A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will

Solution :
Since the reaction is 2nd order $$w.r.t$$  $$CO,$$ Thus, rate law is given as.
$$r = k{\left[ {CO} \right]^2}$$
Let initial concentration of $$CO$$  is a i.e. $$[CO]=a$$
$$\therefore \,\,{r_1} = k{\left( a \right)^2} = k{a^2}$$
when concentration becomes doubled, i.e.$$[CO] =2a$$
$$\therefore \,\,{r_2} = k{\left( {2a} \right)^2} = 4k{a^2}\,\,\,\therefore {r_2} = 4{r_1}$$
So, the rate of reaction becomes 4 times.