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JEE Main 2023 · Chemistry · Chemical Kinetics · medium

Substance A decomposes simultaneously by two parallel first-order paths, A→ B with k1=3×10^-3 min^-1 and A→ C with k2=2×10^-3 min^-1. The time taken…

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Substance A decomposes simultaneously by two parallel first-order paths, A→B\mathrm{A\to B} with k1=3×10−3 min−1k_1=3\times10^{-3}\ \mathrm{min^{-1}} and A→C\mathrm{A\to C} with k2=2×10−3 min−1k_2=2\times10^{-3}\ \mathrm{min^{-1}}. The time taken for 75%75\% of A to be consumed and the fraction of the products that is B are respectively (ln⁡2=0.693\ln2=0.693)
  1. A. 277 min277\ \mathrm{min}, 60%60\%
  2. B. 277 min277\ \mathrm{min}, 40%40\%
  3. C. 139 min139\ \mathrm{min}, 60%60\%
  4. D. 555 min555\ \mathrm{min}, 50%50\%
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Answer: (A) 277 min277\ \mathrm{min}, 60%60\%

Parallel first-order paths add: k=k1+k2=5×10−3 min−1k=k_1+k_2=5\times10^{-3}\ \mathrm{min^{-1}}. For 75%75\% conversion, t=1kln⁡10025=2×0.6935×10−3=277 mint=\frac{1}{k}\ln\frac{100}{25}=\frac{2\times0.693}{5\times10^{-3}}=277\ \mathrm{min} (two half-lives of 138.6138.6 min). The product ratio is fixed by the rate constants at all times: [B]/([B]+[C])=k1/k=3/5=60%[\mathrm B]/([\mathrm B]+[\mathrm C])=k_1/k=3/5=60\%. Using only k1k_1 in the time expression would give 462462 min.
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