Can someone please explain to me how this \"mixed order\" reaction works: Rate E
ID: 836414 • Letter: C
Question
Can someone please explain to me how this "mixed order" reaction works:
Rate Equation: Rate = (k1[E][A]2)/(k2 + k3[A]), where E is a catalyst and A is the single reactant for the reaction.
My book says that when [A] is large at the beginning of the reaction k3[A]>>k2 and the reaction is first-order. However, at the end of the reaction, when [A] is low, k2>>k3[A] and the reaction is second-order. I am having trouble seeing how the rate equation shows this. Can someone please explain to me how the rate equation illustrates this change in order as the reaction progresses? Thanks!
Explanation / Answer
Hi,
The explanation is quite simpleonce you see with the concepts of limits and approximating the denominator.
I'll try to try to get you to 'see' what is happening.
Initially, [A] is high. Now a high number when multiplied to a positive number will be a large number. For example, let us take [A] = 100, k2 = 5, k3 = 1. Then k3[A] + k2 = 100 + 5. you can see that it is approximately equal to 100, so for high concentrations, we can neglect k2 from here and write the denominator approximately as k3[A]
. Then the rate eqn becomes Rate = (k1[E][A]^2)/(k2 + k3[A]) which is approximately equal to (k1[E][A]^2)/ k3[A] = c[E][A] where c is aconstant. (by cancelling [A] from both sides and get the required equation.)
Hence the reaction is first order.
Now we can proceed similarily for low [A], in the same example, say [A] = 0.001. Then again, k3[A] + k2 = 0.001 + 5 . Approximately equal to k2 itself. So we can approximate denominator by k2 by neglecting a small term k3[A].
So rate becomes
Rate = (k1[E][A]^2)/(k2 + k3[A]) = (k1[E][A]^2)/k2 after neglecting k3[A].
So RAte = c[E][A]^2 where c is a constant. Hence reaction is second order.
However, please note that the values of k2 and k3 I took were just to give an example, Similar concept applies for any value of k2 and k3.At low value of [A], it is approximately equal to k2 and at high values of [A] it would be approximately equal to k3[A].
You can compare it to the concept of limit of a functions. As you might remember, similar approximations are made to evaluate limits sometimes.
if you have any doubts, please don't hesitate to ask.
Thanks.