Problem 111
Question
Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}\) Step 2: A \(+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}\) \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).
Step-by-Step Solution
VerifiedKey Concepts
Rate Law
- Rate = k \( [A]^x [B]^y \)
- \( k \) is the rate constant.
- \( [A] \) and \( [B] \) are the concentrations of the reactants.
- \( x \) and \( y \) are the reaction orders with respect to the reactants, usually determined experimentally.
Reaction Mechanism
- Each step in the mechanism can have its own reactants, products, and intermediates.
- Elementary steps are characterized by molecularity, indicating how many molecules participate in that step.
- Step 1: \( A + B \rightarrow C + X \)
- Step 2: \( A + X \rightarrow C + D \)
Rate Determining Step
- If Step 1 is the rate determining step, the reaction rate is dependent on this first step.\[ Rate = k_1 [A][B] \]
- If Step 2 is rate determining, we must consider intermediates, and the expression becomes:\[ Rate = \frac{k_1 k_2}{k_{-1}} [A]^2[B] \]
Intermediate Species
- The intermediate \( X \) is produced in Step 1 and consumed in Step 2.
- Its transient nature makes it important for bridging the two steps together.