Question

Given $2N{O}_{\left(g\right)}+O_{2\left(g\right)}\to 2N{O}_{2\left(g\right)}$; rate =$k\left[NO{]}^2\right[O_2{]}^1$. By how many times of 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 order of the reaction?

Answer 1

Given:

$2NO+O_2\to 2N{O}_2$

Let us assume x moles of $NO$ and y moles of $O_2$ are taken in the vessel of volume v.

Then,

$r_1,=k\left[\frac{x}{v}{]}^2\right[\frac{y}{v}]$

If the volume of vessel is reduced to $\frac{v}{3}$ then,

$r_2=k\left[\frac{x}{\frac{v}{3}}{]}^2\right[\frac{y}{\frac{v}{3}}]$

$=27k\left[\frac{y}{v}{]}^2\right[\frac{y}{v}]$ Where $v=\frac{v}{3}$

So,

$\frac{r_2}{r_1}=27$

Thus, The rate of the reaction becomes 27 times the initial rate and the order of reactionremains unchanged.