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Chemistry · 2023 · 3 marks
CBSE 2023 · Region 2 · Set 1 · Q28
A first order reaction is 50%complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy ( E_a ) for the reaction. [ R…
A first order reaction is $\displaystyle 50$%complete in $\displaystyle 30$ minutes at $\displaystyle 300$ K and in $\displaystyle 10$ minutes at $\displaystyle 320$ K. Calculate activation energy $\displaystyle \left(\mathrm{E}_{\mathrm{a}}\right)$ for the reaction. $\displaystyle \left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$ $\displaystyle 3$ [Given: $\displaystyle \log 2=0.3010, \log 3=0.4771, \log 4=0.6021$ ]
Marking-scheme solution
\[\begin{aligned}
& \mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}} \\
& \mathrm{k}_{1}=\frac{0.693}{30} \mathrm{~min}^{-1} \\
& \mathrm{k}_{2}=\frac{0.693}{10} \mathrm{~min}^{-1} \\
& \log \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{Ea}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right] \\
& \log 3=\frac{\mathrm{Ea}}{2.303 \times 8.314}\left[\frac{1}{300}-\frac{1}{320}\right] \\
& \mathrm{Ea}=\frac{0.4771 \times 19.147 \times 300 \times 320}{20} \\
& \mathrm{Ea}=43848 \mathrm{~J} \mathrm{~mol}^{-1} \text { or } 43.848 \mathrm{KJ} \mathrm{~mol}^{-1} \text { or } 43.85 \mathrm{k} \mathrm{~J} \mathrm{~mol}^{-1} \\
& \text { (Deduct }^{1 / 2} \text { mark for incorrect or no unit) }
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.