Consider the following multistep reaction:
C+D-> <-CB (FAST)
CD+D->CD2 (SLOW)
CD2+D->CD3 (FAST)
C+3D->CD3 (OVERALL)
Based on the mechanism determine the rate law for the overall reaction??
To determine the rate law for the overall reaction, we can start by examining the rate-determining step, which is the slowest step in the mechanism. In this case, the slow step is the second step:
CD + D -> CD2 (SLOW)
The rate of the overall reaction depends on the rate of this slow step. To find the rate law, we look at the reactants in the rate-determining step, which are CD and D. The rate equation for this step can be written as:
rate = k[CD][D]
Now, we need to consider the first step, which is a fast equilibrium reaction:
C + D ⇌ CB (FAST)
Because it is at equilibrium, the forward and reverse reactions occur at the same rate. Therefore, we can write the rate law for this step as:
rate = k1[C][D] - k-1[CB]
However, since this step is fast and at equilibrium, the concentration of the intermediate, CB, remains relatively constant. Hence, we can assume that the rate of the forward and reverse reactions are equal and cancel out. This simplifies the rate law for the first step to:
rate = k1[C][D]
Now, let's combine the rate laws for the two steps. Since the intermediate CB does not appear in the rate law for the slow step, we can express its concentration in terms of the other reactants in the fast equilibrium step. From the fast step, we have:
[CB] = k1[C][D] / k-1
Substituting this expression for [CB] in the rate law for the slow step, we get:
rate = k[CD][D] = k[CD][D] / [CB]
Now, substitute the expression for [CB]:
rate = k[CD][D] / (k1[C][D] / k-1)
Simplifying further:
rate = (k * k-1 * [CD][D]) / (k1[C])
Finally, we can rearrange the rate law to separate the reactants and the rate constant:
rate = kobs[CD][D]
Where kobs is the observed rate constant, given by:
kobs = (k * k-1) / k1
Therefore, the rate law for the overall reaction is:
rate = kobs[CD][D]