One liter of 2 M acetic acid and one liter of 3 Methyl alcohol are mixed to form ester according to the given equation :

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}⟶{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

If each solution is diluted by adding equal volume (1 liter) of water, by how many times the initial forward rate is reduced?

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}⟶{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

Rate $=\mathrm{k}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}^1{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]}^1$

Initial acetic acid concentration (a) = 2M/L
Initial alcohol(b) concentration: 3M/L

Therefore, ${\mathrm{R}}_1=\mathrm{k}\left[\mathrm{a}{]}^1\right[\mathrm{b}{]}^1=\mathrm{k}\left[2{]}^1\right[3{]}^1=6\mathrm{k}\to$ (1)

Each reactant's concentration will be halved if it is diluted with one litre of water.

Therefore,  ${\mathrm{R}}_2=\mathrm{k}{\left[\frac{\mathrm{a}}{2}\right]}^1{\left[\frac{\mathrm{b}}{2}\right]}^1=\mathrm{k}{\left[\frac{2}{2}\right]}^1{\left[\frac{3}{2}\right]}^1$

${\mathrm{R}}_2=\mathrm{k}\left[\frac{3}{2}\right]=\frac{3\mathrm{k}}{2}\to$ (2)

by using equations (1) and (2)

$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{6\mathrm{k}}{\frac{3\mathrm{k}}{2}}=4$

Thus, 

${\mathrm{R}}_1=4{\mathrm{R}}_2$