Q84 · CSIR-NET Chemistry, December 2016
Paper: CSIR-NET December 2016 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Hard
Photochemical decomposition of HI takes place with the following mechanism \[ \begin{array}{l} \mathrm{HI}+\mathrm{hv}\left(\mathrm{I}_{\mathrm{a}}\right) \rightarrow \mathrm{H}+\mathrm{I} \\ \mathrm{H}+\mathrm{HI} \xrightarrow{\mathrm{k}_{1}} \mathrm{H}_{2}+\mathrm{I} \\ \mathrm{I}+\mathrm{I}+\mathrm{M} \xrightarrow{\mathrm{k}_{2}} \mathrm{I}_{2}+\mathrm{M} \end{array} \]
Considering hydrogen (H) and iodine (I) atoms as intermediates, the rate of removal of HI is (a)$\mathrm{I}_{\mathrm{a}} / 2$
(b)$\mathrm{I}_{\mathrm{a}}$
(c)$2 \mathrm{I}_{\mathrm{a}}$
(d)$\mathrm{I}_{\mathrm{a}} 2$
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
Each absorbed photon gives H, which consumes a second HI: −d[HI]/dt = I_a + k₁[H][HI] = 2I_a at steady state (quantum yield 2).
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