Q82 · CSIR NET Chemistry, June 2015
Paper: CSIR NET June 2015 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Medium
In a photochemical reaction, radicals formed according to the equation
\[ \begin{array}{l} \mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{hv} \leftrightharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \\ \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{C}_{2} \mathrm{H}_{5} \xrightarrow{\mathrm{k}_{2}} \mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{C}_{2} \mathrm{H}_{4} \end{array} \]
If I is the intensity of light absorbed, the rate of the overall reaction is proportional to-(a)I
(b)$\mathrm{I}^{1 / 2}$
(c)$\mathrm{I}\left[\mathrm{C}_{4} \mathrm{H}_{10}\right]$
(d)$\mathrm{I}^{1 / 2}\left[\mathrm{C}_{4} \mathrm{H}_{10}\right]^{1 / 2}$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Steady state: initiation rate ∝ I equals termination 2k₂[C₂H₅]², and the product-forming step is that termination, so the overall rate ∝ I (first power).
Study loop for Chemical Kinetics