Question

In which of the following reactions, the increase in volume of the container will favour the formation of products?

• A. $C\left(s\right)+H_{2}O\left(g\right)\rightleftharpoons CO\left(g\right)+H_2\left(g\right)$
• B. $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• C. $4N{H}_3\left(g\right)+5O_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+6H_{2}O\left(l\right)$
• D. $3O_2\left(g\right)\rightleftharpoons 2O_3\left(g\right)$

Answer 1

The correct option is A $C\left(s\right)+H_{2}O\left(g\right)\rightleftharpoons CO\left(g\right)+H_2\left(g\right)$

As per Le - Chatelier principle, If pressure is increased, equilibrium will shift in that direction where number of moles are less that means if no of gaseous moles is more in the product side, the reaction will shift in forward direction. Hence, will favour the formation of products. In option (a), the no of gaseous moles of products are more than reactants. Hence, option (a) will favour the formation of products.

Theory:

Effect of change in Volume :
According to Boyle's law at constant temperature.
$P\propto \frac{1}{V}$
For a reaction, when the volume V increases the concentration $\left(\frac{n_g}{V}\right)$ of the reactants and the products decrease.

Consider $A\left(g\right)\rightleftharpoons B\left(g\right)$ When, $\Delta n_g=0$, we know that even if pressure changes there is no shift in the equilibrium.

Therefore even if volume changes there is no shift in equilibrium.

For a reaction $\Delta n_g>0$
$A\left(g\right)\rightleftharpoons 2B\left(g\right)$ when $\Delta n_g=+ve$
when volume increases then $\frac{n_g}{V}$ decreases, then the reaction goes where $\frac{n_g}{V}$ increases . Reaction goes forward.

For a reaction when $\Delta n_g<0$ $2A\left(g\right)\rightleftharpoons B\left(g\right)$ when $\Delta n_g=-ve$
Volume $V$ increases then $\frac{n_g}{V}$ decreases, then the reaction goes where $\frac{n_g}{V}$ increases . Reaction goes backward.