Question

For a given reaction, the rate of gaseous reaction is given by $k\left[A\right]\left[B\right]$. If the volume of the reaction vessel is reduced to $\frac{1}{4}$, how much will be the new rate compared to the initial rate?

$2A+B\to C+D$

• A. $\frac{1}{16}$ times
• B. $\frac{1}{8}$ times
• C. $4$ times
• D. $16$ times

Answer 1

The correct option is D $16$ times

$c=\frac{n}{v}\Rightarrow c$ $\propto \frac{1}{v}$
$v$ is reduced to $\frac{1}{4}$
Let $r=k\left[A\right]\left[B\right]$
$[A{]}_{new}=4[A]$
$[B{]}_{new}=4[B]$
$r_{new}=k\left[A{]}_{new}\right[B{]}_{new}$
$=16k\left[A\right]\left[B\right]$
$=16r$

Hence, option D is correct.