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Chemistry · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 1 · Q22
For the first order thermal decomposition reaction, following data was obtained: C_2 H_5 Cl ( g ) → C_2 H_4( g )+ HCl ( g ) Calculate rate constant.…
For the first order thermal decomposition reaction, following data was obtained : \[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{HCl}(\mathrm{~g}) \]
Calculate rate constant. [Given : $\displaystyle \log 3=0.48$ ]
| S. No. | Time(s) | Total Pressure (atm) |
| $\displaystyle 1$ | $\displaystyle 0$ | $\displaystyle 0.30$ |
| $\displaystyle 2$ | $\displaystyle 30$ | $\displaystyle 0.50$ |
Marking-scheme solution
\[\mathrm{C_2H_5Cl} \rightarrow \mathrm{C_2H_4} + \mathrm{HCl} \]$\displaystyle t = 0$ $\displaystyle \mathrm{Pi}$ $\displaystyle 0$ $\displaystyle 0$$\displaystyle t = t$ $\displaystyle \mathrm{Pi}-x$ $\displaystyle x$ $\displaystyle x$\[\mathrm{P_t} = \mathrm{Pi} - \cancel{x} + \cancel{x} + x \]\[x = \mathrm{P_t} - \mathrm{Pi} \]\[\mathrm{K} = \frac{2.303}{t} \log \frac{\mathrm{Pi}}{2\mathrm{Pi} - \mathrm{P_t}} \]\[\mathrm{K} = \frac{2.303}{30} \log \frac{0.30}{2 \times 0.30 - 0.50} \]\[\mathrm{K} = \frac{2.303}{30} \log \frac{0.30}{0.10} \]\[\mathrm{K} = \frac{2.303}{30} \times 0.48 \]\[\mathrm{K} = 0.037\ \mathrm{atm}^{-1} \]
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CBSE Class 12 Chemistry past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.