NH\(_3\) \((_g\)) + HCl\((_g\)) → NH\(_4\)Cl \(_(g)\) In the reaction above, increase in pressure will

Answer Details

The reaction is:

\[\text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s)\]

On the reactant side, there are 2 moles of gas (1 mole of NH₃ + 1 mole of HCl). On the product side, NH₄Cl is a solid, so there are effectively 0 moles of gas.

According to Le Chatelier's principle, when the pressure of a gaseous system at equilibrium is increased, the equilibrium shifts towards the side with fewer moles of gas to reduce the pressure.

Since the product side has fewer gaseous moles than the reactant side, increasing the pressure will shift the equilibrium to the right, favouring the product (NH₄Cl).

Note that changing pressure shifts the position of equilibrium but does not change the equilibrium constant (K). The equilibrium constant is only affected by changes in temperature, not pressure or concentration.

Exam tip: When applying Le Chatelier's principle to pressure changes, count only the moles of gaseous species on each side. Solids and liquids are not affected by pressure changes.