11.5: Rotating Frame Analysis
Rotating frame analysis is a specialized part of relative motion analysis. It is typically performed in Cartesian (\(x\)-\(y\)) coordinates for rigid bodies. In the previous section, rotating frames in polar coordinates were used to solve problems with formulae similar to those in particle kinematics. We will adopt a slightly different notation that is specialized to rigid bodies. These formulae are the most general planar kinematics formulae - that is, they can always be used, and will provide the correct answer. However, they are unnecessarily complicated for many types of motion, such as pure rotation.
Rotating frame analysis is really important for cases where objects are not pinned to each other. When do you use rotating frames? When one object is sliding against another object, or two objects are not even in contact, but you want to know something about the motion relationship between them and/or are given information about the motion relationship between them.
Reference Frames:
In most of the preceding material, you have worked with the fixed reference frame \(O_{xyz}\), and a translating (but not rotating) reference frame attached to the rigid body at a point \(A\), frame \(A_{xyz}\).
This has given us relative motion equations:
\begin{align} \text{Position:} \quad &\, \vec{r}_{B / O} = \vec{r}_{A / O} + \vec{r}_{B / A} \\ \text{Velocity:} \quad &\, \vec{v}_{B / O} = \vec{v}_{A / O} + \vec{v}_{B / A} = \vec{v}_{A / O} + \vec{\omega} \times \vec{r}_{B / A} \\ \text{Acceleration:} \quad &\, \vec{a}_{B / O} = \vec{a}_{A / O} + \vec{a}_{B / A} = \vec{a}_{A / O} + \vec{\alpha} \times \vec{r}_{B / A} - \omega^2 \vec{r}_{B / A} \end{align}
Note that the notation \(r_{B / O}\), where \(O\) is the fixed frame, can also be written as \(r_B\). Both indicate an absolute position (or velocity or acceleration) - that is, a value with respect to the fixed frame \(O\).
Rotating Frames:
Now, we will consider a reference frame that is attached to a point on the rigid body and both rotating and translating with the rigid body . We will introduce some extra terms to account for the rotation of the frame.
Consider a rigid body with frame \(x'y'z'\) at point \(A\). \(x'y'z'\) moves and rotates with the body. A bug, \(B\), is crawling along the body.
For an observer sitting at \(A\), the bug appears to be crawling away in a straight line. But for an observer sitting at \(O\), the bug does not look like it is moving in a straight line, because the body it is crawling on is also moving (translating and rotating). We would like to describe the motion of the bug as seen by an observer at \(O\). To remind ourselves that we are dealing with a rotating frame attached to one body, we use capital omega \(\Omega\)) to denote the angular velocity of the object that the rotating frame is attached to, and \(\dot{\Omega}\) for the angular acceleration of the object with the rotating frame attached. Finally, we use brackets and the subscript \(rel\) to denote values of velocity or acceleration expressed with respect to the rotating frame.
Recall that vectors have both magnitude and direction. We can express a vector in components with respect to the \(\hat{i}\) and \(\hat{j}\) unit vectors:
\[ \vec{r}_{B / O} = r_{B/O, x} \hat{i} + r_{B/O, y} \hat{j} \]
Normally, when we take a time derivative of such an expression, the length of the vector (i.e. the \(r_{B/O, x}\) and \(r_{B/O, y}\) terms) change, but the unit vectors do not. In the case of the rotating frame, the unit vectors change. Now, they don't change in length - they remain unit vectors. But they are attached to a rotating body, which means they also rotate. That is, their directions change, and we have to account for that when deriving velocity and acceleration expressions.
You can see we get small vectors \(d \hat{i'}\) and \(d \hat{j'}\) that describe the change in direction of the unit vectors, and the magnitude of that change is directly related to \(\Omega\), the angular velocity of the rigid body and the rotating frame.
\[ \frac{d \hat{i'}}{dt} = \Omega \hat{k} \times \hat{i'} = \Omega \hat{j'} \]
\[ \frac{d \hat{j'}}{dt} = \Omega \hat{k} \times \hat{j'} = -\Omega \hat{i'} \]
When we differentiate the relative position of the bug, \(B\), with respect to the rotating frame, \(A\), we have:
\[\frac{d \vec{r}_{B/A}}{dt} = \left( \dot{r}_{B/A, \ x'} \ \hat{i'} + \dot{r}_{B/A, \ y'} \ \hat{j'} \right) + \vec{\Omega} \times \left( r_{B/A, \ x'} \ \hat{i'} + r_{B/A, \ y'} \hat{j'} \ \right) \]
\begin{align} \text{New term:} \quad &\, \left( \dot{r}_{B/A, \ x'} \ \hat{i'} + \dot{r}_{B/A, \ y'} \ \hat{j'} \right) = \left( \vec{v}_{B/A} \right) _{rel} \\ \text{Same as before:} \quad &\, \vec{\Omega} \times \left( r_{B/A, \ x'} \ \hat{i'} + r_{B/A, \ y'} \ \hat{j'} \right) = \vec{\Omega} \times \vec{r}_{B/A} \end{align}
The new term above describes the motion of the bug due to the rotation of the rigid body (and the \(x'y'z'\) frame), while the familiar term above describes the motion of the bug as seen by an observer fixed to \(x'y'z'\). Our final expression for the velocity of the bug with respect to the fixed frame is:
\[ \vec{v}_{B/O} = \vec{v}_{A/O} + \vec{\Omega} \times \vec{r}_{B/A} + ( \vec{v}_{B/A} )_{rel} \]
This final expression has the new term, \( (\vec{v}_{B/A})_{rel} \), which describes the motion of the bug as viewed from the rotating frame. You can note that if the bug is not moving, \( (\vec{v}_{B/A})_{rel} = 0\), and we get the familiar expression from the translating, non-rotating frames. Thus, this expression is the most general relative velocity expression.
We can similarly derive an acceleration equation by differentiating the above velocity equation. The final expression is:
\[ \vec{a}_{B/O} = \vec{a}_{A/O} + \dot{\vec{\Omega}} \times \vec{r}_{B/A} + \vec{\Omega} \times ( \vec{\Omega} \times \vec{r}_{B/A} ) + 2 \vec{\Omega} \times (\vec{v}_{B/A})_{rel} + (\vec{a}_{B/A})_{rel} \]
| Motion of rotating frame attached to rigid body: | \( \vec{a}_{A/O} + \dot{\vec{\Omega}} \times \vec{r}_{B/A} + \vec{\Omega} \times ( \vec{\Omega} \times \vec{r}_{B/A} ) \) |
| Coriolis acceleration - interaction of object motion with respect to rotating frame and motion of rotating frame: | \( 2 \vec{\Omega} \times (\vec{v}_{B/A})_{rel} \) |
| Motion of object with respect to rotating frame: | \( ( \vec{a}_{B/A} )_{rel} \) |
Note that, for planar (2D) motion:
\[ \vec{\Omega} \times ( \vec{\Omega} \times \vec{r}_{B/A} ) = - \Omega^2 \vec{r}_{B/A} \]
Again, if the bug is sitting still, both \(rel\) terms become zero (it is not moving with respect to the rigid body), and we are left with the same relative motion expression used with translating frames. So, Equation \(\PageIndex{11}\) is the most general form of the relative acceleration equation and can always be used.
When tackling these problems, there are several important things to remember:
- You need to pick the correct body to attach the rotating frame to. The correct body will have another body sliding against it. If the object of interest is pinned to the body you chose, it might not be correct.
- It is often helpful to align your rotating frame such that you end up with a simplified "\(rel\)" term or terms. For example, line up the \(x'\) axis with the direction of movement of the bug, so that \( (\vec{v}_{B/A})_{rel} \) is in the \(x'\)-direction.
- The components of all terms of any one equation must be computed along the same \(\hat{i}\) and \(\hat{j}\) directions. If the frames are not aligned, you must pick ONE FRAME and express all terms along that the directions of the ONE FRAME. (You may be able to switch between frames in different equations, but make sure you are careful with notation!)
- When you are given information in a problem, pay careful attention to what frame is being referred to.
- Make sure you include all the terms in the acceleration equation! It's easy to miss one. You should have five terms.
These are some of the most complex problems in planar kinematics, so take your time!
Exercise \(\PageIndex{1}\)
A camera drone, \(D\), flies over a car race in a curved trajectory (center \(O\)) with a constant ground-speed velocity of \(v_D = 9 \ m/s\). At the moment shown, car \(C\) is traveling with velocity of \(v_c = 12 \ m/s\) and an acceleration of \(a_c = 2 \ m/s^2\) as shown. Assume \(d_1 = 7.5 \ m, \ d_2\) = \ 3 m\).
- Find the velocity of the car as observed by the camera on drone D at this instant.
- Find the acceleration of the car as observed by the camera on drone D at this instant.
- Solution
Example \(\PageIndex{2}\)
A breathing exercise video graphic (somewhat similar to this one ) shows a small circle moving in a constrained circular path (constant radius 80 cm) at a constant angular velocity of 0.4 rad/s around an expanding and contracting inner circle. The inner circle expands and contracts sinusoidally, from a minimum radius of 30 cm to a maximum radius of 60 cm. The distance from the center of the inner circle to a point on the edge of the inner circle can be described by the equation \(r = 0.45 - 0.15 \sin (\theta)\), where \(r\) is in meters and \(\theta\) is the position of the small circle (zero at the \(x\)-axis).
Find the velocity and acceleration of point B on the edge of the inner circle as viewed by an observer on the small circle at point \(A\) (see part C of the figure below). \(\theta = 45\)°, \(\alpha = 45\)°
- Solution


6: Functional Diversity - Life History Traits
Scientists have estimated that there are ~8.7 million species on Earth. How does one planet support such enormous diversity of species? The answer relates to the impressive functional diversity (Fig 6.0.1) that organisms display – that is, the diversity in strategies and adaptations that different organisms have for surviving in their environment. The functional diversity sections address different strategies and adaptations for acquiring nutrients and resources from the environment, dealing with stressful environmental conditions (ex. heat stress), succeeding at or defending against predation, and successfully completing the life cycle.
This chapter focuses on life history strategies and adaptations that plants and animals exhibit, and the trade offs that different life history traits confer.
Figure \(\PageIndex{1}\): Schematic of the magnitude and type of diversity exhibited at different scales. This section focuses on functional group diversity. Figure by L Gerhart-Barley.
Title Image Credit: Pennsylvania Department of Conservation and Natural Resources - Forestry, Bugwood.org; https://www.insectimages.org/browse/...imgnum=5018091


1.9: The ‘state’ of things- epidemiologic comparisons across populations
At the end of this case, students will be able to:
- Apply epidemiologic principles to a public health scenario
- Compare and contrast disease occurrence and health determinants across populations
- Generate conclusions about the health of a population using epidemiologic and pharmacoepidemiologic data
- Explain the dynamic relationship between health data, epidemiology, and public health policies
Introduction
Given pharmacy’s increasing role in research, shaping public policy, and assessing medication use and safety across populations, learning fundamentals of epidemiology and pharmacoepidemiology is a critical component of pharmacy education. This is especially true for pharmacy students interested in pursuing careers in research, industry, administration, or public policy where these skills are consistently required.
Epidemiology is “ the study of the distribution and determinants of health-related states or events in specified populations, and the application of this study to the control of health problems .” 1 Pharmacoepidemiology, a subset of epidemiology, is “ the study of the use and effects/side-effects of drugs in large numbers of people with the purpose of supporting the rational and cost-effective use of drugs in the population thereby improving health outcomes .” 2 As drug experts, pharmacists are already routinely responsible for monitoring drug use and safety across various populations. Additionally, the increasing complexity of health systems and push for a more holistic approach to health – not just drugs – necessitates an increased focus on epidemiology training for pharmacists. This is underscored by the fact that research – and by extension epidemiology and pharmacoepidemiology – serves as the tenth and all-encompassing essential service of public health. 3
In every professional setting, pharmacists are at least in some part responsible for monitoring diseases and drug use. For some pharmacists, the population may be the patients in their community pharmacy, while for other pharmacists, their population may include serving millions of individuals while working for the FDA. Regardless of the setting, you will work with diverse, often ill or at-risk, populations reinforcing the importance of skills and experience in monitoring disease and medication use across populations. To gain further appreciation for epidemiology and its utility, consider the opioid epidemic. Pharmacists lead the charge in tracking opioid utilization, identifying high-risk patients, exploring the risk/benefit of opioids, and designing/assessing various public health policies aimed at mitigating the crisis (e.g., opioid reversal strategies). Examples of pharmacy-related epidemiology and pharmacoepidemiology duties include:
- Monitoring levels of disease and/or drug utilization
- Guiding distribution of resources
- Discovering exposures that facilitate or mitigate patterns in disease and/or drug use
- Providing useful information on the beneficial and harmful effects of drugs, including risk/benefit information.
Case and Case Questions
You have volunteered to serve as a consultant for a new non-profit agency. The agency’s mission is to “improve the health of the population by promoting safe and effective use of medications.” The agency wants to establish roots in a particular state, but isn’t sure where to go. The agency’s board of trustees has asked you to provide a recommendation as to which state they should go to and start their work. Your task is to compare various state pairs and provide a recommendation (with rationale) as to which you think is the “unhealthier” state in regards to a particular health topic. Utilize the various data (outcomes, risk factors, determinants) available on the Henry J Kaiser Family Foundation (KFF) State Health Facts web site ( https://www.kff.org/statedata/ ) and specific to that particular health issue to make your decision. Your recommendation should be written into a clear, focused format that you will present to the board of trustees.
The board of trustees has provided you with several topics of interest and several state pairings for your analysis (see Table ). For each of the eight topics, pick one pairing, and conclude which state is “worse” concerning that particular topic (aka, “which state is in more need of your non-profit agency’s help”). The board has asked that you analyze a different state pair for each topic. Use the outcome data available on KFF State Health Facts website to support your decision.
| Topics | Pairings |
|---|---|
|
Immunizations
Alcohol and drug dependence Opioid epidemic Prescription drugs HIV/AIDS Medicare Medicaid and CHIP Women’s health COVID-19 |
District of Columbia vs. Georgia
Nevada vs. Delaware New Mexico vs. New Jersey Arkansas vs. Illinois Mississippi vs. Utah New York vs. Wyoming Hawaii vs. Ohio Alabama vs. Massachusetts California vs. Kentucky Colorado vs. Pennsylvania Texas vs. Connecticut |
- Topic #1: Immunizations
- Topic #2: Alcohol and Drug Dependence
- Topic #3: Opioid Epidemic
- Topic #4: Prescription Drugs
- Topic #5: HIV/AIDS
- Topic #6: Medicare
- Topic #7: Medicaid and CHIP
- Topic #8: Women’s Health
Author Commentary
Epidemiology and pharmacoepidemiology are extremely broad and complex fields, and this activity is only meant as an introduction into these areas. These concepts are crucial to developing an appreciation for population health, its intricacies, and the many factors that contribute to health. As you delve into these comparisons, you should be careful in how you interpret and present the available data. Also, understand that the available data is limited and that you must make the most informed decision you can with imperfect and incomplete information. This closely reflects what happens in the real world. Lastly, when reviewing topics, you will notice many disparities and inequalities across state populations. As you find these differences, especially large differences, begin to consider how state-level policies and culture may contribute to these found differences. In this way, you will be extending the focus of this activity to include additional related epidemiological concepts such as determinants of health and health disparities.
Patient Approaches and Opportunities
Epidemiology is the “scientific arm” of public health and is essential for assessing trends and patterns of disease and medication use across populations. Pharmacists are responsible for safe and effective medication use, and accordingly, must lead the effort in monitoring medication utilization and safety in populations. A strong foundation in epidemiological concepts is a critical component for pharmacists to have so that they can perform public health research and make sound conclusions when interpreting data. Ultimately, such foundational knowledge will lead to enhanced ability to create effective and meaningful public health programs and policies.
Related chapters of interest:
- More than just diet and exercise: social determinants of health and well-being
- Medication safety: to ‘error’ is human
- The Sustainable Development Goals and pharmacy practice: a blueprint for health
- Open-door policy: a window into creation, implementation, and assessment
- Prescription for change: advocacy and legislation in pharmacy
- A pharmacist’s obligation: advocating for change
External resources:
-
Websites:
- Henry J Kaiser Family Foundation. State Health Facts . https://www.kff.org/statedata/ . Accessed November 30, 2018.
-
Books and Chapters:
- Centers for Disease Control and Prevention. Principles of Epidemiology in Public Health Practice; 3rd Edition . https://www.cdc.gov/ophss/csels/dsepd/ss1978/ss1978.pdf . Accessed November 30, 2018.
- Jean Carter and Marion Slack. Chapter 10: Epidemiology and Disease in Pharmacy in Public Health: Basics and Beyond . Pages 197-226.
References
- Last JM. Dictionary of epidemiology . 4th ed. New York: Oxford University Press; 2001. p. 61.
- World Health Organization. Essential Medicines and Health Products Information Portal. http://apps.who.int/medicinedocs/en/d/Js4876e/2.html . Accessed November 30, 2018.
- The Public Health System & the 10 Essential Public Health Services. https://www.cdc.gov/stltpublichealth/publichealthservices/essentialhealthservices.html . Accessed November 30, 2018.


6.4.5: Combining Like Terms (Part 3)
Lesson
Let's see how we can combine terms in an expression to write it with less terms.
Exercise \(\PageIndex{1}\): Are They Equal?
Select all expressions that are equal to \(8-12-(6+4)\).
- \(8-6-12+4\)
- \(8-12-6-4\)
- \(8-12+(6+4)\)
- \(8-12-6+4\)
- \(8-4-12-6\)
Exercise \(\PageIndex{2}\): X's and Y's
Match each expression in column A with an equivalent expression from column B. Be prepared to explain your reasoning.
A
- \((9x+5y)+(3x+7y)\)
- \((9x+5y)-(3x+7y)\)
- \((9x+5y)-(3x-7y)\)
- \(9x-7y+3x+5y\)
- \(9x-7y+3x-5y\)
- \(9x7y-3x-5y\)
B
- \(12(x+y)\)
- \(12(x-y)\)
- \(6(x-2y)\)
- \(9x+5y+3x-7y\)
- \(9x+5y-3x+7y\)
- \(9x-3x+5y-7y\)
Exercise \(\PageIndex{3}\): Seeing Structure and Factoring
Write each expression with fewer terms. Show or explain your reasoning.
- \(3\cdot 15+4\cdot 15-5\cdot 15\)
- \(3x+4x-5x\)
- \(3(x-2)+4(x-2)-5(x-2)\)
- \(3(\frac{5}{2}x+6\frac{1}{2})+4(\frac{5}{2}x+5\frac{1}{2})-5(\frac{5}{2}x+6\frac{1}{2})\)
Summary
Combining like terms is a useful strategy that we will see again and again in our future work with mathematical expressions. It is helpful to review the things we have learned about this important concept.
- Combining like terms is an application of the distributive property. For example:
\(\begin{array}{c}{2x+9x}\\{(2+9)\cdot x} \\ {11x}\end{array}\)
- It often also involves the commutative and associative properties to change the order or grouping of addition. For example:
\(\begin{array}{c}{2a+3b+4a+5b}\\{2a+4a+3b+5b}\\{(2a+4a)+(3b+5b)}\\{6a+8b}\end{array}\)
- We can't change order or grouping when subtracting; so in order to apply the commutative or associative properties to expressions with subtraction, we need to rewrite subtraction as addition. For example:
\(\begin{array}{c}{2a-3b-4a-5b}\\{2a+-3b+-4a+-5b}\\{2a+-4a+-3b+-5b}\\{-2a+-8b}\\{-2a-8b}\end{array}\)
- Since combining like terms uses properties of operations, it results in expressions that are equivalent.
- The like terms that are combined do not have to be a single number or variable; they may be longer expressions as well. Terms can be combined in any sum where there is a common factor in all the terms. For example, each term in the expression \(5(x+3)-0.5(x+3)+2(x+3)\) has a factor of \((x+3)\). We can rewrite the expression with fewer terms by using the distributive property:
\(\begin{array}{c}{5(x+3)-0.5(x+3)+2(x+3)}\\{(5-0.5+2)(x+3)}\\{6.5(x+3)}\end{array}\)
Glossary Entries
Definition: Expand
To expand an expression, we use the distributive property to rewrite a product as a sum. The new expression is equivalent to the original expression.
For example, we can expand the expression \(5(4x+7)\) to get the equivalent expression \(20x+35\).
Definition: Factor (an expression)
To factor an expression, we use the distributive property to rewrite a sum as a product. The new expression is equivalent to the original expression.
For example, we can factor the expression \(20x+35\) to get the equivalent expression \(5(4x+7)\).
Definition: Term
A term is a part of an expression. It can be a single number, a variable, or a number and a variable that are multiplied together. For example, the expression \(5x+18\) has two terms. The first term is \(5x\) and the second term is 18.
Practice
Exercise \(\PageIndex{4}\)
Jada says, “I can tell that \(\frac{-2}{3}(x+5)+4(x=5)-\frac{10}{3}(x+5)\) equals 0 just by looking at it.” Is Jada correct? Explain how you know.
Exercise \(\PageIndex{5}\)
In each row, decide whether the expression in column A is equivalent to the expression in column B. If they are not equivalent, show how to change one expression to make them equivalent.
A
- \(3x-2x+0.5x\)
- \(3(x+4)-2(x+4)\)
- \(6(x+4)-2(x+5)\)
- \(3(x+4)-2(x+4)+0.5(x+4)\)
- \(20(\frac{2}{5}x+\frac{3}{4}y-\frac{1}{2})\)
B
- \(1.5x\)
- \(x+3\)
- \(2(2x+7)\)
- \(2(2x+7)\)
- \(1.5\)
- \(\frac{1}{2}(16x+30y-20)\)
Exercise \(\PageIndex{6}\)
For each situation, write an expression for the new balance using as few terms as possible.
- A checking account has a balance of -$126.89. A customer makes two deposits, one \(3\frac{1}{2}\) times the other, and then withdraws $25.
- A checking account has a balance of $350. A customer makes two withdrawals, one $50 more than the other. Then he makes a deposit of $75.
(From Unit 6.4.3)
Exercise \(\PageIndex{7}\)
Tyler is using the distributive property on the expression \(9-4(5x-6)\). Here is his work:
\(9-4(5x-6)\)
\(9+(-4)(5x+-6)\)
\(9+-20x+-6\)
\(3-20x\)
Mai thinks Tyler’s answer is incorrect. She says, “If expressions are equivalent then they are equal for any value of the variable. Why don’t you try to substitute the same value for \(x\) in all the equations and see where they are not equal?”
- Find the step where Tyler made an error.
- Explain what he did wrong.
- Correct Tyler's work.
(From Unit 6.4.4)
Exercise \(\PageIndex{8}\)
- If \((11+x)\) is positive, but \((4+x)\) is negative, what is one number that \(x\) could be?
- If \((-3+y)\) is positive, but \((-9+y)\) is negative, what is one number that \(y\) could be?
- If \((-5+z)\) is positive, but \((-6+z)\) is negative, what is one number that \(z\) could be?
(From Unit 6.3.1)


6.2: Slipping vs Tipping
Imagine a box sitting on a rough surface as shown in the figure below. Now imagine that we start pushing on the side of the box. Initially the friction force will resist the pushing force and the box will sit still. However, as we increase the force pushing the box one of two things will occur.
- The pushing force will exceed the maximum static friction force and the box will begin to slide across the surface (slipping).
- Or, the pushing force and the friction force will create a strong enough couple that the box will rotate and fall on its side (tipping).
When we look at cases where either slipping or tipping may occur, we are usually interested in finding which of the two options will occur first. To determine this, we usually determine both the pushing force necessary to make the body slide and the pushing force necessary to make the body tip over. Whichever option requires less force is the option that will occur first.
Determining the Force Required to Make an Object "Slip":
A body will slide across a surface if the pushing force exceeds the maximum static friction force that can exist between the two surfaces in contact. As in all dry friction problems, this limit to the friction force is equal to the static coefficient of friction times the normal force between the body. If the pushing force exceeds this value then the body will slip.
Determining the Force Required to Make an Object "Tip":
The normal forces that support bodies are distributed forces. These forces will not only prevent the body from accelerating into the ground due to gravitational forces, but they can also redistribute themselves to prevent a body from rotating when forces cause a moment to act on the body. This redistribution will result in the equivalent point load for the normal force shifting to one side or the other. A body will tip over when the normal force can no longer redistribute itself any further to resist the moment exerted by other forces (such as the pushing force and the friction force).
The easiest way to think about the shifting normal force and tipping is to imagine the equivalent point load of the distributed normal force. As we push or pull on the body, the normal force will shift to the left or right. This normal force and the gravitational force create a couple that exerts a moment. This moment will be countering the moment exerted by the couple formed by the pushing force and the friction force.
Because the normal force is the direct result of physical contact, we cannot shift the normal force beyond the surfaces in contact (i.e., the edge of the box). If countering the moment exerted by the pushing force and the friction force requires shifting the normal force beyond the edge of the box, then the normal force and the gravity force will not be able to counter the moment and as a result the box will begin to rotate (i.e., tip over).
Example \(\PageIndex{1}\)
The box shown below is pushed as shown. If we keep increasing the pushing force, will the box first begin to slide or will it tip over?
- Solution
Example \(\PageIndex{2}\)
What is the maximum value of \(d\) that will allow the box to slide along the surface before tipping?
- Solution


9.17: Formula for Refutation and Rebuttal
- Describe strategies for effective refutation and rebuttal
Counterargument
An integral part of composing a strong argument is including a counterargument. This can be difficult, especially if a writer is arguing for a position they already agree with. In such cases, writers can sometimes make good points to support their stances; however, their arguments are vulnerable unless they anticipate and address counterarguments. When a writer does this, it is often referred to as rebuttal or refutation . Some scholars of rhetoric differentiate the two words in terms of if you can actually disprove a claim or just argue against it; however, in this section, we will use the terms as basically interchangeable to help get you more used to their function in argument.
When writers are able to skillfully rebut or refute a view that runs counter to their claims, it strengthens their work. Rebuttal and refutation are common in all types of argument, including academic argument. As you complete more advanced work in college, you will be expected to address counterargument often. And while you might not always need to or be able to prove that other points of view are wrong, you may at least need to try to argue against them.
Though writers may handle rebuttal and refutation in different ways, there is a formula for success in academic argument. Here are the key parts of that formula:
1.  Good rebuttal and refutation begin with a solid understanding of all possible points of view on your topic. That may mean you even need to acknowledge and accommodate opposing points of view. Acknowledging other views shows you are aware of ideas that run counter to your claims. You will almost always be expected to at least acknowledge such views in your work. You may also, though, need to accommodate opposing views, especially if many people see them as reasonable. If, for example, you were writing a piece arguing that students should take a gap year between high school and college, it would benefit your work to acknowledge that a gap year isn’t realistic for or even desired by all students. You may further accommodate this other view by explaining how some students may thrive in the structure that school provides and would gain by going directly from high school to college. Remember that even if you cannot prove positions that counter your own are wrong, you can still use rebuttal and refutation to show why they might be problematic, flawed, or just not as good as another possible position for some people.
2. Use a respectful, non-incendiary tone
It doesn’t help the writer’s cause to offend, upset, or alienate potential readers, even those who hold differing views. Treating all potential readers with respect and avoiding words or phrases that belittle people and/or their views will help you get your points across more effectively. For example, if you are writing a paper on why America would benefit from a third viable major political party, it will not help your cause to write that “Republicans are dumb, and Democrats are whiny.” First, those claims are too general. But even if they weren’t, they won’t help your cause. If you choose to break down the perceived problems with members of political parties, you must do so in a way that is as respectful as possible. Calling someone a name or insulting them (directly or indirectly) is very rarely a successful strategy in argument.
3. Use reliable information in your rebuttal/refutation
Always be sure to carefully check the ideas or claims you make in rebutting a counterargument. The brain is not an infallible computer, and there are instances when we think we know information is accurate but it isn’t. Sometimes we know a lot about a particular subject but we get information confused or time has changed things a bit. Additionally, we may be tempted to use a source that backs up our ideas perfectly, but it might not be the most reputable, credible, or up-to-date place for information. Don’t assume you just have all of the information to shoot down counterarguments. Use your knowledge, but also do thorough research, double- and triple-check information, and look for sources that are likely to carry weight with readers. For example, it is widely assumed that bulls are attracted to the color red; however, in reality, bulls are colorblind, so what many people assume as fact is incorrect. Be thorough so you have confidence in your claims when you are rebutting/refuting and likewise when you are attempting to prevent yourself from being open to rebuttal/refutation.
4. Use qualifying words when applicable to help you be more accurate and to avoid locking you into an absolute claim
Qualifying words are terms such as “many,” “most,” “some,” “might,” “rarely,” “doubtful,” “often,” etc. You get the point. These are words that don’t lock you into a claim that could be easily refuted and that can help you more easily rebut counterarguments. For example, if someone says “Nobody dies of tuberculosis anymore” we might get the point that it isn’t as common as it used to be. Still, it isn’t an accurate statement, and a more precise way to phrase such a claim would be to qualify it: “Not many people die each year in America from tuberculosis.” You might not always need to use qualifying terms. If you are making a point that is absolute, feel free to make it strongly; however, if there is a need to give your claim more flexibility, use qualifying words to help you.
refutation : proving a claim is false; arguing against something
rebuttal : contradicting or opposing a claim
Contributors and Attributions
- Rebuttal and Refutation. Provided by : University of Mississippi. Project : PLATO Project. License : CC BY-SA: Attribution-ShareAlike
- Image of the Coexist logo. Authored by : Gerd Altmann. Provided by : Pixabay. Located at : pixabay.com/illustrations/figures-personal-silhouettes-human-1929874/. License : Other . License Terms : pixabay.com/service/terms/#license
- Aristotelian Argument. Authored by : Excelsior OWL. Provided by : https://owl.excelsior.edu/argument-a...-aristotelian/ . License : CC BY: Attribution
- Toulmin's Schema. Provided by : Utah State University. Located at : ocw.usu.edu/English/intermediate-writing/english-2010/-2010/toulmins-schema.html. Project : Intermediate Writing: Research Writing in a Persuasive Mode. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
- Toulmin Argument and Types of Argument. Provided by : Excelsior College OWL. Located at : https://owl.excelsior.edu/argument-and-critical-thinking/organizing-your-argument/organizing-your-argument-toulmin/ . License : CC BY: Attribution


6.7: Persistence During Stressful Times
Stress Management
Being in control of your life and having realistic expectations about your day-to-day challenges are the keys to stress management, which is perhaps the most important ingredient to living a happy, healthy and rewarding life. —Marilu Henner, actress
Stress
As a student, you’re probably plenty familiar with the experience of stress—a condition characterized by symptoms of physical or emotional tension. What you may not know is that it’s a natural response of the mind and body to a situation in which a person feels threatened or anxious. Stress can be positive (the anticipation of preparing for a wedding) or negative (dealing with a natural disaster). As a college student, it may feel like stress is a persistent fact of life. While everyone experiences stress at times, a prolonged bout of it can affect your health and ability to cope with life. That’s why social support and self-care are important. They can help you see your problems in perspective . . . and the stressful feelings ease up.
Sometimes stress can be good. For instance, it can help you develop the skills needed to manage potentially challenging or threatening situations in life. However, stress can be harmful when it is severe enough to make you feel overwhelmed and out of control. Strong emotions like fear, sadness, or other symptoms of depression are normal, as long as they are temporary and don’t interfere with daily activities. If these emotions last too long or cause other problems, it’s a different story.
Signs and Effects of Stress
Physical or emotional tension are often signs of stress. They can be reactions to a situation that causes you to feel threatened or anxious. The following are all common symptoms of stress:
- Disbelief and shock
- Tension and irritability
- Fear and anxiety about the future
- Difficulty making decisions
- Being numb to one’s feelings
- Loss of interest in normal activities
- Loss of appetite (or increased appetite)
- Nightmares and recurring thoughts about a particular event
- Anger
- Increased use of alcohol and drugs
- Sadness and other symptoms of depression
- Feeling powerless
- Crying
- Sleep problems
- Headaches, back pains, and stomach problems
- Trouble concentrating
It’s not only unpleasant to live with the tension and symptoms of ongoing stress; it’s actually harmful to your body, too. Chronic stress can impair your immune system and disrupt almost all of your body’s processes, leading to increased risk of numerous health problems, including anxiety, depression, digestive problems, heart disease, sleep problems, weight gain, and/or memory and concentration impairment. 3
That’s why it’s so important to learn healthy ways of coping with the stressors in your life.
Ways of Managing Stress
The best strategy for managing stress is by taking care of yourself in the following ways:
- Avoid drugs and alcohol. They may seem to be a temporary fix to feel better, but in the long run, they can create more problems and add to your stress instead of taking it away.
- Manage your time. Work on prioritizing and scheduling your commitments. This will help you feel in better control of your life which, in turn, will mean less stress.
- Find support. Seek help from a friend, family member, partner, counselor, doctor, or clergy person. Having a sympathetic listening ear and talking about your problems and stress really can lighten the burden.
- Connect socially. When you feel stressed, it’s easy to isolate yourself. Try to resist this impulse and stay connected. Make time to enjoy being with classmates, friends, and family; try to schedule study breaks that you can take with other people.
- Slow down and cut out distractions for a while . Take a break from your phone, email, and social media.
- Take care of your health. Eat well, exercise regularly, get plenty of sleep, try a relaxation technique (such as meditation or yoga), and maintain a normal, predictable routine.
Not surprisingly, these techniques are similar to those recommended for supporting good mental health in general. If the self-care techniques listed above aren’t enough and stress is seriously interfering with your studies or life, don’t be afraid to get help. The student health center and college counselors are both good resources.
Stress isn’t always all bad. A little bit of stress can motivate you to do well on your exams, assignments, and work. Watch this supplemental video on How to Make Stress Your Friend, a TED Talk by Kelly McGonigal.
Stress Management Plan
Complete the following activity to assess your stress levels and formulate solutions to help reduce it.
ACTIVITY: My Stress Management Plan
My Key Symptoms of Stress
(List major physical, emotional, mental, social, and spiritual symptoms)
My Key Sources of Stress
(Consider internal, external, academic, job, interpersonal, and social stressors)
My Solutions to Stress
(Formulate some clear, specific strategies you will commit to applying to reduce or resolve stressors over the next week)
My Results
(After one week, give a candid, honest assessment of what you did and did not do to implement your stress management plan and the impact this has had on the level of stress in your life)
Journal Assignment
After reading the above content and watching the above video, take some time to consider each of the questions below and document your thoughts:
- What are some stressors that might challenge YOUR ability to maintain a growth mindset?
- How can adopting a growth mindset help you reframe stressful situations as opportunities for learning and development?
- How can viewing mistakes as part of the learning process help to reduce stress in high-pressure situations (like presentations, job interviews, etc.)?
- What role does resilience play in stress management, and how does a growth mindset contribute to building resilience?
- What practical steps will you take to integrate stress management techniques into your daily routine to support a long-term growth mindset?
Licenses and Attribution
" Chapter 16: Managing Your Mental and Physical Health " from Learning Framework: Effective Strategies for College Success 2.0 . Copyright © 2025 by Texas Higher Education Coordinating Board is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.
CC LICENSED CONTENT, ORIGINAL
- Managing Your Mental and Physical Health. Authored by : Laura Lucas. Provided by : Austin Community College. License : CC BY-NC-SA-4.0
CC LICENSED CONTENT, SPECIFIC ATTRIBUTION
- Chapter cover image. Authored by : Eli DeFaria. Provided by : Unsplash. Located at : https://unsplash.com/photos/101unQCLTxQ . License : CC0: No Rights Reserved
- Stress in College Success. Authored by: Unknown. Provided by: Lumen Learning. Located at : https://courses.lumenlearning.com/collegesuccess-lumen/chapter/stress/ . License : CC BY 4.0


10.E: Electromagnetism (Exercises)
Q1
- A parallel-plate capacitor has charge per unit area \(±σ\) on its two plates. Use Gauss’s law to find the field between the plates.
- In the style of Example 10.4.1, transform the field to a frame moving perpendicular to the plates, and verify that the result makes sense in terms of the sources that are present.
- Repeat the analysis for a frame moving parallel to the plates.
Q2
We’ve seen examples such as figure 10.1.1 in which a purely magnetic field in one frame becomes a mixture of magnetic and electric fields in another, and also cases like Example 10.4.1 in which a purely electric field transforms to a mixture. Can we have a case in which a purely electric field in one frame transforms to a purely magnetic one in another? The easy way to do this problem is by using invariants.
Q3
- Starting from Equation 10.3.5 for \(\mathcal{F}^{µν}\), lower an index to find \(\mathcal{F}^µ\: _ν\). Assume Minkowski coordinates and metric signature \(+---\).
- Let \(v = γ(1,u_x,u_y,u_z)\), where \((u_x,u_y,u_z)\) is the velocity threevector. Write out the matrix multiplication \(F^µ = q\mathcal{F}^µ\:_ν\; v^ν\), and show that the result is the Lorentz force law.
Q4
In section 10.6, I presented a list of properties of the electromagnetic stress tensor, followed by an argument in which the tensor is constructed with three unknown constants \(a\), \(b\), and \(c\), to be determined from those properties. The values of \(a\) and \(b\) are derived in the text, and the purpose of this problem is to finish up by proving that \(c = -1\). The idea is to take the field of a point charge, which we know satisfies Maxwell’s equations, and then apply property 8, which requires that the energy-conservation condition \(∂T^{ab}/∂x^a = 0\) hold. This works out nicely if you apply this property to the \(x\) column of \(T\), at a point that lies in the positive \(x\) direction relative to the charge.
Q5
Show that the number of independent conditions contained in equations 10.7.8 and 10.7.9 agrees with the number found in Maxwell's equations.
Q6
Show that
\[\frac{\partial \mathcal{F} ^{\mu\nu}}{\partial x^\lambda} + \frac{\partial \mathcal{F} ^{\nu\lambda}}{\partial x^\mu} + \frac{\partial \mathcal{F} ^{\lambda\mu}}{\partial x^\nu} = 0\]
implies that the magnetic field has zero divergence.
Q7
Write down the fields of an electromagnetic plane wave propagating in the z direction, choosing some polarization. Do not assume a sinusoidal wave. Show that this is a solution of
\[\frac{\partial \mathcal{F} ^{\mu\nu}}{\partial x^\nu} = 0\]


7.8: Gibbs Energy
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Learning Objectives
- To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if \(ΔS_{univ} > 0\) for a process or a reaction, then the process will occur spontaneously as written. Conversely, if \(ΔS_{univ} < 0\), a process cannot occur spontaneously; if \(ΔS_{univ} = 0\), the system is at equilibrium. The sign of \(ΔS_{univ}\) is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using \(ΔS_{univ}\) requires that we calculate \(ΔS\) for both a system and its surroundings . This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Energy and Spontaneity
The Gibbs energy (also known as the Gibbs function ) is defined as
\[G_{sys} = H_{sys} – T S_{sys} \label{23.4.1}\]
in which \(S\) refers to the entropy of the system . Since \(H\), \(T\) and \(S\) are all state functions, so is \(G\). Thus for any change in state, we can expand Equation \ref{23.4.1} to
\[ΔG_{sys} = ΔH_{sys} – T ΔS_{sys} \label{23.4.2}\]
How does this simple equation relate to the entropy change of the universe \(ΔS_{univ}\) that we know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition
\[ΔS_{univ} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}\]
we would first like to get rid of \(ΔS_{surr}\). How can a chemical reaction (a change in the system ) affect the entropy of the surroundings ? Because most reactions are either exothermic or endothermic, they are accompanied by heat \(q_p\) across the system boundary (we are considering constant pressure processes). The enthalpy change of the reaction \(ΔH_{sys}\) is the "flow" of heat into the system from the surroundings under constant pressure, so the heat "withdrawn" from the surroundings will be \(–q_p\).
From the thermodynamic definition of entropy, the change of the entropy of the surroundings will be
\[ΔS_{surr} = –\dfrac{q_p}{T} = –\dfrac{ΔH_{sys}}{T}.\]
We can therefore rewrite Equation \(\ref{23.4.3}\) as
\[ΔS_{univ} = \dfrac{– ΔH_{sys}}{T} + ΔS_{sys} \label{23.4.4}\]
Multiplying through by \(–T\) , we obtain
\[–T ΔS_{univ} = ΔH_{sys} – T ΔS_{sys} \label{23.4.5}\]
which expresses the entropy change of the universe in terms of thermodynamic properties of the system exclusively.
If \(–TΔS_{univ}\) is denoted by \(ΔG\), then we have Equation \ref{23.4.2} which defines the Gibbs energy change for the process.
The criterion for predicting spontaneity is based on (\(ΔG\)), the change in \(G\), at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as \(G\). At constant temperature and pressure,
\[ ΔG = ΔH − TΔS \label{Eq2}\]
where all thermodynamic quantities are those of the system. Recall that at constant pressure, \(ΔH = q\), whether a process is reversible or irreversible, and
\[TΔS = q_{rev}.\]
Using these expressions, we can reduce Equation \(\ref{Eq2}\) to
\[ΔG = q − q_{rev}.\]
Thus \(ΔG\) is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, \(q = q_{rev}\) and \(ΔG = 0\). As we shall soon see, if \(ΔG\) is zero, the system is at equilibrium, and there will be no net change.
What about processes for which \(ΔG ≠ 0\)? To understand how the sign of \(ΔG\) for a system determines the direction in which change is spontaneous, we can rewrite the relationship between \(\Delta S \) and \(q_{rev}\), discussed earlier.
\[ \Delta S= \dfrac{q_{rev}}{T} \nonumber\]
with the definition of \(\Delta H\) in terms of \(q_{rev}\)
\[ q_{rev} = ΔH \nonumber \]
to obtain
\[\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}\]
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, \(ΔS_{univ} > 0\), so substituting we obtain
\[\begin{align*} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \\[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align*}\]
Multiplying both sides of the inequality by \(−T\) reverses the sign of the inequality; rearranging,
\[ΔH_{sys}−TΔS_{sys}<0\]
which is equal to \(ΔG\) (Equation \(\ref{Eq2}\)). We can therefore see that for a spontaneous process, \(ΔG_{sys} < 0\).
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation \(\ref{Eq2}\) allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes (\(ΔH \ll 0\)) that increase the entropy of a system (\(ΔS_{sys} \gg 0\)) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
- If \(ΔG < 0\), the process occurs spontaneously.
- If \(ΔG = 0\), the system is at equilibrium.
- If \(ΔG > 0\), the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of \(ΔG\) dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor.
Example \(\PageIndex{1}\): Vaporizing Water
At Room Temperature (100 °C)
If the conversion of liquid water to water vapor is carried out at 1 atm and the normal boiling point of 100.00 °C (373.15 K), we can calculate \(ΔG\) from the experimentally measured value of \(ΔH_{vap}\) (40.657 kJ/mol). For vaporizing 1 mol of water, \(ΔH = 40,657; J\), so the process is highly endothermic. From the definition of \(ΔS\) (Equation \(\ref{Eq3}\)), we know that for 1 mol of water,
\[ \begin{align*} \Delta S_{\textrm{vap}} &= \dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \\[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \\[4pt] &=\textrm{108.96 J/K} \end{align*}\]
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
\[ \begin{align*} \Delta G_{100^\circ\textrm C} &=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \\[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \\[4pt] &=\textrm{0 J} \end{align*}\]
The energy required for vaporization offsets the increase in entropy of the system. Thus \(ΔG = 0\), and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Above Room Temperature (110 °C)
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of \(ΔG\) for the vaporization of 1 mol of water at 110°C, assuming that \(ΔH\) and \(ΔS\) do not change significantly with temperature, becomes
\[ \begin{align*} \Delta G_{110^\circ\textrm C} &=\Delta H-T\Delta S \\[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \\[4pt] &=-\textrm{1,091 J} \end{align*}\]
Since \(ΔG < 0\), the vaporization of water is predicted to occur spontaneously and irreversibly at 110 °C.
Below Room Temperature (90 °C)
We can also calculate \(ΔG\) for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
\[ \begin{align*}\Delta G_{90^\circ\textrm C} &= \Delta H-T\Delta S \\[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})]\\[4pt] &=\textrm{1,088 J} \end{align*}\]
Since \(ΔG > 0\), water does not spontaneously convert to water vapor at 90 °C. When using all the digits in the calculator display in carrying out our calculations, ΔG 110°C = 1090 J = −ΔG 90°C , as we would predict.
Equilibrium Temperature
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of \(ΔH\) and \(ΔS\) into the definition of \(ΔG\) (Equation \(\ref{Eq2}\)), setting \(ΔG = 0\), and solving for \(T\),
\[\begin{align*} 0 &=40,657\, J−T(108.96\, J/K) \\[4pt] T&=373.15 \,K \end{align*}\]
Thus \(ΔG = 0\) at \(T = 373.15\, K\) and \(1\, atm\), which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water.
At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, \(ΔG\) is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure \(\PageIndex{1}\) shows how the \(ΔH\) and \(TΔS\) terms vary with temperature for the vaporization of water. When the two lines cross, \(ΔG = 0\), and \(ΔH = TΔS.\)
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (\(ΔH > 0\)), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (\(ΔS > 0\)). At temperatures greater than 373 K, the \(TΔS\) term dominates, and \(ΔG < 0\), so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
| \(\Delta{H}\) | \(\Delta{S}\) | \(-T\Delta S\) | \(\Delta{G}\) |
|---|---|---|---|
|
-
(exothermic) |
+
(products more disordered) |
-
(favors spontaneity) |
-
(spontaneous at all T) |
|
-
(exothermic) |
-
(products less disordered) |
+
(opposes spontaneity) |
-
(spontaneous) at low T + (non-spontaneous) at high T "Enthalpically-driven process" |
|
+
(endothermic) |
+
(products more disordered) |
-
(favors spontaneity) |
+
(non-spontaneous) at low T - (spontaneous) at high T "Entropically-driven process" |
|
+
(endothermic) |
-
(products less disordered) |
+
(opposes spontaneity) |
+
(non-spontaneous at all T) |
Enthalpically vs. Entropically Driving Reactions
Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether \(ΔH\) or the \(TΔS\) term dominates. This is technically correct, but misleading because it disguises the important fact that \(ΔS_{total}\), which this equation expresses in an indirect way, is the only criterion of spontaneous change.
Consider the following possible states for two different types of molecules with some attractive force:
There would appear to be greater entropy on the left (state 1) than on the right (state 2). Thus the entropic change for the reaction as written (i.e. going to the right) would be (-) in magnitude, and the energetic contribution to the free energy change would be (+) (i.e. unfavorable) for the reaction as written.
In going to the right, there is an attractive force and the molecules adjacent to each other is a lower energy state (heat energy, \(q\), is liberated). To go to the left, we have to overcome this attractive force (input heat energy) and the left direction is unfavorable with regard to heat energy q. The change in enthalpy is (-) in going to the right (q released), and this enthalpy change is negative (-) in going to the right (and (+) in going to the left). This reaction as written, is therefore, enthalpically favorable and entropically unfavorable . Hence, It is enthalpically driven .
From Table \(\PageIndex{1}\), it would appear that we might be able to get the reaction to go to the right at low temperatures (lower temperature would minimize the energetic contribution of the entropic change). Looking at the same process from an opposite direction:
This reaction as written, is entropically favorable , and enthalpically unfavorable; it is entropically driven . From Table \(\PageIndex{1}\), it would appear that we might be able to get the reaction to go to the right at high temperatures (high temperature would increase the energetic contribution of the entropic change).
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of \(ΔG\) allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of \(ΔG\) for a process provides other important information. The change in Gibbs energy is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure):
\[ΔG = w_{max}.\]
To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation \(\ref{Eq2}\). We can rearrange this equation as follows:
\[ ΔH = ΔG + TΔS \label{Eq4}\]
This equation tells us that when energy is released during an exothermic process (\(ΔH < 0\)), such as during the combustion of a fuel, some of that energy can be used to do work (\(ΔG < 0\)), while some is used to increase the entropy of the universe (\(TΔS > 0\)). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, \(ΔH\) will never equal \(ΔG\) because \(ΔS\) has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table \(\PageIndex{2}\), the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
| Device | Energy Conversion | Approximate Efficiency (%) |
|---|---|---|
| large electrical generator | mechanical → electrical | 99 |
| chemical battery | chemical → electrical | 90 |
| home furnace | chemical → heat | 65 |
| small electric tool | electrical → mechanical | 60 |
| space shuttle engine | chemical → mechanical | 50 |
| mammalian liver cell | chemical → chemical | 30–50 |
| spinach cell | light → chemical | 30 |
| internal combustion engine | chemical → mechanical | 25–30 |
| fluorescent light | electrical → light | 20 |
| solar cell | light → electricity | 10 |
| incandescent light bulb | electricity → light | 5 |
| yeast cell | chemical → chemical | 2–4 |
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation \(\ref{Eq5}\):
\[ΔG° = ΔH° − TΔS° \label{Eq5}\]
If ΔS° and ΔH° for a reaction have the same sign, then the sign of \(ΔG^o\) depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of \(ΔG^o\) for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive \(ΔG^o\) means that the equilibrium constant is less than 1.
Example \(\PageIndex{1}\)
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
\[\ce{ H2(g) + O2(g) <=> H2O2(l)} \nonumber\]
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
- S°(H 2 O 2 ) = 109.6 J/(mol•K),
- S°(O 2 ) = 205.2 J/(mol•K), and
- S°(H 2 ) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given : balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy :
- Calculate ΔS° from the absolute molar entropy values given.
- Use Equation \(\ref{Eq5}\), the calculated value of ΔS°, and other data given to calculate \(ΔG^o\) for the reaction.
- Use the value of \(ΔG^o\) to determine whether the reaction is spontaneous as written.
Solution
A To calculate \(ΔG^o\) for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\[\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})]
\\[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})]
\\[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times 205.2\;\mathrm{J/(mol\cdot K)}] \right \}
\\[4pt] &=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*}\]
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction.
B Substituting the appropriate quantities into Equation \(\ref{Eq5}\),
\[\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \\[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}] \\[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \\[4pt] &=-120.31\textrm{ kJ/mol} \nonumber \end{align*}\]
The negative value of \(ΔG^o\) indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of \(ΔG^o\) depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants.
Exercise \(\PageIndex{1}\)
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
\[\ce{ 2H2(g) + N2(g) <=> N2H4(l)}. \nonumber \]
At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N 2 H 4 ) = 121.2 J/(mol•K), S°(N 2 ) = 191.6 J/(mol•K), and S°(H 2 ) = 130.7 J/(mol•K). Is the reaction spontaneous as written?
- Answer
-
149.5 kJ/mol; no
Tabulated values of standard free energies of formation allow chemists to calculate the values of \(ΔG^o\) for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation (\(ΔG^∘_f\))of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl 2 gas at 298.15 K, for example, has \(ΔG^∘_f = 0\). The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔH ∘ f ) and the standard entropy of formation (ΔS ∘ f ) using the definition of free energy:
\[ΔG^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}\]
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
\[ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum n ΔG^o_{f} (reactants) \label{Eq7a}\]
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative \(ΔG^o\) indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
"Products minus Reactants" Rule
The \(ΔG^o\) of a reaction can be calculated from tabulated \(ΔG^o_f\) values ( Table T1 ) using the “products minus reactants” rule.
Example \(\PageIndex{2}\)
Calculate \(ΔG^o\) for the reaction of isooctane with oxygen gas to give carbon dioxide and water. Use the following data:
- ΔG° f (isooctane) = −353.2 kJ/mol,
- ΔG° f (CO 2 ) = −394.4 kJ/mol, and
- ΔG° f (H 2 O) = −237.1 kJ/mol.
Is the reaction spontaneous as written?
Given : balanced chemical equation and values of ΔG° f for isooctane, CO 2 , and H 2 O
Asked for : spontaneity of reaction as written
Strategy :
Use the “products minus reactants” rule to obtain ΔG ∘ rxn , remembering that ΔG° f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
\[\ce{ C8H18(l) + 25/2 O2(g) -> 8CO2(g) + 9H2O(l)} \nonumber\]
We are given ΔG ∘ f values for all the products and reactants except O 2 (g). Because oxygen gas is an element in its standard state, ΔG ∘ f (O 2 ) is zero. Using the “products minus reactants” rule,
\[\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right]
\\[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})]
\\[4pt] &-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ]
\\[4pt] &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}}) \end{align*}\]
Because \(ΔG^o\) is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of \(ΔG^o\) is largely determined by the ΔG ∘ f of the stable products: water and carbon dioxide.
Exercise \(\PageIndex{2}\)
Calculate \(ΔG^o\) for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
- ΔG ∘ f (benzene) = 124.5 kJ/mol
- ΔG ∘ f (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
- Answer
-
92.8 kJ; no
Calculated values of \(ΔG^o\) are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of \(ΔG^o\) may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having K eq < 1, to one that is, having a K eq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate \(ΔG^o\) at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither \(ΔH\) nor \(ΔS\) vary greatly with temperature.
Example \(\PageIndex{3}\)
Calculate (a) \(ΔG^o\) and (b) ΔG 300°C for the reaction N 2 (g)+3H 2 (g)⇌2NH 3 (g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
- S°(N 2 ) = 191.6 J/(mol•K),
- S°(H 2 ) = 130.7 J/(mol•K),
- S°(NH 3 ) = 192.8 J/(mol•K), and
- ΔH ∘ f (NH 3 ) = −45.9 kJ/mol.
Given : balanced chemical equation, temperatures, S° values, and ΔH ∘ f for NH 3
Asked for : \(ΔG^o\) and ΔG at 300°C
Strategy :
- Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔH ∘ f for any element in its standard state is zero.
- Substitute the appropriate values into Equation \(\ref{Eq5}\) to obtain \(ΔG^o\) for the reaction.
- Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation \(\ref{Eq2}\) to obtain ΔG for the reaction at 300°C.
Solution
A To calculate \(ΔG^o\) for the reaction using Equation \(\ref{Eq5}\), we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\[\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})]
\\[4pt] &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}]
\\[4pt] &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \} \\[4pt] &=-198.1\textrm{ J/K (per mole of N}_2)\end{align*} \]
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔH ∘ f (NH 3 ) is given, and ΔH ∘ f is zero for both N 2 and H 2 :
\[\begin{align*}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})] \\[4pt] &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})] \\[4pt] &=-91.8\textrm{ kJ(per mole of N}_2) \end{align*} \]
B Inserting the appropriate values into Equation \(\ref{Eq5}\)
\[\begin{align*} \Delta G^\circ_{\textrm{rxn}} &=\Delta H^\circ-T\Delta S^\circ \\[4pt] &=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \\[4pt] &=-\textrm{32.7 kJ (per mole of N}_2) \end{align*}\]
C To calculate \(ΔG\) for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH 300°C = H° and ΔS 300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation \(\ref{Eq2}\):
\[\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \\[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ \\[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \\[4pt] &=21.7\textrm{ kJ (per mole of N}_2) \end{align*} \]
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its K eq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise \(\PageIndex{3}\)
Calculate
- \(ΔG°\) and
- \(ΔG_{750°C}\)
for the following reaction
\[\ce{2NO(g) + O2 (g) <=> 2NO2 (g)} \nonumber\]
which is important in the formation of urban smog. Assume that \(ΔH\) and \(ΔS\) do not change between 25.0°C and 750°C and use these data:
- S°(NO) = 210.8 J/(mol•K),
- S°(O 2 ) = 205.2 J/(mol•K),
- S°(NO 2 ) = 240.1 J/(mol•K),
- ΔH ∘ f (NO 2 ) = 33.2 kJ/mol, and
- ΔH ∘ f (NO) = 91.3 kJ/mol.
- Answer a
-
−72.5 kJ/mol of \(O_2\)
- Answer b
-
33.8 kJ/mol of \(O_2\)
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting \(ΔG^o\) = 0 in Equation \(\ref{Eq5}\), as illustrated in Example \(\PageIndex{4}\).
Example \(\PageIndex{4}\)
As you saw in Example \(\PageIndex{3}\), the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example \(\PageIndex{3}\) to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given : ΔH° and ΔS°
Asked for : temperature at which reaction changes from spontaneous to nonspontaneous
Strategy :
Set \(ΔG^o\) equal to zero in Equation \(\ref{Eq5}\) and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example \(\PageIndex{3}\), we calculated that ΔH° is −91.8 kJ/mol of N 2 and ΔS° is −198.1 J/K per mole of N 2 , corresponding to \(ΔG^o\) = −32.7 kJ/mol of N 2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting \(ΔG^o\) equal to zero and rearranging Equation \(\ref{Eq5}\) to solve for T:
\[\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0
\\[4pt] \Delta H^\circ &=T\Delta S^\circ
\\[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*}\]
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise \(\PageIndex{4}\)
As you found in the exercise in Example \(\PageIndex{3}\), ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
- Answer
-
792.6 K
Summary
- The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
- Free-energy change
\[ΔG = ΔH − TΔS \nonumber\]
- Standard free-energy change
\[ΔG° = ΔH° − TΔS° \nonumber\]
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔG ∘ f ), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate \(ΔG^o\) for a reaction.
Contributors and Attributions
-
Stephen Lower, Professor Emeritus ( Simon Fraser U. ) Chem1 Virtual Textbook


15.21D: Arboviral Encephalitis
- Examine the mode of transmission and causes of arboviral encephalitis
Arboviral encephalitis are a group of arthropod-transmitted viruses that cause encephalitis (acute swelling in the brain). The word “arbovirus” directly refers to an ARthropod-BOrne virus. Arthropod vectors transmit the virus upon biting, allowing the virus to enter the circulatory system and replicate and shed additional infection into the bloodstream ( viremia ).
The majority of the arboviruses are spherical in shape although a few are rod-shaped. They are 17-150 nm in diameter and most have an RNA genome (the single exception is African swine fever virus, which has a DNA genome). Many arboviruses (such as African Swine Fever virus) do not infect humans or cause only mild and transient infections characterized by fever, headache, and rash. Those of the arboviral encephalitis group, however, can cause epidemic disease and severe infections that can be fatal. Arboviral encephalitis are found in many places throughout the world, and include California encephalitis, Japanese encephalitis, St. Louis encephalitis, Tick-borne encephalitis, and West Nile fever.
Tick-borne encephalitis (TBE) is an infectious disease of the central nervous system. It can infect a range of hosts including ruminants, birds, rodents, carnivores, horses, and humans. The disease can be zoonotic, with ruminants and dogs providing the principal source of infection for humans. TBE is transmitted through the bite of several species of infected ticks, including Ixodes scapularis, Ixodes ricinus and Ixodes persulcatus, and manifests most often as meningitis, encephalitis, or meningoencephalitis.
TBE Infection can be reliably prevented by vaccination, but is incurable once manifested. Long-lasting or permanent neuropsychiatric sequelae are observed in 10-20% of infected patients; morality occurs in only 1-2% of the infected, with deaths occurring 5 to 7 days after the onset of neurologic symptoms.
TBE and other arboviral encephalitis can be diagnosed through a combination of blood tests, particularly immunologic, serologic, and/or virologic techniques such as ELISA, complement fixation, polymerase chain reaction, Neutralization test, and Hemoagglutination Inhibition test.
Because the arboviral encephalitides are viral diseases, antibiotics are not effective for treatment and no effective antiviral drugs have been discovered yet. Treatment is therefore only supportive, attempting to deal with problems such as swelling of the brain, loss of the automatic breathing, activity of the brain, and other treatable complications like bacterial pneumonia.
Therefore, the immune system plays an important role in defense against arbovirus infections. Arboviruses usually stimulate the production of interferons and antibodies, which help to diminish the extent of viremia. Cell-mediated immunity is also important. Increased immunity is observed with age progression.
Vector control measures, such as habitat control (including the elimination of stagnant water and the spraying of insecticides), are essential to reducing the transmission of disease by arboviruses. People can also reduce the risk of getting bitten by arthropods by employing personal protective measures such as sleeping under mosquito nets, wearing protective clothing, applying insect repellents, tick-checks, and avoiding areas known to harbor high arthropod populations.
Key Points
- Arthropod vectors transmit the virus upon biting, allowing the virus to enter the circulatory system and replicate.
- Arboviral encephalitis are found in many places throughout the world, and include California encephalitis, Japanese encephalitis, St. Louis encephalitis, Tick-borne encephalitis, and West Nile fever.
- TBE Infection can be reliably prevented by vaccination, but is incurable once manifested.
Key Terms
- sequelae : A pathological condition resulting from a disease, injury, or other trauma.
- encephalitis : an inflammation of the brain
- viremia : A medical condition where viruses enter the bloodstream and hence have access to the rest of the body.
- zoonotic : of or relating to zoonosis, the transmission of an infectious disease between species.


2.2B: Ions and Ionic Bonds
Ionic bonds are attractions between oppositely charged atoms or groups of atoms where electrons are donated and accepted.
- Predict whether a given element will more likely form a cation or an anion
Key Points
- Ions form from elements when they gain or lose an electron causing the number of protons to be unequal to the number of electrons, resulting in a net charge.
- If there are more electrons than protons (from an element gaining one or more electrons), the ion is negatively charged and called an anion.
- If there are more protons than electrons (via loss of electrons), the ion is positively charged and is called a cation.
- Ionic bonds result from the interaction between a positively charged cation and a negatively charged anion.
Key Terms
- ion : An atom, or group of atoms, bearing an electrical charge, such as the sodium and chlorine atoms in a salt solution.
- ionic bond : A strong chemical bond caused by the electrostatic attraction between two oppositely charged ions.
Ions and Ionic Bonds
Some atoms are more stable when they gain or lose an electron (or possibly two) and form ions. This results in a full outermost electron shell and makes them energetically more stable. Now, because the number of electrons does not equal the number of protons, each ion has a net charge. Cations are positive ions that are formed by losing electrons (as the number of protons is now greater than the number of electrons). Negative ions are formed by gaining electrons and are called anions (wherein there are more electrons than protons in a molecule ). Anions are designated by their elemental name being altered to end in “-ide.” For example, the anion of chlorine is called chloride, and the anion of sulfur is called sulfide.
This movement of electrons from one element to another is referred to as electron transfer. As illustrated, sodium (Na) only has one electron in its outer electron shell. It takes less energy for sodium to donate that one electron than it does to accept seven more electrons to fill the outer shell. When sodium loses an electron, it will have 11 protons, 11 neutrons, and only 10 electrons. This leaves it with an overall charge of +1 since there are now more protons than electrons. It is now referred to as a sodium ion. Chlorine (Cl) in its lowest energy state (called the ground state) has seven electrons in its outer shell. Again, it is more energy efficient for chlorine to gain one electron than to lose seven. Therefore, it tends to gain an electron to create an ion with 17 protons, 17 neutrons, and 18 electrons. This gives it a net charge of -1 since there are now more electrons than protons. It is now referred to as a chloride ion. In this example, sodium will donate its one electron to empty its shell, and chlorine will accept that electron to fill its shell. Both ions now satisfy the octet rule and have complete outer shells. These transactions can normally only take place simultaneously; in order for a sodium atom to lose an electron, it must be in the presence of a suitable recipient like a chlorine atom.
Electron Transfer Between Na and Cl : In the formation of an ionic compound, metals lose electrons and nonmetals gain electrons to achieve an octet. In this example, sodium loses one electron to empty its shell and chlorine accepts that electron to fill its shell.
Ionic bonds are formed between ions with opposite charges. For instance, positively charged sodium ions and negatively charged chloride ions bond together to form sodium chloride, or table salt, a crystalline molecule with zero net charge. The attractive force holding the two atoms together is called the electromagnetic force and is responsible for the attraction between oppositely charged ions.
Certain salts are referred to in physiology as electrolytes (including sodium, potassium, and calcium). Electrolytes are ions necessary for nerve impulse conduction, muscle contractions, and water balance. Many sports drinks and dietary supplements provide these ions to replace those lost from the body via sweating during exercise.


5.10: Detritus
In biology, detritus is dead particulate organic material, as distinguished from dissolved organic material. Detritus typically includes the bodies or fragments of bodies of dead organisms, and fecal material. Detritus typically hosts communities of microorganisms that colonize and decompose it. In terrestrial ecosystems it is present as leaf litter and other organic matter that is intermixed with soil, which is denominated "soil organic matter". The detritus of aquatic ecosystems is organic substances that is suspended in the water and accumulates in depositions on the floor of the body of water; when this floor is a seabed, such a deposition is denominated "marine snow".
The corpses of dead plants or animals, material derived from animal tissues (e.g. molted skin), and fecal matter gradually lose their form due to physical processes and the action of decomposers, including grazers, bacteria, and fungi. Decomposition, the process by which organic matter is decomposed, occurs in several phases. Micro- and macro-organisms that feed on it rapidly consume and absorb materials such as proteins, lipids, and sugars that are low in molecular weight, while other compounds such as complex carbohydrates are decomposed more slowly. The decomposing microorganisms degrade the organic materials so as to gain the resources they require for their survival and reproduction. Accordingly, simultaneous to microorganisms' decomposition of the materials of dead plants and animals is their assimilation of decomposed compounds to construct more of their biomass (i.e. to grow their own bodies). When microorganisms die, fine organic particles are produced, and if small animals that feed on microorganisms eat these particles they collect inside the intestines of the consumers, and change shape into large pellets of dung. As a result of this process, most of the materials of dead organisms disappear and are not visible and recognizable in any form, but are present in the form of a combination of fine organic particles and the organisms that used them as nutrients. This combination is detritus.
In ecosystems on land, detritus is deposited on the surface of the ground, taking forms such as the humic soil beneath a layer of fallen leaves. In aquatic ecosystems, most detritus is suspended in water, and gradually settles. In particular, many different types of material are collected together by currents, and much material settles in slowly flowing areas.
A large amount of detritus is used as a source of nutrition for animals. In particular, many bottom feeding animals (benthos) living in mud flats feed in this way. In particular, since excreta are materials which other animals do not need, whatever energy value they might have, they are often unbalanced as a source of nutrients, and are not suitable as a source of nutrition on their own. However, there are many microorganisms which multiply in natural environments. These microorganisms do not simply absorb nutrients from these particles, but also shape their own bodies so that they can take the resources they lack from the area around them, and this allows them to make use of excreta as a source of nutrients. In practical terms, the most important constituents of detritus are complex carbohydrates, which are persistent (difficult to break down), and the microorganisms which multiply using these absorb carbon from the detritus, and materials such as nitrogen and phosphorus from the water in their environment to synthesise the components of their own cells.
A characteristic type of food chain called the detritus cycle takes place involving detritus feeders (detritivores), detritus and the microorganisms that multiply on it. For example, mud flats are inhabited by many univalves which are detritus feeders. When these detritus feeders take in detritus with microorganisms multiplying on it, they mainly break down and absorb the microorganisms, which are rich in proteins, and excrete the detritus, which is mostly complex carbohydrates, having hardly broken it down at all. At first, this dung is a poor source of nutrition, and so univalves pay no attention to it, but after several days, microorganisms begin to multiply on it again, its nutritional balance improves, and so they eat it again. Through this process of eating the detritus many times over and harvesting the microorganisms from it, the detritus thins out, becomes fractured and becomes easier for the microorganisms to use, and so the complex carbohydrates are also steadily broken down and disappear over time.
What is left behind by the detritivores is then further broken down and recycled by decomposers, such as bacteria and fungi.
This detritus cycle plays a large part in the so-called purification process, whereby organic materials carried in by rivers is broken down and disappears, and an extremely important part in the breeding and growth of marine resources. In ecosystems on land, far more essential material is broken down as dead material passing through the detritus chain than is broken down by being eaten by animals in a living state. In both land and aquatic ecosystems, the role played by detritus is too large to ignore.
Aquatic ecosystems
In contrast to land ecosystems, dead materials and excreta in aquatic ecosystems are typically transported by water flow; finer particles tend to be transported farther or suspended longer. In freshwater bodies organic material from plants can form a silt known as mulm or humus on the bottom. This material, some called undissolved organic carbon breaks down into dissolved organic carbon and can bond to heavy metal ions via chelation. It can also break down into colored dissolved organic matter such as tannin , a specific form of tannic acid. In saltwater bodies, organic material breaks down and forms a marine snow. This example of detritus commonly consists of organic materials such as dead phytoplankton and zooplankton, the outer walls of diatoms and coccolithophores, dead skin and scales of fish, and fecal pellets. This material will slowly sink to the seafloor, where it makes up the majority of sediment in some areas. Once settled, the material will not only contribute to sediments but will help to feed different species of detritivore, organisms which feed on detritus, such as annelid worms and sea cucumbers, to name a few. The exact composition of this detritus varies based on location and time of year, as it is very closely tied to primary production.
Terrestrial ecosystems
Detritus occurs in a variety of terrestrial habitats including forest, chaparral and grassland. In forests, the detritus is typically dominated by leaf, twig, and bacteria litter as measured by biomass dominance. This plant litter provides important cover for seedling protection as well as cover for a variety of arthropods, reptiles and amphibians. Some insect larvae feed on the detritus. Fungi and bacteria continue the decomposition process after grazers have consumed larger elements of the organic materials, and animal trampling has assisted in mechanically breaking down organic matter. At the later stages of decomposition, mesophilic micro-organisms decompose residual detritus, generating heat from exothermic processes; such heat generation is associated with the well known phenomenon of the elevated temperature of composting.
Consumers
There is an extremely large number of detritus feeders in water. After all, a large quantity of material is carried in by water currents. Even if an organism stays in a fixed position, as long as it has a system for filtering water, it will be able to obtain enough food to get by. Many immobile organisms survive in this way, using developed gills or tentacles to filter the water to take in food, a process known as filter feeding.
Another more widely used method of feeding, which also incorporates filter feeding, is a system where an organism secretes mucus to catch the detritus in lumps, and then carries these to its mouth using an area of cilia. This is called mucus feeding.
Many organisms, including sea slugs and serpent's starfish, scoop up the detritus which has settled on the water bed. Bivalves which live inside the water bed do not simply suck in water through their tubes, but also extend them to fish for detritus on the surface of the bed.
Producers
In contrast, from the point of view of organisms using photosynthesis such as plants and plankton, detritus reduces the transparency of the water and gets in the way of this process. Given that these organisms also require a supply of nutrient salts, in other words fertilizer, for photosynthesis, their relationship with detritus is a complex one.
In land ecosystems, the waste products of plants and animals collect mainly on the ground (or on the surfaces of trees), and as decomposition proceeds, plants are supplied with fertilizer in the form of inorganic salts. In water ecosystems, relatively little waste collects on the water bed, and so the progress of decomposition in water takes a more important role. Investigating the level of inorganic salts in sea ecosystems shows that unless there is an especially large supply, the quantity increases from winter to spring—but is normally extremely low in summer. As such, the quantity of seaweed present reaches a peak in early summer and then decreases. The thinking is that organisms like plants grow quickly in warm periods and thus the quantity of inorganic salts is not enough to keep up with the demand. In other words, during winter, plant-like organisms are inactive and collect fertilizer, but if the temperature rises to some extent they will use this up in a very short period.
It is not entirely true that their productivity falls during the warmest periods. Organisms such as dinoflagellate have mobility, the ability to take in solid food, and the ability to photosynthesise. This type of micro-organism can take in substances such as detritus to grow, without waiting for it to be broken down into fertilizer.
From Wikipedia, November 2023 https://en.Wikipedia.org/wiki/Detritus


3: IONIZATION TECHNIQUES
A variety of ionization techniques are used for mass spectrometry. Most ionization techniques excite the neutral analyte molecule which then ejects an electron to form a radical cation M •+ . This radical cation is the molecular ion and is produced by removing a single electron from from a neutral molecule. Other ionization techniques involve ion molecule reactions between an ion and a neutral molecule that produce an adduct ion like [M+H] + . Many of these reactions cause the addition of a proton H + to the analyte molecule but other ions can also be formed. The most important considerations for selecting an ionization technique are the physical state of the analyte and the ionization energy needed. Electron ionization and chemical ionization are only suitable for gas phase ionization. Fast atom bombardment, secondary ion mass spectrometry, electrospray, and matrix assisted laser desorption are used to ionize condensed phase samples. The ionization energy is significant because it controls the amount of fragmentation observed in the mass spectrum. Although this fragmentation complicates the mass spectrum it provides structural information for the identification of unknown compounds. Some ionization techniques are very soft and only produce molecular ions, the intact ionized analyte molecule, while other techniques are very energetic and cause ions to undergo extensive fragmentation.