How to calculate the concentration after a certain timespan from the specific reaction constant?
I've been going crazy with this question for the past 4 hours. I can't find much help in google because it's not a traditional "find the velocity of the reaction" question. It's fine if you don't give me the answer, just teach me how to get there:
A container at $30~^\circ\mathrm{C}$ contains $\ce{HNC_{(g)}}$ reacting to form $\ce{HCN_{(g)}}$. Knowing that the specific reaction constant is $4.4\cdot10^{-4}~\mathrm{s}^{-1}$ and that you initially have $1~\mathrm{g}$ of $\ce{HNC}$: how many moles of each gas are in the container after one hour and a half?
I'm pretty sure that the reaction is reversible here. However, I have no idea how to calculate how many moles of each gas will there be after $1.5~\mathrm{h}$ (even if it was not reversible).
I have many doubts: Did they skip the concentration because there's only one reactant? Does $K$ take into account the $30~^\circ\mathrm{C}$? What's the equation for "change of concentration over time" here?!
How can I figure out how many moles of $\ce{HNC}$ and $\ce{HCN}$ the container will have after 90 minutes?
EDIT: Thanks for the comments! I get your points and edited the post. However, how can I figure out how many moles of $\ce{HCN}$ I have after 90 minutes ignoring the fact that it's a reversible reaction?.
homework equilibrium gas-laws concentration
Martin - マーチン♦
SebsSebs
$\begingroup$ Where did you get the idea that the reaction formed an equilibrium at equal amounts of HNC and HCN? In truth, the equilibrium is quite favoured towards products; hydrogen isocyanide is a rather unstable molecule. For this problem, you don't really need to think about chemical equilibrium at all (or equivalently, assume the equilibrium constant is infinite). $\endgroup$ – Nicolau Saker Neto May 15 '14 at 3:42
$\begingroup$ I think that because it is such a slow reaction that you don't reach equilibrium in 90 minutes. The provided rate is for the specified temperature of 30°C. For a first approximation, I would treat it as only a forward reaction to see what the concentrations would be. If the HCN concentration is still small (in comparison), I would go with that. $\endgroup$ – LDC3 May 15 '14 at 3:42
$\begingroup$ Okay thanks a lot! I understand. However, how can I figure out how many moles of HCN I have after 90 minutes? $\endgroup$ – Sebs May 15 '14 at 4:16
$\begingroup$ Because the rate constant is $4.4\times 10^{-4} s^{-1}$, it tells us that it is a first order reaction. If you look at this section in Wikipedia, you should see the direction to go. en.wikipedia.org/wiki/Rate_equation#First-order_reactions $\endgroup$ – LDC3 May 15 '14 at 5:19
To answer this, you should know what is the concentration of HNC after 90 minutes. First, you should know the initial concentration of HNC.
After that, look at the rate constant. It tell you if the reaction is a first order. So, you can apply the data in integrated first-order rate law equation:
Don't forget if you must change 90 minutes to seconds :)
Then, now you find the concentration of HNC after 90 minutes. To figure out the moles of HCN after reaction, we know that the initial moles of HNC is the same as the moles of HNC after reaction plus the moles of HNC that changed to HCN.
Then after submit the initial moles of HNC and moles of HNC after reaction, you get the moles of HNC that changed to HCN. This is the same of moles of HCN that present in the container.
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