Question

Given 2NO${}_{\left(g\right)}$ + O${}_{2\left(g\right)}$$\to$ 2NO${}_{2\left(g\right)}$; rate = k$[NO{]}^2$ $[O_2$]${}^1$. By how many times does the rate of the reaction change when the volume of the reaction vessel is reduced to 1/3${}^{rd}$ of its original volume? Will there be any change in the order of the reaction?

Answer 1

The given reaction is :-

$2NO\left(g\right)+O_2\left(g\right)⟶2N{O}_2\left(g\right)$

Rate law is given as:-

Rate=$K\left[NO{]}^2\right[O_2{]}^1$ $-\left(i\right)$

Now, when volume of reaction vessel is reduced to $(1/3{)}^{rd}$ of the original volume, the concentration of each of the reactants will be ${\frac{1}{3}}^{rd}$ of the initial concentration.

So,

$(Rate{)}_{new}=K[NO/3{]}^2[O_2/3{]}^1$

$=K.\frac{[NO{]}^2}{9}.\frac{[O_2{]}^1}{3}$

$\Rightarrow$ $(Rate{)}_{new}$= $\frac{1}{27}.K\left[NO{]}^2\right[O_2{]}^1$ $-\left(ii\right)$

On compaering $\left(i\right)$ & $\left(ii\right)$, the rate of the reaction is $\frac{1}{27}$ times the initial rate.

There will be no change in the order of the reaction because it does not depend on the concentration of the reactants.