The rate law for the given reaction $\text{CH}_3\text{CH}_2\text{Cl}+\text{NaOH}\to\text{CH}_3\text{CH}_2\text{OH}+\text{NaCl}$ is given by $r=k[\text{CH}_3\text{CH}_2\text{Cl}]$. The rate of reaction will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Halved on reducing the concentration of $CH_3CH_2Cl$ to half.
The rate law is $r=k[CH_3CH_2Cl]$, so the reaction is first-order with respect to $CH_3CH_2Cl$ and zero-order with respect to NaOH. Overall order = 1
(1) Doubling temperature changes the rate constant k, so rate will change; hence this option is wrong. ( Rate constant is dependent on Temperature)
(2) Doubling NaOH has no effect because rate only depends on concentration of $CH_3CH_2Cl$
(3) similarly Reducing NaOH concentration also does not affect the rate, as NaOH is not part of the rate law
(4) Halving $CH_3CH_2Cl$ directly halves the rate because it is first-order with respect to this reactant. This statement is correct.