Question

The reaction $C{H}_{3}COO{C}_2{H}_5+NaOH\to C{H}_{3}COONa+C_2{H}_{5}OH$
is allowed to take place with initial concentrations of 0.2 mole/lit of each reactant. If the reaction mixture is diluted with water so that the initial concentration of each reactant becomes 0.1 mole/lit, the rate of the reaction will be:

• A. 1/8 th of the original rate
• B. 1/4 th of the original rate
• C. 1/2 th of the original rate
• D. same as the original rate

Answer 1

The correct option is B 1/4 th of the original rate

Since rate of reaction will depend on the concentration of both $C{H}_{3}COO{C}_2{H}_5$ and NaOH,

rate = k [$C{H}_{3}COO{C}_2{H}_5$] [NaOH].

And since concentration of each reactant reduces to half, original rate of reaction reduces to one-fourth.

Option B is correct.