JEE MAIN - Chemistry (2006 - No. 34)
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
remain unchanged
triple
increase by a factor of 4
double
Explication
Since the reaction is $$2$$nd 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. $$\left[ {CO} \right] = a$$
$$\therefore$$ $$\,\,\,\,{r_1} = k{\left( a \right)^2} = k{a^2}$$
when concentration becomes doubled, i.e. $$\left[ {CO} \right] = 2a$$
$$\therefore$$ $$\,\,\,\,\,{r_2} = k{\left( {2a} \right)^2} = 4k{a^2}$$
$$\therefore$$ $$\,\,\,\,\,{r_2} = 4{r_1}$$
So, the rate of reaction becomes $$4$$ times.
$$r = k{\left[ {CO} \right]^2}$$
Let initial concentration of $$CO$$ is a i.e. $$\left[ {CO} \right] = a$$
$$\therefore$$ $$\,\,\,\,{r_1} = k{\left( a \right)^2} = k{a^2}$$
when concentration becomes doubled, i.e. $$\left[ {CO} \right] = 2a$$
$$\therefore$$ $$\,\,\,\,\,{r_2} = k{\left( {2a} \right)^2} = 4k{a^2}$$
$$\therefore$$ $$\,\,\,\,\,{r_2} = 4{r_1}$$
So, the rate of reaction becomes $$4$$ times.