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Chemistry · 2023 · 3 marks
CBSE 2023 · Region 4 · Set 1 · Q30
The following data were obtained during the first order thermal decomposition of C_2 H_5 Cl at a constant volume: C_2 H_5 Cl ( g ) → C_2 H_4( g )+…
The following data were obtained during the first order thermal decomposition of $\displaystyle \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}$ at a constant volume : \[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{HCl}(\mathrm{~g}) \]
Calculate the rate constant. $\displaystyle 3$ (Given : $\displaystyle \log 2=0 \cdot 3010, \quad \log 3=0 \cdot 4771, \quad \log 4=0 \cdot 6021$ )
| Experiment | Time $\displaystyle \left(\mathrm{s}^{-1}\right)$ | Total pressure (atm) |
| $\displaystyle 1$ | $\displaystyle 0$ | $\displaystyle 0.4$ |
| $\displaystyle 2$ | $\displaystyle 100$ | $\displaystyle 0.6$ |
Marking-scheme solution
\(\displaystyle \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{p_{\mathrm{i}}}{\left(2 p_{\mathrm{i}}-p_{\mathrm{t}}\right)}\)
\[\begin{aligned}
& \mathrm{k}=\frac{2.303}{100} \log \frac{0.4}{0.8-0.6} \\
& =\frac{2.303}{100} \log 2 \\
& =\frac{2.303 \times 0.3010}{100} \\
& =0.0069 \mathrm{~s}^{-1} \text { or } 0.007 \mathrm{~s}^{-1}
\end{aligned}
\]
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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.