An elementary reaction is given as $2\mathrm{p}+\mathrm{Q}\to \mathrm{products}.$If concentration of Q is kept constant and concentration of P is doubled the rate of reaction is

Elementary reaction is

$2\mathrm{P}+\mathrm{Q}\to \mathrm{Products}$

From rate law , $\mathrm{r}=\mathrm{K}{\left[\mathrm{P}\right]}^2\left[\mathrm{Q}\right]\to \mathrm{eqn}\left(1\right)$

${\mathrm{r}}_1=\mathrm{K}{\left[\mathrm{x}\right]}^2\left[\mathrm{y}\right]\to \left(2\right)$

Where x and y are initial concentration of the reactants P and Q respectively. When we double the concentration of P by keeping Q as constant, then we will get 

${\mathrm{r}}_2=\mathrm{K}{\left[2\mathrm{x}\right]}^2\left[\mathrm{y}\right]\to \left(2\right)$

From equation 2 and 3 ,

$\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{K{\left[x\right]}^2\left[Y\right]}{K\left[4x^2\right]\left[Y\right]}$

$\Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{1}{4}$

And ${\mathrm{r}}_2=4{\mathrm{r}}_1$

Therefore, the rate of reaction becomes quadruple.