Question

For a reaction, $2NO\left(g\right)+O_2\left(g\right)\to 2{NO}_2\left(g\right)$
Rate $=k{\left[NO\right]}^2\left[O_2\right]$ if the volume of the reaction vessel is doubled, then the rate of the reaction:

• A. will diminish to $1/4$ of initial value
• B. will diminish to $1/8$ of initial value
• C. will grow $4$ times
• D. will grow $8$ times

Answer 1

The correct option is B will diminish to $1/8$ of initial value

${2NO}_{\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons {2NO}_{2\left(g\right)}$

The above reaction take place in one step, thus rate of reaction is,

$r=k{\left[NO\right]}^2\left[O_2\right]$

Suppose $x$ moles of $NO$ and $y$ mole of $O_2$ are taken in the vessle of volume $vlitre$,

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

If the volume of vessel is doubled to $2v$ then for the same moles of ${NO}_2$ and $O_2$.

$r_2=k{\left[\frac{x}{2v}\right]}^2\left[\frac{y}{2v}\right]=\frac{k}{8}{\left[\frac{x}{v}\right]}^2\left[\frac{y}{v}\right]\therefore \frac{r_1}{r_2}=\frac{1}{8}\Rightarrow r_2=\frac{r_1}{8}$

The rate of reaction becomes ${\frac{1}{8}}^{th}times$ the initial rate.