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Chemistry · 2024 · 2 marks
CBSE 2024 · Region 2 · Set 1 · Q18
A first order reaction has a rate constant 1.25 × 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 2.5 g? [ log 2=0 · 301, log 3=0 ·…
A first order reaction has a rate constant $\displaystyle 1.25 \times 10^{-3} \mathrm{~s}^{-1}$. How long will $\displaystyle 5$ g of this reactant take to reduce to $\displaystyle 2.5$ g ? $\displaystyle [\log 2=0 \cdot 301, \log 3=0 \cdot 4771, \log 4=0 \cdot 6021]$
Marking-scheme solution
\[\begin{aligned}
& \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{[\mathrm{R}] \text { o }}{[\mathrm{R}]} \\
& \begin{aligned}
1.25 \times 10^{-3}=\frac{2.303}{\mathrm{t}} \log \left(\frac{5}{2.5}\right) & \\
& \mathrm{t}=\frac{2.303}{1.25 \times 10^{-3}} \log 2 \\
& \mathrm{t}=\frac{2 \cdot 303 \times 0.301}{1.25 \times 10^{-3}} \\
\mathrm{t}=554.5 \mathrm{~s} & \text { or } 5.54 \times 10^{2} \mathrm{~s}
\end{aligned}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.