Chemistry - Chemical Kinetics Question with Solution | TestHub

$2S{O}_2+O_2⟶2S{O}_3$

Rate law expression for the above reaction is given by $R=k[S{O}_2][O_2]$.

Let $[S{O}_2]=xM$ and $[O_2]=yM$ at any time. Then, the rate $R_{}$ is given by $R_1=k(xy)$.

If the volume of the reaction vessel is tripled, then $[S{O}_2]=\frac{x}{3}M$ and $[O_2]=\frac{y}{3}M$.

Therefore, the new rate, $R_2$, is:

$R_2=k\left(\frac{x}{3}\right)\left(\frac{y}{3}\right)=\frac{kxy}{9}=\frac{R_1}{9}$