Question

For the reaction,
$A{B}_2\left(g\right)+A\left(s\right)\rightleftharpoons B_2\left(g\right)+A_2\left(g\right)$ following graph is obtained.

where, $K\to \text{Equilibrium Constant}$
$T\to \text{Temperature in Kelvin}$
Which of the following will increase the concentration of $A{B}_2$ at equilibrium?

• A. Adding more of $A\left(s\right)$
• B. Decreasing temperature
• C. Adding inert gas at constant volume
• D. Increasing the volume of container

Answer 1

The correct option is B Decreasing temperature

We know,
$\ln K=-\frac{\Delta H^{\circ }}{RT}+\frac{\Delta S^{\circ }}{R}$
Here slope is negative.
$\Rightarrow -\frac{\Delta H^{\circ }}{R}=-ve$
$\Rightarrow \Delta H=+ve$
$\therefore$ Reaction is endothermic.
So, on decreasing temperature, reaction will shift backward and concentration of $A{B}_2$ will increase.
On increasing volume, concentration of every species will decrease.
By adding more $A\left(s\right)$, there will be no effect on equilibrium, as it is pure solid.
By adding inert gas at constant volume, total pressure will increase but there will be no effect on partial pressure.