Question

One litre of $2M$ acetic acid and one litre of $3M$ ethyl alcohol were mixed. The esterification takes place according to the reaction:
$C{H}_{3}COOH+C_2{H}_{5}OH\to C{H}_{3}COO{C}_2{H}_5+H_{2}O$
If each solution is diluted by one litre water, the rate would become:

• A. $4$ times
• B. $2$ times
• C. $0.5$ times
• D. $0.25$ times

Answer 1

The correct option is A $4$ times

$C{H}_{3}COOH+C_2{H}_{5}OH\to C{H}_{3}COO{C}_2{H}_5+H_{2}O$

$Rate=k\left[C{H}_{3}COOH{]}^1\right[C_2{H}_{5}OH{]}^1$

Initial concentration of acetic acid (a) = $2M/L$

Initial concentration of alcohol(b) = $3M/L$

Therefore, $R_1=k\left[a{]}^1\right[b{]}^1=k\left[2{]}^1\right[3{]}^1=6k\to \left(1\right)$

If each reactant is diluted by one litre water, concentration will become half.

Therefore, $R_2=k\left[\frac{a}{2}{]}^1\right[\frac{b}{2}{]}^1=k\left[\frac{2}{2}{]}^1\right[\frac{3}{2}{]}^1$

$R_2=k\left[\frac{3}{2}\right]=\frac{3k}{2}\to \left(2\right)$

From equations (1) and (2)

$\frac{R_1}{R_2}=\frac{6k}{\frac{3k}{2}}=4$

Thus,

$R_1=4R_2$