Question

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

• A. will diminish to $(\frac{1}{4}{)}^{th}$ of inital value
• B. will diminish to $(\frac{1}{8}{)}^{th}$ of inital value
• C. will increase to 4 times
• D. will increase to 8 times

Answer 1

The correct option is B will diminish to $(\frac{1}{8}{)}^{th}$ of inital value

$2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)$
$Rate=K\left[NO{]}^2\right[O_2]$
$R=K\left[NO{]}^2\right[O_2]$

Suppose 'x' moles of $NO$ and 'y' mole of $O_2$ are taken in the vessel of volume 'V' litre, then,

$r_1=K^{\prime }{\left[\frac{x}{V}\right]}^2$ $\left[\frac{y}{V}\right]$

When the volume of reaction vessel is doubled,
$r_2=K^{\prime }{\left[\frac{x}{2V}\right]}^2$ $\left[\frac{y}{2V}\right]........eqn.1$

$r_2=K^{\prime }\frac{1}{8}{\left[\frac{x}{V}\right]}^2$ $\left[\frac{y}{V}\right].......eqn.2$

Dividing equation 1 by 2,

$\frac{r_1}{r_2}=\frac{8}{1}$
$\Rightarrow r_2=\frac{r_1}{8}$

Thus, the rate of reaction become $\frac{1}{8}th$ times the intial rate.