Question

The rate of a gaseous reaction is given by the expression, Rate = $k\left[A\right]\left[B\right]$. If the volume of the reaction vessel is suddenly reduced to one-fourth of the initial volume, the reaction rate relatively the original rate will be:

• A. $1/16$ times
• B. $1/8$ times
• C. $8$ times
• D. $16$ times

Answer 1

The correct option is D $16$ times

Volume of both $A$ and $B$ have fallen so
$[A{]}_{new}=\left[\frac{A}{4}\right]$ and $\left[B_{new}\right]=\left[\frac{B}{4}\right]$

$\therefore V_{new}=\frac{\left[A\right]\left[B\right]}{16}=\frac{V_{old}}{16}$

Rate(new) = 16R