Question

Given $2NO+O_2\to 2N{O}_2$ rate$=K{\left[NO\right]}^2\left[O_2\right]$ by how many times does the rate of 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

Here rate law is given as:-
rate$=k\left[NO{]}^2\right[O_2]$........(1)
$\left[NO\right]$(concentration of $NO$) can be written as $\frac{n}{V}$
where $n$=number of moles and $V$=volume
From (1)
rate=$K{\left[\frac{n}{V}\right]}^2\left[\frac{n}{V}\right]$......(2)
Now,
Volume is related to ${1/3}^{r}d$
From $V$ to $\frac{V}{3}$
put new volume in (2)
New rate will be
$\begin{array}{c}{r}^{\prime }=K{\left[\frac{3n}{V}\right]}^2\left[\frac{3n}{V}\right]=K(3{)}^2{\left[\frac{n}{V}\right]}^2\left(2\right)\left[\frac{n}{V}\right]=27K{\left[\frac{3n}{V}\right]}^2\left[\frac{3n}{V}\right]\\ so,\\ \frac{r^{\prime }}{r}=\frac{27K{\left[\frac{3n}{V}\right]}^2\left[\frac{3n}{V}\right]}{K{\left[\frac{n}{V}\right]}^2\left[\frac{n}{V}\right]}=27\end{array}$
Rate of reaction changes by 27 times.