Q79 · CSIR NET Chemistry, December 2016
Paper: CSIR NET December 2016 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Steady State Approximation · Marks: 2 · Difficulty: Medium
A reaction goes through the following elementary steps
\[ \begin{array}{l} \mathrm{A}+\mathrm{B} \xrightarrow[\mathrm{k}_{-1}]{\mathrm{k}_{1}} 2 \mathrm{C} \text { (fast) } \\ \mathrm{A}+\mathrm{C} \xrightarrow{\mathrm{k}_{2}} \mathrm{D} \text { (slow) } \end{array} \]
Assuming that steady state approximation can be applied to C, on doubling the concentration of A, the rate of production of D will increase by (assume $\mathrm{K}_{2}[\mathrm{A}] \ll \mathrm{K}_{-1}[\mathrm{C}]$ )(a)2 times
(b)4 times
(c)8 times
(d)$2 \sqrt{2}$ times
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
With the fast pre-equilibrium, [C] = √(K[A][B]), so rate = k₂[A][C] ∝ [A]^(3/2). Doubling [A] multiplies the rate by 2^(3/2) = 2√2 (d).
Study loop for Chemical Kinetics