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JEE Main 2022 · Chemistry · Chemical Kinetics · hard

A first-order reaction is 75% complete in 40 min at 300 K. Its activation energy is 55 kJ mol^-1 (R=8.314 J mol^-1K^-1). The time needed for 87.5%…

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A first-order reaction is 75%75\% complete in 4040 min at 300 K300\ \mathrm K. Its activation energy is 55 kJ mol−155\ \mathrm{kJ\,mol^{-1}} (R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}). The time needed for 87.5%87.5\% completion at 310 K310\ \mathrm K is closest to
  1. A. 59.059.0 min
  2. B. 29.529.5 min
  3. C. 19.619.6 min
  4. D. 14.714.7 min
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Answer: (B) 29.529.5 min

75%75\% completion is two half-lives, so t1/2(300)=20t_{1/2}(300)=20 min. Arrhenius: ln⁡k310k300=EaR(1300−1310)=6615×1.075×10−4=0.711\ln\dfrac{k_{310}}{k_{300}}=\dfrac{E_a}{R}\Big(\dfrac1{300}-\dfrac1{310}\Big)=6615\times1.075\times10^{-4}=0.711, giving k310/k300=2.04k_{310}/k_{300}=2.04. Hence t1/2(310)=20/2.04=9.82t_{1/2}(310)=20/2.04=9.82 min. 87.5%87.5\% completion is three half-lives: 3×9.82=29.53\times9.82=29.5 min. Using two half-lives gives 19.619.6 min; forgetting the temperature change gives 6060 min.
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