A reaction A+B+C → D follows the mechanism A+B ⇌ AB AB+C → D in which first step remains essentially in equilibrium. If Δ H is the enthalpy change for the first reaction and E₀ is the activation energy for the second reaction, the activation energy of the overall reaction will be given by

Q87 · CSIR-NET Chemistry, December 2014

Paper: CSIR-NET December 2014 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Arrhenius Activation · Marks: 2 · Difficulty: Hard

A reaction $\mathrm{A}+\mathrm{B}+\mathrm{C} \rightarrow \mathrm{D}$ follows the mechanism

\[ \begin{array}{l} \mathrm{A}+\mathrm{B} \leftrightharpoons \mathrm{AB} \\ \mathrm{AB}+\mathrm{C} \rightarrow \mathrm{D} \end{array} \]

in which first step remains essentially in equilibrium. If $\Delta \mathrm{H}$ is the enthalpy change for the first reaction and $\mathrm{E}_{0}$ is the activation energy for the second reaction, the activation energy of the overall reaction will be given by

(a)$\mathrm{E}_{0}$

(b)$\mathrm{E}_{0}-\Delta \mathrm{H}$

(c)$\mathrm{E}_{0}+\Delta \mathrm{H}$

(d)$\mathrm{E}_{0}+2 \Delta \mathrm{H}$

Answer

Answer: C ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation

k_obs = K₁k₂, so E_a = ΔH₁ + E₀ (van 't Hoff for the pre-equilibrium plus Arrhenius for the slow step).

Study loop for Chemical Kinetics

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