JEE Advanced 2025 Paper 2 · Chemistry · Chemical Kinetics · medium

A first-order reaction A→ products has a half-life of 20 min at 300 K. At 320 K it is 80% complete in 10 min. Taking R=8.314 J mol^-1K^-1, the…

Numerical answer

A first-order reaction

\mathrm{A\rightarrow products}

has a half-life of

20\ \mathrm{min}

300\ \mathrm{K}

. At

320\ \mathrm{K}

it is

80\%

complete in

10\ \mathrm{min}

. Taking

R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}

, the activation energy (in

\mathrm{kJ\,mol^{-1}}

) is ____. Give the answer correct to two decimal places.

Show answer and solution

Answer: 61.28

300\ \mathrm{K}

k_1=\frac{\ln2}{20}=0.034657\ \mathrm{min^{-1}}

. At

320\ \mathrm{K}

80\%

conversion leaves

20\%

, so

k_2=\frac{1}{10}\ln\frac{100}{20}=\frac{\ln5}{10}=0.160944\ \mathrm{min^{-1}}

. Using the Arrhenius equation,

\ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac1{T_1}-\frac1{T_2}\right)

. Here

\frac{k_2}{k_1}=4.644

\ln4.644=1.5357

and

\frac1{300}-\frac1{320}=2.0833\times10^{-4}\ \mathrm{K^{-1}}

. Hence

E_a=\frac{8.314\times1.5357}{2.0833\times10^{-4}}=6.128\times10^{4}\ \mathrm{J\,mol^{-1}}=61.28\ \mathrm{kJ\,mol^{-1}}

. The common slip is using

k=\ln5/(0.8\times10)

or treating

80\%

as a half-life-type fraction.

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