Reaction Mechanisms

Question

Consider the following elementary steps.

Step 1: A  ⇌ C (--> is k1, <--- is k-1)

Step 2: C+B --> D (--> is k2)

Assume that both reaction have similar rates. That is, no slow or fast steps. You will need to use the steady state approximation. The rate law is...

The answer provided on the worksheet is simply k2[C][B]. How can it be that simple? If we don't know that step 2 is slow or fast, why is the rate law simply equal to its reactants?

Answer 1

A little explanation: For a steady state reaction, the concentration of the intermediate [C] will be much smaller than A or D. You can imagine that is being consumed in more than one direction: back to A and forward to D. It also means that A forming into C is much slower than B+C forming into D, because as A forms into C, C forms into A. The reason these types of reactions occur is that [C] is an unstable intermediate (part of why the reaction is reversible), and so much of [C] produced converts back to A rather than continuing the reaction. The small portion of [C] that do encounter B will have a fast forward reaction due to C's desire to achieve stability - this leads to a steady state in which the concentration of C remains roughly the same (very slowly decreases over time). The irreversible second step of this steady-state reaction is considered to determine the rate because the small quantity of C must be formed and then collide with B in the short time before it reverts to A. In general, k2/k1 > 10 for a steady state reaction, but the first step is like a dripping faucet controlling a steady amount of C flowing into the second step.

Therefore, answer k2[B][C] is not entirely wrong, as this step determines the formation of products; however, it is a simplified result, as technically the intermediate should not appear in the overall rate law.

First, use the approximation that the rate of consumption of C is equal to the rate of production of C:

rate of production = k1[A]

rate or consumption = k2[B][C] + k-1[C]

(Remember, [C] is also being "consumed" by the reverse reaction)

rate of consumption = rate of production

k1[A] = k2[B][C]+k-1[C]

Isolate [C]. Factor [C[ out of the right side. Then, divide the remaining expression to the other side.

k1[A] = [C] * (k2[B] + k-1)

[C] = k1[A] / (k2[B] + k-1)

Finally, this expression will get substituted into the expression for the production of D. This gives a direct relationship between the consumption of [A] and [B] and the production of [D], without the intermediate which is assumed to be of roughly equal concentration throughout the reaction.

production of D = consumption of B and C

rate of production of reaction = k2[B][C] //This is where your teacher left the answer

Substitute the expression for [C] into this rate law.

rate = k2[B] * k1[A] / (k2[B] + k-1)

This would be a more appropriate answer but the teacher is simply leaving it in the original form to save you some math. You are correct that it seemed too simple because the intermediate should not be in the final answer. However, the answer k2[B][C] is technically still correct.