Chemistry - Chemical Equilibrium Question with Solution | TestHub

Solution:

(P) $2{\mathrm{NH}}_3(\mathrm{g})\rightleftharpoons {\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})$: $\Delta n_g=(1+3)-2=2$.

(Q) ${\mathrm{PCl}}_5(\mathrm{g})\rightleftharpoons {\mathrm{PCl}}_3(\mathrm{g})+{\mathrm{Cl}}_2(\mathrm{g})$: $\Delta n_g=(1+1)-1=1$.

(R) $2\mathrm{HI}(\mathrm{g})\rightleftharpoons {\mathrm{H}}_2(\mathrm{g})+{\mathrm{I}}_2(\mathrm{g})$: $\Delta n_g=(1+1)-2=0$.

(S) ${\mathrm{NH}}_4\mathrm{HS}(\mathrm{s})\rightleftharpoons {\mathrm{NH}}_3(\mathrm{g})+{\mathrm{H}}_2\mathrm{S}(\mathrm{g})$: $\Delta n_g=(1+1)-0=2$.

(1) Catalyst does not affect equilibrium position for any reaction. So, 1 applies to P, Q, R, S.

(2) Addition of more reactant shifts equilibrium to the right (forward direction) for all reactions, but in (S) the reactant is solid hence it will not shift the reaction forward.

(3) Adding inert gas at constant volume does not change partial pressures, so equilibrium is unaffected. Applies to P, Q, R, S.

(4) Adding inert gas at constant pressure increases volume, shifting equilibrium towards more moles of gas. Applies to P, Q, S (where $\Delta n_g>0$).

(5) $\alpha$ is independent of pressure when $\Delta n_g=0$. Applies to R.

Final matching:

P $\to$ 1, 3, 4

Q $\to$ 1, 3, 4

R $\to$ 1, 3, 5

S $\to$ 1, 2, 3, 4

The option D matches this combination.