CBSE 2022 · Region 2 · Set 1 · Q7 · 3 marks
A first order reduction takes $\displaystyle 30$ minutes for $\displaystyle 75$% decomposition. Calculate $\displaystyle \mathrm{t}_{1 / 2}$. Given : $\displaystyle [\log 2=0 \cdot 3, \quad \log 3=0 \cdot 48, \quad \log 4=0 \cdot 6, \quad \log 5=0 \cdot 7]$
Marking-scheme solution
\[k=\frac{2 \cdot 303}{t} \log \frac{(R)_{0}}{(R)}
\]
\[k=\frac{2 \cdot 303}{30} \log \frac{100}{25}
\]
\[k=\frac{2 \cdot 303}{30} \log 4=0 \cdot 046 \mathrm{~min}^{-1}
\]
\[t_{1 / 2}=\frac{0 \cdot 693}{k}=\frac{0 \cdot 693}{0 \cdot 046}
\]
\[t_{1 / 2}=15 \mathrm{~min}
\]
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CBSE Class 12 Chemistry past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.