Question

The rate of gaseous reaction is given by the expression $k\left[A\right]\left[B\right]$. If the volume of the reaction vessel is suddenly reduced to $1/4^{th}$ of the initial volume, the reaction rate relating to original rate will be:

• A. $1/10$
• B. $1/8$
• C. $8$
• D. $16$

Answer 1

The correct option is D $16$

Concentration is number of moles per liter, $n/V$. it is proportional to volume.
Rate of reaction is $k\left[A\right]\left[B\right]$ where $\left[A\right]\text{and}\left[B\right]$ are concentration of $A\text{and}B$.
When the volume is suddenly reduced to $1/4$ times the original, the concentration changes 4 times the original concentration.
The new rate will be $k\left[4A\right]\left[4B\right]=16k\left[A\right]\left[B\right]$ The new rate will be 16 times the original rate.