CUET · CHEMISTRY · PYQ PAPER 2025
The rate law for the given reaction
CH\(_3\)\(\mathrm{CH}_2\)Cl + NaOH \(\rightarrow\) CH\(_3\)\(\mathrm{CH}_2\)OH + NaCl
is given by r = k[CH\(_3\)\(\mathrm{CH}_2\)Cl]. The rate of reaction will be:
- A Unaffected by doubling the temperature of the reaction.
- B Doubled on doubling the concentration of NaOH
- C Halved on reducing the concentration of NaOH
- D Halved on reducing the concentration of CH\(_3\)\(\mathrm{CH}_2\)Cl to half.
Answer & Solution
Correct Answer
(D) Halved on reducing the concentration of CH\(_3\)\(\mathrm{CH}_2\)Cl to half.
Step-by-step Solution
Detailed explanation
Given: \( \mathrm{r} = \mathrm{k}[\mathrm{CH_3CH_2Cl}] \) New rate \( \mathrm{r'} = \mathrm{k} \left( \frac{1}{2}[\mathrm{CH_3CH_2Cl}] \right) = \frac{1}{2} \mathrm{r} \) Halved on reducing the concentration of \( \mathrm{CH_3CH_2Cl} \) to half.
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