Only $35.99/year

Terms in this set (34)

Solution
For mechanism A, the rate law is
rate=𝑘1[H2][I2]=𝑘[H2][I2]
This is consistent with the experimentally determined rate law.
For mechanism B, the rate law is given by the rate-determining (slow) step
rate=𝑘2[H2][I]2
A rate law cannot contain an intermediate. Step 1 can be used to eliminate the intermediate [I] from the rate law. In step 1, the forward and reverse reactions are at equilibrium, therefore
𝑘1[I2]=𝑘−1[I]2
Rearrnge this expression to solve for [I]2. The rate law can be written as
rate=𝑘1𝑘2𝑘−1[H2][I2]=𝑘[H2][I2]
which is also consistent with the experimentally determined rate law.
For mechanism C, the rate law is given by step 2, the rate-determining step. Step 3 is a fast reaction after the rate-determining step (step 2), and it will not influence the rate of reaction.
rate=𝑘2[H2][I]
As with mechanism B, the intermediate must be removed from the rate law. Eliminate the intermediate [I] from this expression using step 1. In step 1, the forward and reverse reactions are at equilibrium, therefore
𝑘1[I2]=𝑘−1[I]2
Rearrnge this expression to solve for [I]. The rate law can be written as
rate=𝑘2(𝑘1/𝑘−1)1/2[H2][I2]1/2=𝑘[H2][I2]1/2
Therefore, only the first two mechanisms are consistent with the observed rate law.
The second and third mechanisms involve an I(g) intermediate, meaning both are consistent with the additional observation. However, only the second mechanism is consistent with the rate law. Therefore, only mechanism B is consistent with the experimental observations.