Question 12.WKC.12: The relative rates of the reaction A + B → AB in vessels (1)......
The relative rates of the reaction A + B \rightarrow AB in vessels (1)–(4) are 1:2:1:2. Red spheres represent A molecules, blue spheres represent B molecules, and yellow spheres represent molecules of a third substance C.
(a) What is the order of the reaction in A, B, and C?
(b) Write the rate law.
(c) Write a mechanism that agrees with the rate law.
(d) Why doesn’t C appear in the equation for the overall reaction?
(a) Count the number of molecules of each type in vessels (1)–(4), and compare the relative rates with the relative number of molecules. The concentration of A mole-cules in vessel (2) is twice that in vessel (1) while the concentrations of B and C remain constant. Because the reaction rate in vessel (2) is twice that in vessel (1), the rate is proportional to [A], and so the reaction is first order in A. When [B] is dou-bled [compare vessels (1) and (3)], the rate is unchanged, so the reaction is zero order in B. When [C] is doubled [compare vessels (1) and (4)], the rate doubles, so the reaction is first order in C.
(b) The rate law can be written as where the rate = k[A]^{m}[B]^{n}[C]^{p} exponents m, n, and p specify the reaction orders in A, B, and C, respectively. Since the reaction is first order in A and C, and zero order in B, the rate law is rate = k[A][C].
(c) The rate law tells us that A and C collide in the rate-determining step because the rate law for the overall reaction is the rate law for the rate-determining step. Subse-quent steps in the mechanism are faster than the rate-determining step, and the var-ious steps must sum up to give the overall reaction. Therefore, a plausible mechanism is
A + C \rightarrow AC \quad\quad \text{Slower, rate-determining} \\ AC + B \rightarrow AB + C \quad \ \text{Faster}A + B \rightarrow AB \quad\quad \text{Overall reaction}
(d) C doesn’t appear in the overall reaction because it is consumed in the first step and regenerated in the second step. C is therefore a catalyst. AC is an intermediate because it is formed in the first step and consumed in the second step.