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JEE Main 2019 · Chemistry · Chemical Kinetics · hard

A first-order reaction has activation energy 50 kJ mol^-1 and rate constant 1.0×10^-3 s^-1 at 300 K. Assuming the Arrhenius equation with constant A…

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A first-order reaction has activation energy 50 kJ mol−150\ \mathrm{kJ\,mol^{-1}} and rate constant 1.0×10−3 s−11.0\times10^{-3}\ \mathrm{s^{-1}} at 300 K300\ \mathrm{K}. Assuming the Arrhenius equation with constant AA and R=8.314 J K−1 mol−1R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}, the time needed for 87.5%87.5\% of the reactant to be consumed at 320 K320\ \mathrm{K} is nearest to
  1. A. 2080 s2080\ \mathrm{s}
  2. B. 590 s590\ \mathrm{s}
  3. C. 1040 s1040\ \mathrm{s}
  4. D. 300 s300\ \mathrm{s}
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Answer: (B) 590 s590\ \mathrm{s}

ln⁡k320k300=EaR(1300−1320)=500008.314(2.083×10−4)=1.253\ln\dfrac{k_{320}}{k_{300}} = \dfrac{E_a}{R}\left(\dfrac1{300}-\dfrac1{320}\right) = \dfrac{50000}{8.314}(2.083\times10^{-4}) = 1.253, so k320=1.0×10−3×3.50=3.50×10−3 s−1k_{320} = 1.0\times10^{-3}\times 3.50 = 3.50\times10^{-3}\ \mathrm{s^{-1}}. For 87.5%87.5\% completion, [A]0/[A]=8[A]_0/[A] = 8, i.e. three half-lives: t=ln⁡8/k=2.079/3.50×10−3≈594 st = \ln 8/k = 2.079/3.50\times10^{-3} \approx 594\ \mathrm{s}. Using the 300 K300\ \mathrm{K} constant by mistake gives 2080 s2080\ \mathrm{s}.
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Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake