[Big-Daddy]

Problem:
Desorption of CH from the IIG surface follows first-order kinetics. Calculate the average residence time of CH on the surface at 20 K. Assume that A = 1 x 10¹² s^-1 and E_des = 12 kJ mol^-1.

Comments:
Seems really simple - k (s^-1) = A·e^(-E_des/(R·T)) = 1·10¹² · exp(-12000/(8.314·293.15)) = 7.273 · 10⁹ s^-1
From which it follows that t(average)=1/k = 1.3749 · 10^-10 s.

But the solution booklet evaluates e^(-E_des/(R·T)) differently it would seem - as 5·10^-32? What's going on?

Also, I've made the intuitive note that inversing the rate constant will find the average time. But really, we should be inversing rate itself to find the average time, no? The only reason I didn't do this was because I couldn't (there's no concentration or activity data to calculate rate with, only the rate constant).

[Corribus]

(1) You've made a very simple error, one which will likely induce forehead slapping when you discover it. Here's a hint: be careful with your temperature units and read the problem carefully.

(2) The rate constant (for first order reaction) is the (average) number of reaction events per unit time.  Therefore the inverse of the rate constant is the (average) amount of time for one reaction event to occur. So your intuition is correct. For a first order reaction, the amount of time it takes for a reaction event to occur is independent of the amount of the substance in the system.

http://en.wikipedia.org/wiki/Rate_equation#First-order_reactions

This should make sense, because there is no collision required in a first order reaction. In a second order reaction, for instance, the probability of a reaction event depends on the probability of two molecules colliding. Here, the reactants can be considered "independent of each other". That is, the probability of a molecule undergoing a reaction is the same if it's the only molecule in the system, or if it's highly concentrated.

For other reaction orders, things are more complicated, because the rate constant changes as a function of concentration.

[Big-Daddy]

(1) You've made a very simple error, one which will likely induce forehead slapping when you discover it. Here's a hint: be careful with your temperature units and read the problem carefully.

I see now. T=20 K. Not 20°C. That's annoying.

(2) The rate constant (for first order reaction) is the (average) number of reaction events per unit time.  Therefore the inverse of the rate constant is the (average) amount of time for one reaction event to occur. So your intuition is correct. For a first order reaction, the amount of time it takes for a reaction event to occur is independent of the amount of the substance in the system.

http://en.wikipedia.org/wiki/Rate_equation#First-order_reactions

This should make sense, because there is no collision required in a first order reaction. In a second order reaction, for instance, the probability of a reaction event depends on the probability of two molecules colliding. Here, the reactants can be considered "independent of each other". That is, the probability of a molecule undergoing a reaction is the same if it's the only molecule in the system, or if it's highly concentrated.

Understood.

For other reaction orders, things are more complicated, because the rate constant changes as a function of concentration.

So from the rate constant alone it is impossible to find the average number of reaction events per unit time for a non-first order reaction. Is it possible to find this average number per unit time some other way?

[Corribus]

So from the rate constant alone it is impossible to find the average number of reaction events per unit time for a non-first order reaction. Is it possible to find this average number per unit time some other way?

Good question. What do you think?

Here is a revealing exercise for you, if you're game for a little math.

Start by finding a way to prove that the intuition in your first part is correct: that the average time it takes for a reaction event to occur in a first order reaction is independent of the initial concentration of reactant. Once you are convinced you can do this, try to apply it to a second order reaction, such as 2A products.

See what you get and think about what it means.

[Big-Daddy]

Start by finding a way to prove that the intuition in your first part is correct: that the average time it takes for a reaction event to occur in a first order reaction is independent of the initial concentration of reactant.

(M could be pressure, molarity, whatever is being transferred and measured)
d[A]/dt (M·s^-1) = k (s^-1) · [A] (M)
If k is something per second, then this must be events per second where each event transfers a certain amount of M. Then taking the inverse of k reverses the units ((event·s^-1)^-1=s·event^-1) which is the average time taken for a single event.

Not sure where to go from here; this is still a rate-constant based approach so I am probably missing the point. Can I have a hint?

[Corribus]

Ok, maybe that was a little vague.

Start with the integrated expression for [A] as a function of t (time) for a first order reaction.  Now, this equation can be reformulated for t as a function of [A]. Now calculate the average value of t, <t>, over the bounds of interest (which are what?), for the function t([A]).  This gives you the average value of time per reaction event. If your intuition is right, it should give an answer of 1/k - that is, independent of the reaction concentration.

One you do this, try it for the analogous integrated expression for the second order reaction.

If you still need more guidance, let me know.

[Big-Daddy]

Ok then let me give it a shot. The general integration for the first-order differential:

[tex]t = t_0 + \frac{log_e(\frac{c_{A,0}}{c_A})}{k}[/tex]

Now we have t([A]), the average of which according to this page (discussing averages of functions: http://www.math.northwestern.edu/~mlerma/courses/math214-2-03f/notes/c2-mvt.pdf) is found by 1/(-[A]₀) multiplied by the integral from [A]₀ to 0 of our t([A]) with respect to [A]. I tried evaluating some of this but I'm sure it won't drop out as 1/k ... should I be using something else for the average?

Edit: Haha wow. I finished the integral and look what turns out:

[tex]\int^0_{c_{A,0}}( t (c_A)) d(c_A) = \frac{-c_{A,0} \cdot (k \cdot t_0 + 1)}{k}[/tex]

Therefore

[tex]t_{average}=\frac{-1}{c_{A,0}} \cdot \frac{-c_{A,0} \cdot (k \cdot t_0 + 1)}{k}[/tex]

So that

[tex]t_{average}=t_0+\frac{1}{k}[/tex]

Just like you said. t₀=0 for our case since we are trying to measure an average time between events and so the starting off-set doesn't exist.

And this is the average time taken for the loss of one mole of A? Or is it one molecule? The units here don't seem to indicate either way ...

[Corribus]

Well done, BD.  Before analyzing too much, try something else. What happens if you set initial bound as [A]₀ and the final bound as [A]₀/2? And what about [A]₀/2 and 0 as the bounds? Anything jump out at you? You might want to consider how your answers relate to the half life of a first order reaction.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Reaction_Rates/First-Order_Reactions

A more interesting case is to try to follow the same procedure with the second order reaction 2A products. Try it and see what happens.

Regarding your question about what the units are: it is irrelevant. The rate constant for a first order reaction is the amount of reaction events (conversions of A to B) per unit time.  It doesn't matter whether you express the amount of 'stuff' as absolute molecules or molar concentrations because the amount of stuff doesn't matter as far as the kinetics are concerned.  This is, you will find, not the case for a second order reaction, where the rate constant depends on the amount of stuff.  It is convention to express the rate constant as the amount of conversions per unit time (s) per unit concentration (M).

[Big-Daddy]

Ok, the completely general form of the first-order integral is

[tex]t_{average} = \frac{A_{final} \cdot log_e( \frac{A_0}{A_{final}} ) - A_{initial} \cdot log_e( \frac{A_0}{A_{initial}} ) + (A_{final}-A_{initial}) \cdot (k \cdot t_0 + 1)}{k \cdot ( A_{final) - A_{initial} ) }[/tex]

In the previous case we had A_final=0 and A_initial=A₀. Now if I try out A_initial=A₀ and A_final=(1/2)A₀ I get

[tex]t_{average} = -t_0 - \frac{1}{k} + \frac{log_e(2)}{k}[/tex]

So we're ending up with that same ln(2)/k that we get for the half-life in first-order reactions.

I'll get to work on the non-1st order integral then.