CBSE 2024 · Region 5 · Set 3 · Q28 · 3 marks
The rate constant of a reaction quadruples when the temperature changes from $\displaystyle 700$ K to $\displaystyle 720$ K . Calculate the activation energy for this reaction. $\displaystyle \left[\log 2=0.30, \log 4=0.60,2.303 \mathrm{R}=19.15 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
Marking-scheme solution
\[\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 \mathrm{R}}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)
\]
It is given that, \(\displaystyle k_{2}=4 k_{1}\)
Therefore, \(\displaystyle \log \frac{4 k_{1}}{\dot{k}_{1}}=\frac{E_{a}}{19.15}\left(\frac{720-700}{700 \times 720}\right)\)
Chemical KineticsTemperature Dependence of the Rate of a ReactionApplyshort_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.