O KINETICS AND EQUILIBRIUM Writing the rate law implied by a simple mechanism Suppose the reaction between nitric oxide and bromine proceeds by the following mechanism: 1. NO(g) + Br2(g) ? NOBr2(g) 2. 2 NOBr2(g) + NO(g) ? 2 NOBr(g) Suppose also k1 > k2. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally-observable rate law for the overall chemical reaction. rate = k Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k?1 and k?2 for the reverse of the two elementary reactions in the
The Correct Answer and Explanation is:
To solve this kinetics and equilibrium problem, we must identify the overall balanced chemical equation, the rate law, and then derive the rate constant expression eliminating intermediates.
🔹 Given Mechanism:
Step 1 (Fast Equilibrium):
NO(g) + Br₂(g) ⇌ NOBr₂(g) (k₁ forward, k₋₁ reverse)
Step 2 (Slow):
2 NOBr₂(g) + NO(g) → 3 NOBr(g) (k₂)
🔹 Step 1: Overall Balanced Equation
To find the overall reaction, add both elementary steps and cancel any intermediates:
Step 1: NO + Br₂ → NOBr₂
Step 2: 2 NOBr₂ + NO → 3 NOBr
Add:
NO + Br₂ + 2 NOBr₂ + NO → NOBr₂ + 3 NOBr
Cancel NOBr₂ (intermediate):
Overall Reaction:
3 NO(g) + Br₂(g) → 3 NOBr(g)
🔹 Step 2: Rate Law
Since Step 2 is the rate-determining step, the rate law is based on that step: rate=k2[NOBr2]2[NO]\text{rate} = k_2[\text{NOBr}_2]^2[\text{NO}]
However, NOBr₂ is an intermediate, so we need to eliminate it.
From the fast equilibrium (Step 1), we assume: k1[NO][Br2]=k−1[NOBr2]k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2]
Solve for [NOBr₂]: [NOBr2]=k1k−1[NO][Br2][\text{NOBr}_2] = \frac{k_1}{k_{-1}}[\text{NO}][\text{Br}_2]
Substitute into the rate law: rate=k2(k1k−1[NO][Br2])2[NO]\text{rate} = k_2 \left( \frac{k_1}{k_{-1}}[\text{NO}][\text{Br}_2] \right)^2 [\text{NO}]
Simplify: rate=k2(k12k−12)[NO]2[Br2]2[NO]\text{rate} = k_2 \left( \frac{k_1^2}{k_{-1}^2} \right)[\text{NO}]^2[\text{Br}_2]^2[\text{NO}] rate=(k12k2k−12)[NO]3[Br2]2\text{rate} = \left( \frac{k_1^2 k_2}{k_{-1}^2} \right)[\text{NO}]^3[\text{Br}_2]^2
🔹 Final Answers
- Overall Balanced Reaction:
3NO(g)+Br2(g)→3NOBr(g)\boxed{3 \text{NO}(g) + \text{Br}_2(g) \rightarrow 3 \text{NOBr}(g)}
- Rate Law:
rate=k[NO]3[Br2]2\boxed{\text{rate} = k[\text{NO}]^3[\text{Br}_2]^2}
- Rate Constant Expression:
k=k12k2k−12\boxed{k = \frac{k_1^2 k_2}{k_{-1}^2}}
🔍 Explanation (300+ words)
In reaction mechanisms, the rate-determining step (RDS) controls the observed kinetics. Here, we are given a two-step mechanism: the first step is a fast equilibrium, while the second is slow. When a fast step precedes the slow step, and the slow step involves an intermediate, we use the fast step to express the intermediate in terms of observable species.
Step 1 shows nitric oxide (NO) reacting with bromine (Br₂) to form an intermediate, NOBr₂. Because this step is fast and reversible, we assume it quickly reaches equilibrium. Step 2 is slower and thus dictates the overall rate. However, its rate law includes the concentration of the intermediate NOBr₂, which is not directly measurable. Therefore, we solve for [NOBr₂] using the equilibrium condition from Step 1.
By substituting [NOBr₂] in terms of [NO] and [Br₂], we eliminate the intermediate from the rate expression, resulting in a rate law expressed entirely in terms of reactants. The final rate law—rate = k[NO]³[Br₂]²—shows the reaction is third order in NO and second order in Br₂.
Lastly, to relate the overall rate constant to the elementary constants, we substitute the expressions, revealing how microscopic kinetics (k₁, k₂, k₋₁) combine to define the macroscopic rate constant k. This approach illustrates how mechanisms predict kinetics consistent with experiment while connecting observed behavior with molecular steps.