Question

The reaction \( \text{A}_2 + \text{B}_2 \to 2\text{AB} \) follows the mechanism: \[ \text{A}_2 \xrightarrow{k_1} \text{A} + \text{A} \, (\text{fast}), \quad \text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B} \, (\text{slow}), \quad \text{A} + \text{B} \to \text{AB} \, (\text{fast}). \] The overall order of the reaction is:

• A. 3
• B. 1.5
• C. 2.5
• D. 2

Hint:

The overall order of a reaction is determined by the rate-determining step and the exponents of the reactants in that step.

Answer 1

The Correct Option is B

The reaction mechanism consists of three steps:

• \(\text{A}_2 \xrightarrow{k_1} \text{A} + \text{A}\) (fast)
• \(\text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B}\) (slow, rate-determining step)
• \(\text{A} + \text{B} \to \text{AB}\) (fast)

To determine the overall order, focus on the rate-determining step, as it controls the reaction rate. The slow step rate law is expressed as: \[\text{Rate} = k_2[\text{A}][\text{B}_2]\] The breakdown of \(\text{A}_2\) in the fast step achieves a rapid equilibrium, such that \( [\text{A}] = K_1^{0.5} [\text{A}_2]^{0.5} \). Substitute \([\text{A}]\) in the rate law: \[\text{Rate} = k_2(K_1^{0.5}[\text{A}_2]^{0.5})[\text{B}_2]\] \[\text{Rate} = k'[\text{A}_2]^{0.5}[\text{B}_2]\] The overall order is determined by the sum of the exponents: \[0.5 (\text{from } [\text{A}_2]) + 1 (\text{from } [\text{B}_2]) = 1.5\] Hence, the overall order of the reaction is 1.5.

Answer 2

Step 1: Write the given mechanism.
The overall reaction is:
\[ \text{A}_2 + \text{B}_2 \rightarrow 2\text{AB} \] The given mechanism is:
1. \( \text{A}_2 \xrightarrow{k_1} 2\text{A} \) (fast)
2. \( \text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B} \) (slow, rate-determining step)
3. \( \text{A} + \text{B} \xrightarrow{k_3} \text{AB} \) (fast)

Step 2: Identify the rate-determining step (RDS).
The second step is the slow step, so the rate law is determined by this step:
\[ \text{Rate} = k_2[\text{A}][\text{B}_2] \] However, [A] is an intermediate, so we need to express it in terms of the stable species (\( \text{A}_2 \) and \( \text{B}_2 \)).

Step 3: Apply the steady-state approximation for [A].
For intermediate A, the rate of formation = rate of consumption.
\[ \text{Rate of formation of A} = \text{Rate of consumption of A} \] From step 1 (formation): \( \text{Rate} = 2k_1[\text{A}_2] \)
From steps 2 and 3 (consumption): \( \text{Rate} = k_2[\text{A}][\text{B}_2] + k_3[\text{A}][\text{B}] \)

Equating both rates:
\[ 2k_1[\text{A}_2] = k_2[\text{A}][\text{B}_2] + k_3[\text{A}][\text{B}] \] \[ [\text{A}] = \frac{2k_1[\text{A}_2]}{k_2[\text{B}_2] + k_3[\text{B}]} \]

Step 4: Express [B] in terms of [B₂].
In the slow step, B is formed, and from step 3 (fast equilibrium), we can assume:
\[ [\text{B}] \propto [\text{B}_2]^{1/2} \] Hence, approximately: \[ [\text{B}] = k'[\text{B}_2]^{1/2} \]

Step 5: Substitute and simplify the rate law.
Rate law from step 2:
\[ \text{Rate} = k_2[\text{A}][\text{B}_2] \] Substitute [A]: \[ \text{Rate} = k_2 \left( \frac{2k_1[\text{A}_2]}{k_2[\text{B}_2] + k_3[\text{B}]} \right)[\text{B}_2] \] Since [B] ∝ [B₂]¹ᐟ², this implies that the effective dependence on [B₂] will be \( [\text{B}_2]^{1/2} \).

Therefore, the overall rate law is approximately: \[ \text{Rate} \propto [\text{A}_2]^1 [\text{B}_2]^{1/2} \] Hence, the overall order of reaction = 1 + ½ = 1.5.

Final Answer:
\[ \boxed{\text{Overall order of the reaction} = 1.5} \]