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JEE Main 2018 · Chemistry · Chemical Kinetics · hard

A first-order reaction has a half-life of 20 min at 300 K and an activation energy of 50 kJ mol^-1 (assume it is independent of temperature). The…

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A first-order reaction has a half-life of 20 min20\ \mathrm{min} at 300 K300\ \mathrm{K} and an activation energy of 50 kJ mol−150\ \mathrm{kJ\,mol^{-1}} (assume it is independent of temperature). The time required for 87.5%87.5\% completion at 320 K320\ \mathrm{K} is closest to (R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}})
  1. A. 5.7 min5.7\ \mathrm{min}
  2. B. 17.1 min17.1\ \mathrm{min}
  3. C. 60 min60\ \mathrm{min}
  4. D. 210 min210\ \mathrm{min}
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Answer: (B) 17.1 min17.1\ \mathrm{min}

Arrhenius: ln⁡k2k1=EaR(1300−1320)=6014×2.083×10−4=1.253\ln\dfrac{k_2}{k_1}=\dfrac{E_a}{R}\left(\dfrac1{300}-\dfrac1{320}\right)=6014\times2.083\times10^{-4}=1.253, so k2/k1=3.50k_2/k_1=3.50. Hence t1/2(320 K)=20/3.50=5.71t_{1/2}(320\ \mathrm{K})=20/3.50=5.71 min. 87.5%87.5\% completion needs three half-lives (1→12→14→181\to\frac12\to\frac14\to\frac18 remaining): t=3×5.71=17.1t=3\times5.71=17.1 min. The value 5.75.7 is a single half-life, 6060 ignores the temperature change, and 210210 results from inverting the rate ratio.
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