Question

A open ended mercury manometer is used to measure the pressure exerting by a trapped gas as shown in the figure. Initially manometer shows no difference in mercury level in both column as shown in diagram.
After sparkling ${NH}_3$ dissociated according to following given reaction
${NH}_3\left(g\right)⟶\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$
If pressure of ${NH}_3$ decreases to 0.9 atm. Then difference in mercury level is (assume temperature is constant during entire process.

• A. 228 mm Hg
• B. 38 mm Hg
• C. 76 mm Hg
• D. 152 mm Hg

Answer 1

The correct option is C 76 mm Hg

Solution:- (C) $76\text{mm Hg}$

Initially, the pressure of $N{H}_3$ is equal to the atmospheric pressure which is 1 \; atm$.

As the pressure of $N{H}_3$ is decreased to $0.9atm$.

Therefore, at equilibrium,

$P_{N{H}_3}=0.9atm$

$P_{H_2}=0.15atm$

$P_{N_2}=0.05atm$

$\therefore$ Total pressure at equilibrium $=P_{N{H}_3}+P_{H_2}+P_{N_2}=0.9+0.15+0.05=1.1$

Thus the total pressure is increased by $0.1atm$.

Therefore,

Corresponding difference in mercury level $=0.1\times 760=76\text{mm Hg}$

Hence the difference in mercurey level is $76\text{mm Hg}$.