CBSE 2023 · Region 3 · Set 3 · Q20 · 2 marks
The half-life of a first order reaction is $\displaystyle 60$ minutes. How long will it take to consume $\displaystyle 90$%of the reactant? [Given: $\displaystyle \log 2=0 \cdot 3010, \log 3=0 \cdot 4771, \log 10=1$ ]
Marking-scheme solution
\[\begin{aligned}
\mathrm{t}_{1 / 2} & =\frac{0.693}{\mathrm{k}} \\
\mathrm{k} & =\frac{0.693}{60} \mathrm{~min}^{-1} \\
\mathrm{k} & =\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\
\frac{0.693}{60} & =\frac{2.303}{\mathrm{t}} \log \frac{100}{10} \\
\mathrm{t} & =\frac{2.303 \times 60}{0.693} \mathrm{~min} \\
\mathrm{t} & =199.3 \mathrm{~min}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.