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Chemistry · 2024 · 3 marks
CBSE 2024 · Region 3 · Set 1 · Q24
A first-order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% complete? 3 [Given:…
A first-order reaction is $\displaystyle 25$% complete in $\displaystyle 40$ minutes. Calculate the value of rate constant. In what time will the reaction be $\displaystyle 80$% complete ? $\displaystyle 3$ [Given : $\displaystyle \log 2=0 \cdot 30, \log 3=0 \cdot 48, \log 4=0 \cdot 60, \log 5=0 \cdot 69$ ]
Marking-scheme solution
\[\begin{aligned}
\mathrm{k}= & \frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}]_{\mathrm{s}}}{[\mathrm{R}]} \\
\mathrm{k}= & \frac{2.303}{40 \mathrm{~min}} \log \frac{4}{3} \\
\mathrm{k}= & \frac{2.303}{40 \mathrm{~min}} \times 0.12 \\
= & 0.0069 \mathrm{~min}^{-1}
\end{aligned}
\]
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