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Chemistry · 2023 · 3 marks
CBSE 2023 · Region 5 · Set 1 · Q26
(a) For the reaction 2 N_2 O_5( g ) → 4 NO_2( g ) + O_2( g ) at 318 K calculate the rate of reaction if rate of disappearance of N_2 O_5( g ) is 1.4…
(a)
For the reaction $\displaystyle 2 \mathrm{~N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ at $\displaystyle 318$ K calculate the rate of reaction if rate of disappearance of $\displaystyle \mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})}$ is $\displaystyle 1.4 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}$.
(b)
For a first order reaction derive the relationship $\displaystyle \mathrm{t}_{99 \%}=2 \mathrm{t}_{90 \%}$
Marking-scheme solution
(a) \[\begin{aligned}
\text { Rate of reaction } & =-\frac{1}{2} \frac{\Delta\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{\Delta \mathrm{t}} \\
& =\frac{1}{2} \times 1 \cdot 4 \times 10^{-3}=0 \cdot 7 \times 10^{-3} \mathrm{M} \mathrm{~s}^{-1} \text { or } 7 \times 10^{-4} \mathrm{M} \mathrm{~s}^{-1}
\end{aligned}
\]
(Unit may be ignored)
(b) \[\begin{aligned}
\mathrm{t} & =\frac{2 \cdot 303}{k} \log \frac{[R]_{0}}{[R]} \\
\mathrm{t}_{99 \%} & =\frac{2 \cdot 303}{k} \log \frac{100}{1}=\frac{2 \cdot 303}{k} \log 100 \\
\mathrm{t}_{90 \%} & =\frac{2 \cdot 303}{k} \log \frac{100}{10}=\frac{2 \cdot 303}{k} \log 10 \\
\mathrm{t}_{90 \%} & =\frac{\log 100}{\log 10}=2
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.