CBSE 2024 · Region 5 · Set 3 · Q21 · 2 marks
A first order reaction takes $\displaystyle 40$ min for $\displaystyle 75$ % decomposition. Calculate rate constant. [Given : $\displaystyle \log 2=0.30, \log 4=0.60$ ]
Marking-scheme solution
\[t=\frac{2.303}{\mathrm{k}} \log \frac{[A]_{0}}{A}
\]
\(\displaystyle 40 \mathrm{~min}=2.303 / \mathrm{k} \times \log (100 / 25)\)
\[\begin{aligned}
& \mathrm{k}=2.303 / 40 \mathrm{~min}^{2} \times \log 4 \\
& \mathrm{k}=2.303 / 40 \mathrm{~min}^{2} \times 0.60 \\
& \mathrm{k}=0.0345 \mathrm{~min}^{-1}
\end{aligned}
\]
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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.