Question

The rate of a gaseous reaction :
$A\left(g\right)+B\left(g\right)\to \text{product}$, is given by
$K\left[A{]}^2\right[B]$. If the volume (V) of reaction vessel is suddenly doubled, which of the followings will happen?

• A. The rate w.r.t 'A' decreases by four times
• B. The rate w.r.t 'B' decreases by two times
• C. The overall rate decreases by eight times
• D. All of the above

Answer 1

The correct option is D All of the above

$A\left(g\right)+B\left(g\right)\to \text{Product}$
Rate for the above reaction,
$R=K\left[A{]}^2\right[B]$

Suppose 'x' moles of $A$ and 'y' mole of $B$ 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.

The rate of reaction with respect to 'A' is
$r_1^{\prime }=K^{\prime \prime }{\left[\frac{x}{V}\right]}^2$
$r_2^{\prime }=K^{\prime \prime }{\left[\frac{x}{2V}\right]}^2$
$\frac{r_1^{\prime }}{r_2^{\prime }}=4$
$r_2^{\prime }=\frac{r_1^{\prime }}{4}$
Hence, if we double the volume of vessel A, the rate w.r.t 'A' will decrease by 4 times.

The rate of reaction with respect to 'B' is
$r_1^{\prime \prime }=K^{\prime \prime \prime }\left[\frac{x}{V}\right]$
$r_2^{\prime \prime }=K^{\prime \prime \prime }\left[\frac{x}{2V}\right]$
$\frac{r_1^{\prime \prime }}{r_2^{\prime \prime }}=2$
$r_2^{\prime \prime }=\frac{r_1^{\prime \prime }}{2}$
Hence, if we double the volume of vessel B the rate w.r.t 'B' will decrease by 2 times.