From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
- Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant ( k ).
Solution
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO 2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10 −5 ) ÷ (1.35 × 10 −5 ) = 4.0], which means that the reaction rate is proportional to [NO 2 ] 2 . Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO 2 ] 2 . This behavior is characteristic of a second-order reaction.
B We have rate = k [NO 2 ] 2 . We can calculate the rate constant ( k ) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
\[\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2
\\5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2
\\0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber \]
When the highly reactive species HO 2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
\[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber \]
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
| Experiment | [HO 2 ] 0 (M) | Initial Rate (M/s) |
|---|---|---|
| 1 | 1.1 × 10 −8 | 1.7 × 10 −7 |
| 2 | 2.5 × 10 −8 | 8.8 × 10 −7 |
| 3 | 3.4 × 10 −8 | 1.6 × 10 −6 |
| 4 | 5.0 × 10 −8 | 3.5 × 10 −6 |
Determine the reaction order and the rate constant.
- Answer
- second order in HO 2 ; k = 1.4 × 10 9 M −1 ·s −1
If a plot of reactant concentration versus time is not linear, but a plot of 1/(reactant concentration) versus time is linear, then the reaction is second order.
If a flask that initially contains 0.056 M NO 2 is heated at 300°C, what will be the concentration of NO 2 after 1.0 h? How long will it take for the concentration of NO 2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction ( Equation \ref{14.4.9} ) and the rate constant calculated above.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
- Given k , t , and [A] 0 , use the integrated rate law for a second-order reaction to calculate [A].
- Setting [A] equal to 1/10 of [A] 0 , use the same equation to solve for \(t\).
Solution
A We know k and [NO 2 ] 0 , and we are asked to determine [NO 2 ] at t = 1 h (3600 s). Substituting the appropriate values into Equation \ref{14.4.9} ,
\[\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \\[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \\[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber \]
Thus [NO 2 ] 3600 = 5.1 × 10 −4 M.
B In this case, we know k and [NO 2 ] 0 , and we are asked to calculate at what time [NO 2 ] = 0.1[NO 2 ] 0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation \(\ref{14.4.9}\) for t , using the concentrations given.
\[ \begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \\[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \\[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber \]
NO 2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
In the previous exercise, you calculated the rate constant for the decomposition of HO 2 as k = 1.4 × 10 9 M −1 ·s −1 . This high rate constant means that HO 2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO 2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO 2 , calculate the concentration of HO 2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO 2 to decompose? Use the integrated rate law for a second-order reaction ( Equation \(\ref{14.4.9}\) ) and the rate constant calculated in the exercise in Example \(\PageIndex{3}\).
- Answer
- 2.0 × 10 −13 M; 6.4 × 10 −6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form \(A + B \rightarrow products\), in which the reaction is first order in \(A\) and first order in \(B\). The differential rate law for this reaction is as follows:
\[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \nonumber \]
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.
Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
- zeroth-order reaction: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k \nonumber \] \[[A] = [A]_0 − kt \nonumber \]
- first-order reaction : \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \nonumber \] \[[A] = [A]_0e^{−kt} \nonumber \] \[\ln[A] = \ln[A]_0 − kt \nonumber \]
- second-order reaction: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2 \nonumber \] \[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \nonumber \]


6.3.2: Introduction
Like other cells, the osmotic environment surrounding microorganisms can affect them. Osmosis is the net movement of water molecules across a semi-permeable membrane, from an area of low solute (high water activity) concentration to higher solute concentration (lower water activity). Water molecules want to balance the solute concentration on either side of the membrane to a more isotonic state. Osmotic pressure is the physical force to draw water to one side of the membrane or other. In a hypertonic solution (more solutes outside of the cell, higher osmotic pressure) there will be net movement of water out of the cell (in order to create equal osmotic pressure on both sides of the membrane). Plasmolysis , cell shrinkage, may occur. In a hypotonic solution the opposite will happen. There are more solutes inside the cell relative to the solution and the net movement of water will be into the cell, which may result in lysis of the cell.
It’s important to remember that osmosis is relative to the two side of the membrane. When considering a microorganism and its environment, one should consider the inside of the cell relative to the environment. Is the environment hypertonic (higher solutes) or hypotonic (lower solutes) relative to the cell? This will determine the net movement of water.
Net movement of water
The bacterial cell wall helps those (with a cell wall) maintain their osmotic pressure in a fairly stable environment so that neither plasmolysis nor lysis will occur. Most bacteria live in a narrow range of salinity, usually less than 3% \(\ce{NaCl}\). Halophiles will grow in higher than 3% \(\ce{NaCl}\). Extreme halophiles require a very high salt concentration (up to 30%) and can be found in places like the Dead Sea, Great Salt Lake, etc. These organisms have proteins, enzymes, membranes, etc., that function in the high osmotic environment. In addition, some bacteria have adapted mechanisms to survive in a wider range of osmotic environments and are osmotolerant .


13.3: What Are The Roles of The Design Team?
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The project management approach to developing and delivering technology-based teaching and learning ensures that resources are used efficiently and that individual team members contribute appropriate skills and knowledge to the project”. (Bates, 2000, p. 68).
Overview of The Design Team
Online courses are designed using a variety of configurations. For quite some time, a very common approach focused on the single instructor acting as content expert, course writer, and designer. This approach is what has been popularly called the “Lone Ranger” or “laissezfaire” style (Bates, 2000). “Certainly, there is a time in an organization when the laissez-faire or Lone Ranger approach may be suitable, and that is when a university or college is just beginning to commit to the use of new technologies” (p. 66).
A number of factors favoured this approach to design, most notably, cost and workload issues. The ‘going it alone’ approach is still alive and well in the e-learning landscape, but some experts stress that the disadvantages of this method far outweigh the benefits. “It is too hit and miss. It wastes resources, ignores the experience and many lessons that have been learned outside the higher education sector about how to design and develop creative media products and services, and above all fails to ensure high-quality, technology-based teaching in any consistent or widespread form” (Bates, 2000, p. 66). On the other hand, there are expert instructors who do have the pedagogical, technical, and content expertise to create viable and high quality courses on their own (Struthers, 2002). However, in reality, there are several different configurations adopted by various institutions, ranging from the single-course author supported by information technology experts to the extensive project team approach described in this section.
Current instructional design and e-learning research and practice usually stress the need for a project team approach, where a diverse variety of experts work together to create high quality, pedagogically sound courses and programs. This project team can be made up of a number of people filling specific team roles, the most common include a project manager, content or subject matter expert, a content writer, a multimedia developer, an editor, and an instructional designer. Often, a concurrent instructional design approach is used, where each member works on their portion of the project simultaneously or “as needed”, creating a modulated, synergistic milieu for designing the course or program. For instance, once the content expert and writer have determined the desired topics and inherent content, the multimedia and/or graphic designer can begin to work on the supportive visual and multi-sensory content or learning objects to augment the foundational content.
There are some drawbacks to using the project team approach to course design. The biggest hurdle may well be teacher buy-in. Most faculty, especially in higher education, are used to functioning autonomously, and may be resistant to sharing the design of a course because of intellectual property considerations. “The project management approach is often seen as a bureaucratic, expensive, and unnecessarily complicated process, and a process that restricts the freedom and autonomy of the teacher” (Bates, 2000, p. 72).
Another possible drawback is the notion that project management can restrict the creativity and/or originality of the course designer. Obviously, there needs to be open communication between administration and the various members of the project team to be able to design a top quality course together successfully. As long as each member of the team is respected for their own expertise and contribution, and the issues of ownership and copyright are amicably decided, most teachers feel some relief that creative and knowledgeable team members support their efforts. Unless an individual course designer is multi-talented, with skills in content writing, editing, multimedia design, and so on, it is unlikely that a truly interactive, original, dynamic course can be created all alone.
Human Infrastructure
Four levels of human infrastructure support are fundamental to the development of any course or program, especially when done at an across-institutional, regional or national level (Bates, 2001). These include:
- technology infrastructure support people (design, maintain the learning network)
- educational technology infrastructure support people (design, maintain the learning interface structure such as navigation, screen components)
- instructional design infrastructure support people (coordinate the actual online course components and structure such as structure of learning activities or modules)
- subject expert infrastructure support people (design content, provide instruction).
Instructional Design Team Roles
Often, the human infrastructure needed to design a high quality course is best achieved by appointing a diverse instructional design team. Each member of the instructional design team fulfills specific roles.
Project manager
The project manager or leader often applies project management methodology to organize the project plan in conjunction with the rest of the design team. Often, the project manager liaisons with the instructional designer to set project start and end dates, determine what resources are needed to fulfill each project task, and set the project goals, challenges, milestones, and needs. The project manager is also responsible for ensuring that all team members are able to fulfill their tasks on time, and responds to challenges as they occur across the project timeline. The manager also coordinates copyright adherence and final details of the course project.
Instructional designer
The instructional designer is responsible for the course layout, branching, and positioning the written content within the online environment. Often the designer is involved with determining the course module or lesson objectives, the evaluative components, and may help the content writer and/or expert to develop the content. The instructional designer also works with the multimedia/graphics designer to determine the specific graphics, audio, video, movie and other multi-sensory components to augment the content. The role of coordination is often shared between the instructional designer and the project manager, to ensure consistency across the team, and to help identify problems and obstacles that emerge as the design process progresses.
Content or subject matter expert
The content expert is the team member who has well developed knowledge about the subject content. The content expert usually works very closely with the writer to ensure that the core essentials of the determined content are current, accurate, and meet the learning objectives of the course or program. The content expert also assesses the written content to verify that it addresses the intended audience, and, in conjunction with the instructional designer, helps to decide what multimedia and graphical objects are required to make the learning experience rich and meaningful for the learners.
Content writer
The content writer is the member who brings expertise in writing content for the course. Sometimes, one team member serves as both the content writer and subject matter expert. Their role entails researching the content, incorporating the input from the subject matter expert into the written component of the course (or sometimes, rewriting and editing existing content), and fashioning the content so that it suits the online course environment. The content writer works with the rest of the team to determine course and individual lesson objectives and other components, and selects the supportive materials such as text books and readings, usually with the content expert, instructional designer, and project manager.
Multimedia and graphics designer/technologist
The multimedia designer is responsible for designing the animations, visual graphics, audio segments, and other multi-sensory objects that will support the instructional requirements of the course. Working with all members of the team, especially the course writer, expert, and instructional designer, the multimedia designer helps to bring the course to life, providing a robustness and aesthetic appeal to the course design.
Editor or technical writer
The editor is responsible for ensuring that the content is well written and meets quality standards. The editor edits the course content for spelling, grammar, tone, and general usability, usually working closely with the content writer and the instructional designer.
“Communication is human nature. Knowledge sharing is human nurture”. – Alison Tucker, Buckman Laboratories.


25.0: Introduction
The human musculoskeletal system is composed of muscles, bones, and joints, and is most recognized for the framework and shape of the body. This system offers the ability to stand upright and to move. Protection is another factor offered by the musculoskeletal system, as critical organs and structures like the heart and lungs are enclosed within the thoracic cage, the brain inside the skull, and the spinal cord sheathed within the vertebral column. Hematopoiesis, or production of red and white blood cells and platelets, is another function of the musculoskeletal system. It is within the bone marrow where the production of these cells occurs. And finally, this body system supplies certain minerals as needed, as bones are the primary storage facility for the electrolytes calcium and phosphorous.
While diagnosis of musculoskeletal problems—injuries and disease processes—is within the realm of diagnosticians—physicians, nurse practitioners, and physician’s assistants, generalist nurses are expected to assess the musculoskeletal system and to recognize normal versus abnormal findings. As with other body systems, abnormal assessment results should be brought to the attention of the advanced practitioner for further diagnostic testing, a definitive diagnosis, and creation of the treatment plan.
There are problems that plague the musculoskeletal system in a variety of ways. Fractures, for example, may be problematic at any age, especially in consideration of traumatic accident or injury. Certain issues arise from birth and may be either congenital or have a genetic component; childbirth itself may lead to compression injuries or fractures of the skeletal system or certain muscles. The aging process impacts musculoskeletal structures, as degenerative changes cause deterioration of bones or atrophy of muscles. Generalist nurses are expected to use clinical judgment to assess and act on common musculoskeletal disorders and, once again, refer such patients, as well as those with unusual, rare conditions, to diagnosticians for diagnosis and treatment planning.


Ionization Constants of Weak Acids
| Ionization Constants of Weak Acids | |||
|---|---|---|---|
| Acid | Formula | K a at 25 °C | Lewis Structure |
| acetic | CH 3 CO 2 H | 1.8 × 10 −5 | |
| arsenic | H 3 AsO 4 | 5.5 × 10 −3 | |
| 1.7 × 10 −7 | |||
| 5.1 × 10 −12 | |||
| arsenous | H 3 AsO 3 | 5.1 × 10 −10 | |
| boric | H 3 BO 3 | 5.4 × 10 −10 | |
| carbonic | H 2 CO 3 | 4.3 × 10 −7 | |
| 5.6 × 10 −11 | |||
| cyanic | H CNO | 2 × 10 −4 | |
| formic | HCO 2 H | 1.8 × 10 −4 | |
| hydrazoic | H N 3 | 2.5 × 10 −5 | |
| hydrocyanic | H CN | 4.9 × 10 −10 | |
| hydrofluoric | H F | 3.5 × 10 −4 | |
| hydrogen peroxide | H 2 O 2 | 2.4 × 10 −12 | |
| hydrogen selenide | H 2 Se | 1.29 × 10 −4 | |
| H Se – | 1 × 10 −12 | ||
| hydrogen sulfate ion | 1.2 × 10 −2 | ||
| hydrogen sulfide | H 2 S | 8.9 × 10 −8 | |
| H S – | 1.0 × 10 −19 | ||
| hydrogen telluride | H 2 Te | 2.3 × 10 −3 | |
| H Te – | 1.6 × 10 −11 | ||
| hypobromous | H BrO | 2.8 × 10 −9 | |
| hypochlorous | H ClO | 2.9 × 10 −8 | |
| nitrous | H NO 2 | 4.6 × 10 −4 | |
| oxalic | H 2 C 2 O 4 | 6.0 × 10 −2 | |
| 6.1 × 10 −5 | |||
| phosphoric | H 3 PO 4 | 7.5 × 10 −3 | |
| 6.2 × 10 −8 | |||
| 4.2 × 10 −13 | |||
| phosphorous | H 3 PO 3 | 5 × 10 −2 | |
| 2.0 × 10 −7 | |||
| sulfurous | H 2 SO 3 | 1.6 × 10 −2 | |
| 6.4 × 10 −8 |


1.3: Modularizing the Code
1.3.1: Designing a Potential Function
We could continue altering the above code in a straightforward way. For example, we could add more particles by adding variables
x1
,
x2
,
q1
,
q2
, and so forth, and altering our formula for computing
phi
. However, this is not very satisfactory: each time we want to consider a new collection of particle positions or charges, or change the number of particles, we would have to re-write the program's internal "logic"—i.e., the part that computes the potentials. In programming terminology, our program is insufficiently "modular". Ideally, we want to isolate the part of the program that computes the potential from the part that specifies the numerical inputs to the calculation, like the positions and charges.
To modularize the code, let's define a function that computes the potential of an arbitrary set of charged particles, sampled at an arbitrary set of positions. Such a function would need three sets of inputs:
- An array of particle positions \(\vec{x}\equiv [x_{0}, \cdots, x_{N-1}]\). (Don't get confused, by the way: we are using these \(N\) numbers to refer to the positions of \(N\) particles in a 1D space, not the position of a single particle in an \(N\)-dimensional space.)
- An array of particle charges \(\vec{q}\equiv [q_{0}, \cdots, q_{N-1}]\).
- An array of sampling points \(\vec{X}\equiv [X_{0}, \cdots, X_{M-1}]\), which are the points where we want to know \(\phi(X)\).
The number of particles, \(N\), and the number of sampling points, \(M\), should be arbitrary positive integers. Furthermore, \(N\) and \(M\) need not be equal.
The function we intend to write must compute the array
\[\begin{bmatrix}\phi(X_{0})\\ \phi(X_{1}) \\ \vdots \\ \phi(X_{M-1})\end{bmatrix}\]
which contains the value of the total electric potential at each of the sampling points. The total potential can be written as the sum of contributions from all particles. Let us define \(\phi_{j}(x)\) as the potential produced by particle \(j\):
\[\phi_{j(x)}\equiv\frac{q_{j}}{|x-x_{j}|}\]
Then the total potential is
\[\begin{bmatrix}\phi(X_0)\\ \phi(X_1) \\ \vdots \\ \phi(X_{M-1})\end{bmatrix} = \begin{bmatrix}\phi_0(X_0)\\ \phi_0(X_1) \\ \vdots \\ \phi_0(X_{M-1})\end{bmatrix} + \begin{bmatrix}\phi_1(X_0)\\ \phi_1(X_1) \\ \vdots \\ \phi_1(X_{M-1})\end{bmatrix} + \cdots + \begin{bmatrix}\phi_{N-1}(X_0)\\ \phi_{N-1}(X_1) \\ \vdots \\ \phi_{N-1}(X_{M-1})\end{bmatrix}.\]
1.3.2 Writing the Program
Let's code this up. Return to the file
potentials.py
, and
delete the entire contents of the file
. Then replace it with the following:
from scipy import * import matplotlib.pyplot as plt ## Return the potential at measurement points X, due to particles ## at positions xc and charges qc. xc, qc, and X must be 1D arrays, ## with xc and qc of equal length. The return value is an array ## of the same length as X, containing the potentials at each X point. def potential(xc, qc, X): M = len(X) N = len(xc) phi = zeros(M) for j in range(N): phi += qc[j] / abs(X - xc[j]) return phi charges_x = array([0.2, -0.2]) charges_q = array([1.5, -0.1]) xplot = linspace(-3, 3, 500) phi = potential(charges_x, charges_q, xplot) plt.plot(xplot, phi) pmin, pmax = -50, 50 plt.ylim(pmin, pmax) plt.show()
When typing or pasting the above into your file, be sure to preserve the indentation (i.e., the number of spaces at the beginning of each line). Indentation is important in Python; as we'll see, it's used to determine program structure. Now save and run the program again:
-
In the Windows GUI, type
F5
in the editing window showingpotentials.py
. -
On GNU/Linux, type
python -i potentials.py
from the command line. -
Alternatively, from the Python command line, type
import potentials
, which will load and run yourpotentials.py
file.
You should now see a figure like the one on the right, showing the electric potential produced by two particles, one at position \(x_{0}=0.2\) with charge \(q_{0}=1.5\) and the other at position \(x_{1}=-0.2\) with charge \(q_{1}=-0.1\).
There are less than \(20\) lines of actual code in the above program, but they do quite a lot of things. Let's go through them in turn:
Module Imports
The first two lines import the Scipy and Matplotlib modules, for use in our program. We have not yet explained how importing works, so let's do that now.
from scipy import * import matplotlib.pyplot as plt
Each Python module, including Scipy and Matplotlib, defines a variety of functions and variables. If you use multiple modules, you might have a situation where, say, two different modules each define a function with the same name, but doing entirely different things. That would be Very Bad. To help avoid this, Python implements a concept called a namespace . Suppose you import a module (say Scipy) like this:
import scipy
One of the functions defined by Scipy is
linspace
, which we have already seen. This function was defined by the
scipy
module, and lies inside the
scipy
namespace. As a result, when you import the Scipy module using the
import scipy
line, you have to call the
linspace
function like this:
x = scipy.linspace(-3, 3, 500)
The
scipy.
in front says that you're referring to the
linspace
function that was defined in the
scipy
namespace. (Note: the online documentation for
linspace
refers to it as
numpy.linspace
, but the exact same function is also present in the
scipy
namespace. In fact, all
numpy.*
functions are replicated in the
scipy
namespace. So unless stated otherwise, we only have to import
scipy
.)
We will be using a lot of functions that are defined in the
scipy
namespace. Since it would be annoying to have to keep typing
scipy.
all over the place, we opt to use a slightly different import statement:
from scipy import *
This imports all the functions and variables in the
scipy
namespace directly into your program's namespace. Therefore, you can just call
linspace
, without the
scipy.
prefix. Obviously, you don't want to do this for every module you use, otherwise you'll end up with the name-clashing problem we alluded to earlier! The only module we'll use this shortcut with is
scipy
.
Another way to avoid having to type long prefixes is shown by this line:
import matplotlib.pyplot as plt
This imports the
matplotlib.pyplot
module (i.e., the
pyplot
module which is nested inside the
matplotlib
module). That's where
plot
,
show
, and other plotting functions are defined. The
as plt
in the above line says that we will refer to the
matplotlib.pyplot
namespace as the short form
plt
instead. Hence, instead of calling the
plot
function like this:
matplotlib.pyplot.plot(x, y)
we will call it like this:
plt.plot(x, y)
Comments
Let's return to the program we were looking at earlier. The next few lines, beginning with
#
, are "comments". Python ignores the
#
character and everything that follows it, up to the end of the line. Comments are very important, even in simple programs like this.
When you write your own programs, please remember to include comments. You don't need a comment for every line of code—that would be excessive—but at a minimum, each function should have a comment explaining what it does, and what the inputs and return values are.
Function Definition
Now we get to the
function definition
for the function named
potential
, which is the function that computes the potential:
def potential(xc, qc, X): M = len(X) N = len(xc) phi = zeros(M) for j in range(N): phi += qc[j] / abs(X - xc[j]) return phi
The first line, beginning with
def
, is a
function header
. This function header states that the function is named
potential
, and that it has three inputs. In computing terminology, the inputs that a function accepts are called
parameters
. Here, the parameters are named
xc
,
qc
and
X
. As explained by the comments, we intend to use these for the positions of the particles, the charges of the particles, and the positions at which to measure the potential, respectively.
The function definition consists of the function header, together the rest of the indented lines below it. The function definition terminates once we get to a line which is at the same indentation level as the function header. (That terminating line is considered a separate line of code, which is not part of the function definition).
By convention, you should use 4 spaces per indentation level.
The indented lines below the function header are called the function body . This is the code that is run each time the function is called. In this case, the function body consists of six lines of code, which are intended to compute the total electric potential, according to the procedure that we have outlined in the preceding section:
-
The first two lines define two helpful variables,
M
andN
. Their values are set to the lengths of theX
andxc
arrays, respectively. -
The next line calls the
zeros
function. The input tozeros
isM
, the length of theX
array (i.e., our function's third parameter). Therefore,zeros
returns an array, of the same same length asX
, with every element set to 0.0. For now, this represents the electric potential in the absence of any charges. We give this array the namephi
. -
The function then iterates over each of the particles and add up its contribution to the potential, using a construct known as a
for
loop . The codefor j in range(N):
is the loop's "header line", and the next line, indented 4 spaces deeper than the header line, is the "body" of the loop.The header line states that we should run the loop body several times, with the variable
j
set to different values during each run. The values ofj
to loop over are given byrange(N)
. This is a function call to the range function, withN
(the number of electric charges) as the input. Therange(N)
function call returns a sequence specifyingN
successive integers, from 0 toN-1
, inclusive. (Note that the last value in the sequence isN-1
, notN
. Because we start from 0, this means that there is a total ofN
integers in the sequence. Also, callingrange(N)
is the same as callingrange(0,N)
.) -
For each
j
, we computeqc[j] / abs(X - xc[j])
. This is an array whose elements are the values of the electric potential at the set of positionsX
, arising from the individual particle\(j\). In mathematical terms, we are calculating
\[\phi_j(X) \equiv \frac{q_j}{|X - x_j|}\]using the array of positions
X
. We then add this array tophi
. Once this is done for allj
, the arrayphi
will contain the desired total potential,
\[\phi(X) = \sum_{j=0}^{N-1}\phi_j(X).\] -
Finally, we call
return
to specify the function's output, or return value . This is the arrayphi
.
Top-Level Code: Numerical Constants
After the function definition comes the code to use the function:
charges_x = array([0.2, -0.2]) charges_q = array([1.5, -0.1]) xplot = linspace(-3, 3, 500)
Like the import statements at the beginning of the program, these lines of code lie at
top level
, i.e., they are not indented. The function header which defines the
potential
function is also at top level. Running a Python program consists of running its top level code, in sequence.
The above lines define variables to store some numerical constants. In the first two lines,
charges_x
and
charges_q
variables store the numerical values of the positions and charges we are interested in. These are initialized using the
array
function. You may be wondering why the
array
function call has square brackets nested in commas. We'll explain later, in part 2 of the tutorial.
On the third line, the
linspace
function call returns an array, whose contents are initialized to the 500 numbers between -3 and 3 (inclusive).
Next, we call the
potential
function, passing
charges_x
,
charges_q
and
xplot
as the inputs:
phi = potential(charges_x, charges_q, xplot)
These inputs provide the values of the function definition's parameters
xc
,
qc
, and
X
respectively. The return value of the function call is an array containing the total potential, evaluated at each of the positions specified in
xplot
. This return value is stored as the variable named
phi
.
Plotting
Finally, we create the plot:
plt.plot(xplot, phi) pmin, pmax = -50, 50 plt.ylim(pmin, pmax) plt.show()
We have already seen how the
plot
and
show
functions work. Here, prior to calling
plt.show
, we have added two extra lines to make the potential curve is more legible, by adjust the plot's y-axis bounds. The
ylim
function accepts two parameters, the lower and upper bounds of the y-axis. In this case, we set the bounds to -50 and 50 respectively. There is an
xlim
function to do the same for the x-axis.
Notice that in the line
pmin, pmax = -50, 50
, we set two variables (
pmin
and
pmax
) on the same line. This is a little "syntactic sugar" to make the code a little easier to read. It's equivalent to having two separate lines, like this:
pmin = -50 pmax = 50
We'll explain how this construct works in the next part of the tutorial.


11: Motivation Last updated Save as PDF Page ID 47570 Lumen Learning Lumen Learning 11.1: Introduction to Goal-Setting Theory 11.2: Goal-Setting Theory 11.3: Introduction to Reinforcement Theory 11.4: Reinforcement Theory 11.5: Introduction to a Manager’s Role in Motivating Employees 11.6: Manager’s Role in Promoting Motivation 11.7: Putting It Together- Motivation 11.8: Assignment- The “Chief Motivator” at Sun City Boards 11.9: Assignment- Motivation 11.10: Discussion- Sun City Boards Employee Rewards 11.11: Why It Matters- Motivation 11.12: Discussion- Rethinking Motivation 11.13: Introduction to the Importance of Employee Motivation 11.14: The Importance of Employee Motivation 11.15: Introduction to Needs-Based Theories of Motivation 11.16: Needs-Based Theories of Motivation 11.17: Introduction to Process-Based Theories of Motivation 11.18: Process-Based Theories of Motivation 11.19: Introduction to Job Characteristics that Affect Motivation 11.20: Job Characteristics that Affect Motivation


8.4: Meteorites
Meteorites
I would more easily believe that (a) Yankee professor would lie than that stones would fall from heaven. — Thomas Jefferson, 1808
Most meteoritic material that strikes the Earth’s atmosphere does not make it to the surface to be recovered. The meteor burns up, or if it makes it to the surface, falls into water, since Earth is a water surface planet.
There is more gold mined each year than the total amount of meteorites found in all collections – museums, universities, and private collections. So, how does one recognize if a rock is potentially a meteorite?
- The rock attracts a magnet — most meteorites have some iron in them. Not all, though.
- The rock looks burned , called a fusion crust. Yet, if the meteorite has been sitting in a desert region for a while, the fusion crust might be weathered away.
- The rock looks shaped , like it’s been oriented as it came through the atmosphere. Yet again, this aerodynamic shaping might have been weathered away.
- The rock looks like it has thumbprint-like impressions , called regmaglypts. These impressions are formed as the meteorite comes through the atmosphere, and material is ablated away from the meteorite.
The three major groups or classes of meteorites, based on their make-up:
- Stones or Stony Meteorites
- These are composed of silicates and other materials. Most stony meteorites contain metal in the form of iron and nickel. Plus, stone meteorites are by far the most common type; some 94% of recovered meteorites are stony. Stone meteorites also include very rare types of meteorites, including lunar, Martian, and possibly even comets.
- Iron meteorites, or Irons
- These are composed of iron (Fe) and nickel (Ni), with some cobalt (Co) and other trace elements and minerals. Iron is the predominant metal, and the type of Iron meteorite depends on the ratio of iron to nickel. No Earth rock has iron and nickel in combination. Thus a ‘rock’ recovered with iron and nickel together is a meteorite. Irons are very dense meteorites, in comparison to stony meteorites.
- Stony Irons
- As the name indicates, they are a combination of iron and stony meteorite properties and the rarest type or class of meteorite.
- Tektites
- Very dry glasses, and are hypothesized to be meteoritic splash from a large impact. The meteorite strikes Earth, and soil is blown up into the air, almost back into space. Yet, the escape velocity is not high enough, so the now-molten material re-enters and lands, found as the glassy material found in specific areas on Earth.
Click on the image to view the video.
Content by Florida State College at Jacksonville is licensed under a Creative Commons Attribution 4.0 International License. ]
Rocks which are not meteorites are referred to as meteorwrongs by those who work in the field. People often find an unusual-looking rock and think it is a meteorite. After testing, which will include the four steps noted and/or chemical testing, many of these are meteorwrongs.


17.4: The Loss of American Indian Life and Culture
As American settlers pushed westward, they inevitably came into conflict with Indian tribes that had long been living on the land. Although the threat of Indian attacks was quite slim and nowhere proportionate to the number of U.S. Army actions directed against them, the occasional attack—often one of retaliation—was enough to fuel the popular fear of the “savage” Indians. The clashes, when they happened, were indeed brutal, although most of the brutality occurred at the hands of the settlers. Ultimately, the settlers, with the support of local militias and, later, with the federal government behind them, sought to eliminate the tribes from the lands they desired. The result was devastating for the Indian tribes, which lacked the weapons and group cohesion to fight back against such well-armed forces. The Manifest Destiny of the settlers spelled the end of the Indian way of life.
CLAIMING LAND, RELOCATING LANDOWNERS
Back east, the popular vision of the West was of a vast and empty land. But of course this was an exaggerated depiction. On the eve of westward expansion, as many as 250,000 Indians, representing a variety of tribes, populated the Great Plains. Previous wars against these tribes in the early nineteenth century, as well as the failure of earlier treaties, had led to a general policy of the forcible removal of many tribes in the eastern United States. The Indian Removal Act of 1830 resulted in the infamous “Trail of Tears,” which saw nearly fifty thousand Seminole, Choctaw, Chickasaw, and Creek Indians relocated west of the Mississippi River to what is now Oklahoma between 1831 and 1838. Building upon such a history, the U.S. government was prepared, during the era of western settlement, to deal with tribes that settlers viewed as obstacles to expansion.
As settlers sought more land for farming, mining, and cattle ranching, the first strategy employed to deal with the perceived Indian threat was to negotiate settlements to move tribes out of the path of white settlers. In 1851, the chiefs of most of the Great Plains tribes agreed to the First Treaty of Fort Laramie. This agreement established distinct tribal borders, essentially codifying the reservation system. In return for annual payments of $50,000 to the tribes (originally guaranteed for fifty years, but later revised to last for only ten) as well as the hollow promise of noninterference from westward settlers, Indians agreed to stay clear of the path of settlement. Due to government corruption, many annuity payments never reached the tribes, and some reservations were left destitute and near starving. In addition, within a decade, as the pace and number of western settlers increased, even designated reservations became prime locations for farms and mining. Rather than negotiating new treaties, settlers—oftentimes backed by local or state militia units—simply attacked the tribes out of fear or to force them from the land. Some Indians resisted, only to then face massacres.
In 1862, frustrated and angered by the lack of annuity payments and the continuous encroachment on their reservation lands, Dakota Sioux Indians in Minnesota rebelled in what became known as the Dakota War, killing the white settlers who moved onto their tribal lands. Over one thousand white settlers were captured or killed in the attack, before an armed militia regained control. Of the four hundred Sioux captured by U.S. troops, 303 were sentenced to death, but President Lincoln intervened, releasing all but thirty-eight of the men. The thirty-eight who were found guilty were hanged in the largest mass execution in the country’s history, and the rest of the tribe was banished. Settlers in other regions responded to news of this raid with fear and aggression. In Colorado, Arapahoe and Cheyenne tribes fought back against land encroachment; white militias then formed, decimating even some of the tribes that were willing to cooperate. One of the more vicious examples was near Sand Creek, Colorado, where Colonel John Chivington led a militia raid upon a camp in which the leader had already negotiated a peaceful settlement. The camp was flying both the American flag and the white flag of surrender when Chivington’s troops murdered close to one hundred people, the majority of them women and children, in what became known as the Sand Creek Massacre.  Subsequent investigations by the U.S. Army condemned Chivington’s tactics and their results; however, the raid served as a model for some settlers who sought any means by which to eradicate the perceived Indian threat.
Hoping to forestall similar uprisings and all-out Indian wars, the U.S. Congress commissioned a committee to investigate the causes of such incidents. The subsequent report of their findings led to the passage of two additional treaties: the Second Treaty of Fort Laramie and the Treaty of Medicine Lodge Creek, both designed to move the remaining tribes to even more remote reservations. The Second Treaty of Fort Laramie moved the remaining Sioux to the Black Hills in the Dakota Territory and the Treaty of Medicine Lodge Creek moved the Cheyenne, Arapaho, Kiowa, and Comanche to “Indian Territory,” later to become the State of Oklahoma.
The agreements were short-lived, however. With the subsequent discovery of gold in the Black Hills, settlers seeking their fortune began to move upon the newly granted Sioux lands with support from U.S. cavalry troops. By the middle of 1875, thousands of white prospectors were illegally digging and panning in the area. The Sioux protested the invasion of their territory and the violation of sacred ground. The government offered to lease the Black Hills or to pay $6 million if the Indians were willing to sell the land. When the tribes refused, the government imposed what it considered a fair price for the land, ordered the Indians to move, and in the spring of 1876, made ready to force them onto the reservation.
In the Battle of Little Bighorn, perhaps the most famous battle of the American West, a Sioux chieftain, Sitting Bull, urged Indians from all neighboring tribes to join his men in defense of their lands (Figure 17.4.1). At the Little Bighorn River, the U.S. Army’s Seventh Cavalry, led by Colonel George Custer, sought a showdown. Driven by his own personal ambition, on June 25, 1876, Custer foolishly attacked what he thought was a minor Indian encampment. Instead, it turned out to be the main Sioux force. The Sioux warriors—nearly three thousand in strength—surrounded and killed Custer and 262 of his men and support units, in the single greatest loss of U.S. troops to an Indian attack in the era of westward expansion. Eyewitness reports of the attack indicated that the victorious Sioux bathed and wrapped Custer’s body in the tradition of a chieftain burial; however, they dismembered many other soldiers’ corpses in order for a few distant observers from Major Marcus Reno’s wounded troops and Captain Frederick Benteen’s company to report back to government officials about the ferocity of the Sioux enemy.
AMERICAN INDIAN SUBMISSION
Despite their success at Little Bighorn, neither the Sioux nor any other Plains tribe followed this battle with any other armed encounter. Rather, they either returned to tribal life or fled out of fear of remaining troops, until the U.S. Army arrived in greater numbers and began to exterminate Indian encampments and force others to accept payment for forcible removal from their lands. Sitting Bull himself fled to Canada, although he later returned in 1881 and subsequently worked in Buffalo Bill’s Wild West show. In Montana, the Blackfoot and Crow were forced to leave their tribal lands. In Colorado, the Utes gave up their lands after a brief period of resistance. In Idaho, most of the Nez Perce gave up their lands peacefully, although in an incredible episode, a band of some eight hundred Indians sought to evade U.S. troops and escape into Canada.
Chief Joseph, known to his people as “Thunder Traveling to the Loftier Mountain Heights,” was the chief of the Nez Perce tribe, and he had realized that they could not win against the whites. In order to avoid a war that would undoubtedly lead to the extermination of his people, he hoped to lead his tribe to Canada, where they could live freely. He led a full retreat of his people over fifteen hundred miles of mountains and harsh terrain, only to be caught within fifty miles of the Canadian border in late 1877. His speech has remained a poignant and vivid reminder of what the tribe had lost.
Tell General Howard I know his heart. What he told me before, I have it in my heart. I am tired of fighting. Our Chiefs are killed; Looking Glass is dead, Ta Hool Hool Shute is dead. The old men are all dead. It is the young men who say yes or no. He who led on the young men is dead. It is cold, and we have no blankets; the little children are freezing to death. My people, some of them, have run away to the hills, and have no blankets, no food. No one knows where they are—perhaps freezing to death. I want to have time to look for my children, and see how many of them I can find. Maybe I shall find them among the dead. Hear me, my Chiefs!  From where the sun now stands I will fight no more forever.
—Chief Joseph, 1877
The final episode in the so-called Indian Wars occurred in 1890, at the Battle of Wounded Knee in South Dakota. On their reservation, the Sioux had begun to perform the “Ghost Dance,” which told of an Indian Messiah who would deliver the tribe from its hardship, with such frequency that white settlers began to worry that another uprising would occur. The militia prepared to round up the Sioux. The tribe, after the death of Sitting Bull, who had been arrested, shot, and killed in 1890, prepared to surrender at Wounded Knee, South Dakota, on December 29, 1890. Although the accounts are unclear, an apparent accidental rifle discharge by a young male Indian preparing to lay down his weapon led the U.S. soldiers to begin firing indiscriminately upon the Indians. What little resistance the Indians mounted with a handful of concealed rifles at the outset of the fight diminished quickly, with the troops eventually massacring between 150 and 300 men, women, and children. The U.S. troops suffered twenty-five fatalities, some of which were the result of their own crossfire. Captain Edward Godfrey of the Seventh Cavalry later commented, “I know the men did not aim deliberately and they were greatly excited. I don’t believe they saw their sights. They fired rapidly but it seemed to me only a few seconds till there was not a living thing before us; warriors, squaws, children, ponies, and dogs . . . went down before that unaimed fire.” With this last show of brutality, the Indian Wars came to a close. U.S. government officials had already begun the process of seeking an alternative to the meaningless treaties and costly battles. A more effective means with which to address the public perception of the “Indian threat” was needed.Americanization provided the answer.
AMERICANIZATION
Through the years of the Indian Wars of the 1870s and early 1880s, opinion back east was mixed. There were many who felt, as General Philip Sheridan (appointed in 1867 to pacify the Plains Indians) allegedly said, that the only good Indian was a dead Indian. But increasingly, several American reformers who would later form the backbone of the Progressive Era had begun to criticize the violence, arguing that the Indians should be helped through “Americanization” to become assimilated into American society. Individual land ownership, Christian worship, and education for children became the cornerstones of this new, and final, assault on Indian life and culture.
Beginning in the 1880s, clergymen, government officials, and social workers all worked to assimilate Indians into American life. The government permitted reformers to remove Indian children from their homes and place them in boarding schools, such as the Carlisle Indian School or the Hampton Institute, where they were taught to abandon their tribal traditions and embrace the tools of American productivity, modesty, and sanctity through total immersion. Such schools not only acculturated Indian boys and girls, but also provided vocational training for males and domestic science classes for females. Adults were also targeted by religious reformers, specifically evangelical Protestants as well as a number of Catholics, who sought to convince Indians to abandon their language, clothing, and social customs for a more Euro-American lifestyle (Figure 17.4.2).
A vital part of the assimilation effort was land reform. During earlier negotiations, the government had respected that the Indian tribes used their land communally. Most Indian belief structures did not allow for the concept of individual land ownership; rather, land was available for all to use, and required responsibility from all to protect it. As a part of their plan to Americanize the tribes, reformers sought legislation to replace this concept with the popular Euro-American notion of real estate ownership and self-reliance. One such law was the Dawes Severalty Act of 1887, named after a reformer and senator from Massachusetts, which struck a deadly blow to the Indian way of life. In what was essentially an Indian version of the original Homestead Act, the Dawes Act permitted the federal government to divide the lands of any tribe and grant 160 acres of farmland or 320 acres of grazing land to each head of family, with lesser amounts to others. In a nod towards the paternal relationship with which whites viewed Indians—similar to the justification of the previous treatment of African American slaves—the Dawes Act permitted the federal government to hold an individual Indian’s newly acquired land in trust for twenty-five years. Only then would he obtain full title and be granted the citizenship rights that land ownership entailed. It would not be until 1924 that formal citizenship was granted to all Native Americans. Under the Dawes Act, Indians were given the most arid, useless land. Further, inefficiencies and corruption in the government meant that much of the land due to be allotted to Indians was simply deemed “surplus” and claimed by settlers. Once all allotments were determined, the remaining tribal lands—as much as eighty million acres—were sold to white American settlers.
The final element of “Americanization” was the symbolic “last arrow” pageant, which often coincided with the formal redistribution of tribal lands under the Dawes Act. At these events, Indians were forced to assemble in their tribal garb, carrying a bow and arrow. They would then symbolically fire their “last arrow” into the air, enter a tent where they would strip away their Indian clothing, dress in a white farmer’s coveralls, and emerge to take a plow and an American flag to show that they had converted to a new way of life. It was a seismic shift for the Indians, and one that left them bereft of their culture and history.
Take a look at the Carlisle Industrial Indian School where Indian students were “civilized” from 1879 to 1918. It is worth looking through the photographs and records of the school to see how this well-intended program obliterated Indian culture.
Section Summary
The interaction of the American Indians with white settlers during the western expansion movement was a painful and difficult one. For settlers raised on the notion of Manifest Destiny and empty lands, the Indians added a terrifying element to what was already a difficult and dangerous new world. For the Indians, the arrival of the settlers meant nothing less than the end of their way of life. Rather than cultural exchange, contact led to the virtual destruction of Indian life and culture. While violent acts broke out on both sides, the greatest atrocities were perpetrated by whites, who had superior weapons and often superior numbers, as well as the support of the U.S. government.
The death of the Indian way of life happened as much at the hands of well-intentioned reformers as those who wished to see the Indians exterminated. Individual land ownership, boarding schools, and pleas to renounce Indian gods and culture were all elements of the reformers’ efforts. With so much of their life stripped away, it was ever more difficult for the Indians to maintain their tribal integrity.
Review Questions
Which of the following was not a primary method by which the American government dealt with American Indians during the period of western settlement?
relocation
appeasement
extermination
assimilation
B
What did the Last Arrow pageant symbolize?
the continuing fight of the Indians
the total extermination of the Indians from the West
the final step in the Americanization process
the rebellion at Little Bighorn
C
Glossary
- Americanization
- the process by which an Indian was “redeemed” and assimilated into the American way of life by changing his clothing to western clothing and renouncing his tribal customs in exchange for a parcel of land
- Battle of Wounded Knee
- an attempt to disarm a group of Lakota Sioux Indians near Wounded Knee, South Dakota, which resulted in members of the Seventh Cavalry of the U.S. Army opening fire and killing over 150 Indians
- Sand Creek Massacre
- a militia raid led by Colonel Chivington on an Indian camp in Colorado, flying both the American flag and the white flag of surrender; over one hundred men, women, and children were killed


2.6: Algebra of Vectors
- Apply analytical methods of vector algebra to find resultant vectors and to solve vector equations for unknown vectors.
- Interpret physical situations in terms of vector expressions.
Vectors can be added together and multiplied by scalars. Vector addition is associative ( Equation 2.2.8 ) and commutative ( Equation 2.2.7 ), and vector multiplication by a sum of scalars is distributive ( Equation 2.2.9 ). Also, scalar multiplication by a sum of vectors is distributive:
\[\alpha(\vec{A} + \vec{B}) = \alpha \vec{A} + \alpha{B} \ldotp \label{2.22}\]
In this equation, \(\alpha\) is any number (a scalar). For example, a vector antiparallel to vector \(\vec{A}\) = A x \(\hat{i}\) + A y \(\hat{j}\) + A z \(\hat{k}\) can be expressed simply by multiplying \(\vec{A}\) by the scalar \(\alpha\) = −1:
\[- \vec{A} = A_{x} \hat{i} - A_{y} \hat{j} - A_{z} \hat{k} \ldotp \label{2.23}\]
In a Cartesian coordinate system where \(\hat{i}\) denotes geographic east, \(\hat{j}\) denotes geographic north, and \(\hat{k}\) denotes altitude above sea level, a military convoy advances its position through unknown territory with velocity \(\vec{v}\) = (4.0 \(\hat{i}\) + 3.0 \(\hat{j}\) + 0.1 \(\hat{k}\)) km/h. If the convoy had to retreat, in what geographic direction would it be moving?
Solution
The velocity vector has the third component \(\vec{v}_{z}\) = (+ 0.1 km/h) \(\hat{k}\), which says the convoy is climbing at a rate of 100 m/h through mountainous terrain. At the same time, its velocity is 4.0 km/h to the east and 3.0 km/h to the north, so it moves on the ground in direction tan −1 (3 /4) ≈ 37° north of east. If the convoy had to retreat, its new velocity vector \(\vec{u}\) would have to be antiparallel to \(\vec{v}\) and be in the form \(\vec{u} = - \alpha \vec{v}\), where \(\alpha\) is a positive number. Thus, the velocity of the retreat would be \(\vec{u}\) = \(\alpha\)(−4.0 \(\hat{i}\) − 3.0 \(\hat{j}\) − 0.1 \(\hat{k}\)) km/h. The negative sign of the third component indicates the convoy would be descending. The direction angle of the retreat velocity is tan −1 (−3 \(\alpha\) − 4 \(\alpha\)) ≈ 37° south of west. Therefore, the convoy would be moving on the ground in direction 37° south of west while descending on its way back.
The generalization of the number zero to vector algebra is called the null vector , denoted by \(\vec{0}\). All components of the null vector are zero, \(\vec{0}\) = 0 \(\hat{i}\)+ 0 \(\hat{j}\) + 0 \(\hat{k}\), so the null vector has no length and no direction.
Two vectors \(\vec{A}\) and \(\vec{B}\) are equal vectors if and only if their difference is the null vector: \(\vec{0}\) = \(\vec{A}\) − \(\vec{B}\) = (A x \(\hat{i}\) + A y \(\hat{j}\) + A z \(\hat{k}\)) − (B x \(\hat{i}\) + B y \(\hat{j}\) + B z \(\hat{k}\)) = (A x − B x ) \(\hat{i}\) + (A y − B y ) \(\hat{j}\) + (A z − B z ) \(\hat{k}\). This vector equation means we must have simultaneously A x − B x = 0, A y − B y = 0, and A z − B z = 0. Hence, we can write \(\vec{A} = \vec{B}\) if and only if the corresponding components of vectors \(\vec{A}\) and \(\vec{B}\) are equal:
\[ \vec{A} = \vec{B} \Leftrightarrow \begin{cases} A_{x} = B_{x} \\ A_{y} = B_{y} \\ A_{z} = B_{z} \end{cases} \ldotp \label{2.24}\]
Two vectors are equal when their corresponding scalar components are equal. Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in vector component form (given by Equation 2.5.4 ) allows us to use vector algebra to find sums or differences of many vectors analytically (i.e., without using graphical methods). For example, to find the resultant of two vectors \(\vec{A}\) and \(\vec{B}\), we simply add them component by component, as follows:
\[\vec{R} = \vec{A} + \vec{B} = (A_{x}\; \hat{i} + A_{y}\; \hat{j} + A_{z}\; \hat{k}) + (B_{x}\; \hat{i} + B_{y}\; \hat{j} + B_{z}\; \hat{k}) = (A_{x} + B_{x})\; \hat{i} + (A_{y} + B_{y})\; \hat{j} + (A_{z} + B_{z})\; \hat{k} \ldotp\]
In this way, using Equation \ref{2.24}, scalar components of the resultant vector \(\vec{R}\) = R x \(\hat{i}\) + R y \(\hat{j}\) + R z \(\hat{k}\) are the sums of corresponding scalar components of vectors \(\vec{A}\) and \(\vec{B}\):
\[ \begin{cases} R_{x} = A_{x} + B_{x}, \\ R_{y} = A_{y} + B_{y}, \\ R_{z} = A_{z} + B_{z} \end{cases} \ldotp\]
Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up N vectors \(\vec{F}_{1}\), \(\vec{F}_{2}\), \(\vec{F}_{3}\), … , \(\vec{F}_{N}\), where each vector is \(\vec{F}_{k}\) = F kx \(\hat{i}\) + F ky \(\hat{j}\) + F kz \(\hat{k}\), the resultant vector \(\vec{F}_{R}\) is
\[ \vec{F}_{R} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} + \ldots + \vec{F}_{N} = \sum_{k = 1}^{N} \vec{F}_{k} = \sum_{k = 1}^{N} \big(F_{kx} \hat{i} + F_{ky} \hat{j} + F_{kz} \hat{k}\big) = \bigg(\sum_{k = 1}^{N} F_{kx}\bigg) \hat{i} + \bigg(\sum_{k = 1}^{N} F_{ky}\bigg) \hat{j} + \bigg(\sum_{k = 1}^{N} F_{kz}\bigg) \hat{k} \ldotp\]
Therefore, scalar components of the resultant vector are
\[ \begin{cases} F_{Rx} = \sum_{k = 1}^{N} F_{kx} = F_{1x} + F_{2x} + \ldots + F_{Nx} \\ F_{Ry} = \sum_{k = 1}^{N} F_{ky} = F_{1y} + F_{2y} + \ldots + F_{Ny} \\ F_{Rz} = \sum_{k = 1}^{N} F_{kz} = F_{1z} + F_{2z} + \ldots + F_{Nz} \ldotp \end{cases} \label{2.25}\]
Having found the scalar components, we can write the resultant in vector component form:
\[\vec{F}_{R} = F_{Rx}\; \hat{i} + F_{Ry}\; \hat{j} + F_{Rz}\; \hat{k} \ldotp\]
Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.
In many physical situations, we often need to know the direction of a vector. For example, we may want to know the direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector of direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector \(\vec{d}\) = −5 m \(\hat{i}\) is unit vector \(\hat{d}\) = − \(\hat{i}\). The general rule of finding the unit vector \(\hat{V}\) of direction for any vector \(\vec{V}\) is to divide it by its magnitude V:
\[\hat{V} = \frac{\vec{V}}{V} \ldotp \label{2.26}\]
We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the denominator in Equation \ref{2.26} have the same physical unit. In this way, Equation \ref{2.26} allows us to express the unit vector of direction in terms of unit vectors of the axes. Example 2.7.6 illustrates this principle.
Contributors
-
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .


27: Wave Optics
-
- 27.2: Huygens's Principle - Diffraction
- An accurate technique for determining how and where waves propagate is given by Huygens’s principle: Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets. Diffraction is the bending of a wave around the edges of an opening or other obstacle.
-
- 27.3: Young’s Double Slit Experiment
- Young’s double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference and destructive interference depending on the angle of probing.
-
- 27.4: Multiple Slit Diffraction
- A diffraction grating is a large collection of evenly spaced parallel slits that produces an interference pattern similar to but sharper than that of a double slit. There is constructive interference for a diffraction grating when \(d\sin{\theta} = m \lambda \left( for m = 0,1,-1,2,-2,...\right)\), where \(d\) is the distance between slits in the grating, \(\lambda\) is the wavelength of light, and \(m\) is the order of the maximum.
-
- 27.5: Single Slit Diffraction
- A single slit produces an interference pattern characterized by a broad central maximum with narrower and dimmer maxima to the sides. There is destructive interference for a single slit when \(D \sin{\theta} = m \lambda,~ \left(for~m = 1, -1, 2, -2, 3, ...\right)\) where \(D\) is the slit width, \(\lambda\) is the light's wavelength, \(\theta\) is the angle relative to the original direction of the light, and \(m\) is the order of the minimum. Note that there is no \(m = 0\) minimum.
-
- 27.6: Limits of Resolution- The Rayleigh Criterion
- Diffraction limits resolution. For a circular aperture, lens, or mirror, the Rayleigh criterion states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. This occurs for two point objects separated by the angle \(\theta = 1.22 \frac{\lambda}{D}\), where \(\lambda\) is the wavelength of light (or other electromagnetic radiation) and \(D\) is the diameter of the aperture, lens, mirror, et
-
- 27.7: Thin Film Interference
- The bright colors seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colors are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; this effect is known as thin film interference. Interference effects are most prominent when light interacts with something having a size similar to its wavelength. A thin film is one having a thickness smaller than a few times the wavelengt
-
- 27.8: Polarization
- Polarization is the attribute that a wave’s oscillations have a definite direction relative to the direction of propagation of the wave. (This is not the same type of polarization as that discussed for the separation of charges.) Waves having such a direction are said to be polarized. For an EM wave, we define the direction of polarization to be the direction parallel to the electric field. Thus we can think of the electric field arrows as showing the direction of polarization.
-
- 27.9: Microscopy Enhanced by the Wave Characteristics of Light
- Physics research underpins the advancement of developments in microscopy. As we gain knowledge of the wave nature of electromagnetic waves and methods to analyze and interpret signals, new microscopes that enable us to “see” more are being developed. It is the evolution and newer generation of microscopes that are described in this section.
Thumbnail: Physical optics is used to explain effects such as diffraction; t his photo shows diffraction from a single pinhole . (CC-SA-BY-3.0; Wisky).


10.1: Integumentary Introduction
- Identify the patients at risk for impaired skin integrity
- Identify factors related to alterations in the integumentary system across the life span
- Assess a patient’s skin integrity
- Note normal from abnormal findings
- Assess the characteristics of the wound
- Apply correct terminology in the description of wounds
- Adapt care based on integumentary assessment data gathered
- Identify evidence-based practices
The integumentary system includes skin, hair, and nails. The skin is the largest organ of the body and has many purposes. Our skin keeps us warm and contains nerve endings that control the ability to feel the sensations of hot, cold, pain, and pressure. Our skin also keeps harmful things out of the body, such as dirt, bacteria, and viruses, and keeps helpful things in like moisture. Maintaining intact skin is important to prevent infection and maintain health. This chapter will review the anatomy and physiology of the integumentary system, factors that affect healthy skin and healing, and interventions that nurses perform to repair and protect this vital organ.


7: Electricity
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- 7.1: Prelude to Electric Charge and Electric Field
- Franklin demonstrated a connection between lightning and static electricity. Sparks were drawn from a key hung on a kite string during an electrical storm. These sparks were like those produced by static electricity, such as the spark that jumps from your finger to a metal doorknob after you walk across a wool carpet.
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- 7.2: Static Electricity and Charge - Conservation of Charge
- When various materials are rubbed together in controlled ways, certain combinations of materials always produce one type of charge on one material and the opposite type on the other. By convention, we call one type of charge “positive”, and the other type “negative.” E.g., when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Since the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer.
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- 7.3: Conductors and Insulators
- Some substances, such as metals and salty water, allow charges to move through them with relative ease. Some of the electrons in metals and similar conductors are not bound to individual atoms or sites in the material. These free electrons can move through the material much as air moves through loose sand.
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- 7.4: Coulomb's Law
- Through the work of scientists in the late 18th century, the main features of the electrostatic force—the existence of two types of charge, the observation that like charges repel, unlike charges attract, and the decrease of force with distance—were eventually refined, and expressed as a mathematical formula. The mathematical formula for the electrostatic force is called Coulomb’s law after the French physicist Charles Coulomb.
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- 7.5: Introduction to Electric Potential and Electric Energy
- Two of the most familiar aspects of electricity are its energy and voltage. But energy and voltage are not the same thing. In this chapter, we shall examine the relationship between voltage and electrical energy and begin to explore some of the many applications of electricity.
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- 7.6: Electric Potential Energy- Potential Difference
- Electric potential is potential energy per unit charge. The potential difference between points A and B, VB−VA , defined to be the change in potential energy of a charge qmoved from A to B, is equal to the change in potential energy divided by the charge, Potential difference is commonly called voltage, represented by the symbol ΔV .
Contributors and Attributions
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.


9.3: Stop
The first move is the simplest. Stop reminds you of two things.
First, when you find an article and start to read it — Stop .  What is their reputation? Did you find the article in a library database (which is vetted)? Is the website reputable (e.g. American Heart Association vs. John Smith’s Heart Page)? Can you trust the social media site? If you don’t have that information, use the other moves to get a sense of what you’re looking at. You can read it but don’t share it until you know what it is.
Second, after you begin to use the other moves it can be easy to go down a rabbit hole, going off on tangents only distantly related to your original task. If you feel yourself getting overwhelmed in your fact-checking efforts, Stop and take a second to remember your purpose. If you just want to repost the article, read an interesting story, or get a high-level explanation of a concept, it’s probably good enough to find out whether the publication is reputable. If you are doing deep research of your own, you may want to chase down individual claims on a webpage or in a journal or newspaper article and independently verify them.
Relevant sources are those that pertain to your research question. You’ll be able to figure that out fairly quickly by reading or skimming particular parts of sources and maybe jotting down notes in little tables that help you keep track. We’ll show you how below, including where to look in specific kinds of sources and what questions to ask yourself as you do.
Please keep in mind that both sorts of investigations are equally useful. Quick and shallow investigations will form most of what we do on the web. We get quicker with the simple stuff in part so we can spend more time on the stuff that matters to us. But in either case, stopping periodically and reevaluating your search strategy is key.
For instance, if your research question is about the life sciences, you probably should consider only the most recent sources relevant because the life sciences change so quickly. Anything but the most recent sources may be out of date. So aim for sources no more than 5 years old. An example discipline that calls for even newer sources is computer security.
But suppose your research question is about the Edo Period in Japan (1603-1868) or about Robert Falcon Scott, who explored the Antarctic from 1901-1913. In these cases, an item from 1918 might be just as useful as an item from 2020, although new information may have been found in the 100+ year gap. But something from 1899 about Antarctica or from 1597 about Japan would NOT be current enough for these research questions.
To be considered relevant to your research question, a source doesn’t necessarily have to cover all of your main concepts, but finding sources that do is the ideal. Otherwise, you just have to make do with what you’ve got. Don’t forget that each source will have to pass the currency test, too, if the currency is important to your research question. So it’s wise to record your decisions about the sources’ currency on your tables, too.
Example
EXAMPLE: Sources’ Main Concepts and Currency
Research question: How does having diverse members in a group increase the critical thinking of the group?
| Currency Okay | Group Diversity | Critical Thinking | |
| Source A title | X | X | |
| Source B title | X | ||
| Source C title | X | X | X |
The table in this hypothetical example indicates that both Sources A and C are relevant because each pertains to at least one main concept from the research question. Currency doesn’t seem to matter much to our research question, so all three sources were marked current. But since the currency is all that Source B has to offer, it is not relevant for this project.
If you do make little tables for relevance, it’s probably a good idea to hang on to them. You might find them helpful later in your research process.
Exercise: Evaluate the Currency of Sources
An interactive H5P element has been excluded from this version of the text. You can view it online here:
https://minnstate.pressbooks.pub/ctar/?p=171#h5p-36