The rate law for the reaction $\mathrm{RCl}+{\mathrm{NaOH}}_{\left(\mathrm{aq}\right)}\to \mathrm{ROH}+\mathrm{NaCl}$

$\mathrm{rate} =\mathrm{K}\left[\mathrm{RCl}\right]$ The rate of reaction will be

For the reaction $\mathrm{RCl}+{\mathrm{NaOH}}_{\left(\mathrm{aq}\right)}\to \mathrm{ROH}+\mathrm{NaCl}$

$\mathrm{rate} =\mathrm{K}\left[\mathrm{RCl}\right]$

Here, the rate of reaction depends on concentration of RCl only but not on NaOH solution. The rate of reaction depends on temperature. If we increase temperature then rate of reaction is increased. 1 Initial rate ${\mathrm{r}}_1=\mathrm{K}\left[\mathrm{x}\right]\to \mathrm{eqn}\left(1\right)$  If reducing the concentration of RCl to half, then the rate will Become 

Final rate ${\mathrm{r}}_2=\mathrm{K}\left[\mathrm{x}/2\right]\to \mathrm{eqn}\left(2\right)$

From equations 1 and 2,

$\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{K\left[x\right]}{K[x/2]}$

$\Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=2$

$\Rightarrow {\mathrm{r}}_2=\frac{{\mathrm{r}}_1}{2}$

Therefore, on reducing the concentration of RCl to half the rate of reaction would become halved.