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JEE Advanced 2021 Paper 2 · Chemistry · Chemical Kinetics · medium

A substance A decomposes by two parallel first-order paths, A→ B (rate constant k1=3.0×10^-3 s^-1, activation energy 60 kJ mol^-1) and A→ C (rate…

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A substance A\mathrm{A} decomposes by two parallel first-order paths, A→B\mathrm{A\to B} (rate constant k1=3.0×10−3 s−1k_1=3.0\times10^{-3}\ \mathrm{s^{-1}}, activation energy 60 kJ mol−160\ \mathrm{kJ\,mol^{-1}}) and A→C\mathrm{A\to C} (rate constant k2=1.0×10−3 s−1k_2=1.0\times10^{-3}\ \mathrm{s^{-1}}, activation energy 90 kJ mol−190\ \mathrm{kJ\,mol^{-1}}), both values of kk referring to 300 K300\ \mathrm{K}. Take ln⁡2=0.693\ln 2=0.693. The correct statement(s) is/are
  1. A. The half-life of A\mathrm{A} at 300 K300\ \mathrm{K} is about 173 s173\ \mathrm{s}
  2. B. At complete reaction, 75%75\% of the A consumed ends up as B\mathrm{B} at 300 K300\ \mathrm{K}
  3. C. The ratio [B]/[C][\mathrm{B}]/[\mathrm{C}] in the product mixture increases when the temperature is raised
  4. D. The apparent activation energy for the overall disappearance of A\mathrm{A} at 300 K300\ \mathrm{K} is 75 kJ mol−175\ \mathrm{kJ\,mol^{-1}}
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Answer: A, B

For parallel first-order paths, −d[A]/dt=(k1+k2)[A]-d[\mathrm{A}]/dt=(k_1+k_2)[\mathrm{A}], so t1/2=0.693/(4.0×10−3)=173t_{1/2}=0.693/(4.0\times10^{-3})=173 s: A is correct. The products form in the ratio of the rate constants at every instant, so the final fraction of A converted into B is k1/(k1+k2)=0.75k_1/(k_1+k_2)=0.75: B is correct. For C, k1/k2∝exp⁡[−(E1−E2)/RT]=exp⁡[+30000/RT]k_1/k_2\propto\exp[-(E_1-E_2)/RT]=\exp[+30000/RT], which decreases as TT rises, so the ratio [B]/[C][\mathrm{B}]/[\mathrm{C}] falls (the higher-activation path C gains). C is incorrect. For D, the apparent activation energy of the overall disappearance is the rate-weighted mean Ea=(k1E1+k2E2)/(k1+k2)=(3×60+1×90)/4=67.5 kJ mol−1E_a=(k_1E_1+k_2E_2)/(k_1+k_2)=(3\times60+1\times90)/4=67.5\ \mathrm{kJ\,mol^{-1}}, not the simple average 7575. D is incorrect. Answer: A, B.
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