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 CH3CH2Cl to half.
The rate law shows how the rate depends on the concentration of each reactant, and it is found by experiment, not from the balanced equation. A reactant that does not appear in the rate law does not affect the rate: the reaction is of zero order with respect to it.
Here the rate law is
\( \displaystyle r = k[\ce{CH3CH2Cl}] \)
The reaction is first order in \( \ce{CH3CH2Cl} \) and zero order in NaOH. The rate is directly proportional to the concentration of chloroethane, so halving it halves the rate:
\( \displaystyle r' = k\left(\frac{1}{2}[\ce{CH3CH2Cl}]\right) = \frac{1}{2}\,r \)
Why the other options are incorrect
Option (1) Unaffected by doubling the temperature of the reaction. — Incorrect. Temperature changes the rate constant k; for most reactions k is nearly doubled by a rise of only 10 K, so the rate cannot stay the same.
Option (2) Doubled on doubling the concentration of NaOH — Incorrect. NaOH does not appear in the rate law, so changing its concentration does not change the rate.
Option (3) Halved on reducing the concentration of NaOH — Incorrect. For the same reason, the rate does not depend on [NaOH].