MHT CET · Chemistry · Chemical Kinetics
Find the rate law for the reaction, \(\mathrm{CHCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{CCl}_{4(\mathrm{~g})}+\mathrm{HCl}_{(\mathrm{g})}\) if order of reaction with respect to \(\mathrm{CHCl}_{\mathrm{a}(\mathrm{g})}\) is one and \(\frac{1}{2}\) with \(\mathrm{Cl}_{2(\mathrm{~g})}\).
- A Rate \(=\mathrm{k}\left[\mathrm{CHCl}_3\right]\left[\mathrm{Cl}_2\right]^{1 / 2}\)
- B Rate \(=\mathrm{k}\left[\mathrm{CHCl}_3\right]^2\left[\mathrm{Cl}_2\right]^{1 / 2}\)
- C Rate \(=\mathrm{k}\left[\mathrm{CHCl}_3\right]^{3 / 2}\left[\mathrm{Cl}_2\right]\)
- D Rate \(=\mathrm{k}\left[\mathrm{CHCl}_3\right]^{1 / 2}\left[\mathrm{Cl}_2\right]\)
Answer & Solution
Correct Answer
(A) Rate \(=\mathrm{k}\left[\mathrm{CHCl}_3\right]\left[\mathrm{Cl}_2\right]^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Std. \(12 \mid\) Ch-6 | Subtopic-6.3
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