CBSE 2025 · Region 1 · Set 3 · Q24 · 3 marks
The rate of a reaction quadruples when the temperature changes from $\displaystyle 293$ K to $\displaystyle 313$ K . Calculate the energy of activation of the reaction assuming that it does not change with temperature. $\displaystyle 3$ [Given : $\displaystyle \log 4=0.602, \log 2=0.301, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
Marking-scheme solution
log K = - Ea /$\displaystyle 2.303$ RT
log
𝐸𝑎
$\displaystyle 2.303$ ×$\displaystyle 8.314$
[$\displaystyle 313$−$\displaystyle 293$]
[$\displaystyle 313$ ×$\displaystyle 293$]
log $\displaystyle 4$ =
𝐸𝑎
[$\displaystyle 20$]
[$\displaystyle 313$ ×$\displaystyle 293$]
Ea=
$\displaystyle 0.602$ ×$\displaystyle 19.147$ ×$\displaystyle 313$ ×$\displaystyle 293$
=$\displaystyle 52850$ J mol-$\displaystyle 1$ / $\displaystyle 52.85$ kJ mol-$\displaystyle 1$
Chemical KineticsTemperature Dependence of the Rate of a ReactionApplyshort_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.