Question

$C{H}_{3}COO{C}_2{H}_5+NaOH\to C{H}_{3}COONa+C_2{H}_{5}OH$

Rate $=K\left[C{H}_{3}COO{C}_2{H}_5\right]\left[NaOH\right]$

If three times water is added to the reaction mixture, the rate of the reaction compared to the original rate will be:

• A. $\frac{1}{3}rd$
• B. $\frac{1}{9}th$
• C. $\frac{1}{16}th$
• D. $16times$

Answer 1

The correct option is C $\frac{1}{16}th$

The rate law expression is $r=K\left[C{H}_{3}COO{C}_2{H}_5\right]\left[NaOH\right]$

When three times water is added, the volume becomes four times the original volume.

Hence, the concentration of each reactant becomes one fourth.

The rate law expression becomes $r^{\prime }=K.\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]}{4}\frac{\left[NaOH\right]}{4}=\frac{r}{16}$.

Hence, option C is correct.