Question

For a reaction between gaseous compounds
$2A+B\to C+D$
the reaction rate $=k\left[A\right]\left[B\right]$. If the volume of the container is made $\frac{1}{4}$ of the initial, then what will be the rate of reaction as compared to the initial rate?

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

Answer 1

The correct option is A $16$ times

When volume is reduced to one fourthconcentrations become four times. Hence, the rate of reaction becomes $16$ times as compared to the initial rate.