Question

Assuming that $2NO+O_2⟶2{NO}_2$ is a single step reaction, what will be the rate of reaction when the volume of the reaction vessel is reduced to $\frac{1}{4}$th of the initial value? The original rate of reaction is $64$ mole $l^{-1}{s}^{-1}$.

Answer 1

$2NO+O_2\to 2N{O}_2$
The rate of reaction is $r=k{\left[NO\right]}^2\left[O_2\right]$, as it is a single step reaction. when the volume of the reaction vessel reduced to 1/4th of the initial volume, the concentration of each reactant is increased by 4 times. Since the reaction is 2nd order with respect to $NO$ and 1st order with respect to $O_2$, the rate of reaction increases by 16 times for $NO$ and 4 times for $O_2$. On the whole, the rate of reaction increases by 64 times (${\left[NO\right]}^2\times \left[O_2\right]=[4{]}^2\times [4\left]\right)$. The original rate of reaction is 64 $mole{l}^{-1}{s}^{-1}$ and the new rate of reaction is 64 times more than this, thus
$r_2=64\times 64=4096=4.096\times {10}^{3}mole{l}^{-1}{s}^{-1}$.