Chemistry

Problem

For the reaction

\mathrm{N_2(g)+3\,H_2(g)\;\rightleftharpoons\;2\,NH_3(g)}

with

\Delta H = -92\,\text{kJ mol}^{-1}

, compression of the reaction vessel at constant temperature increases the total pressure. In one or two sentences, explain why this compression increases the equilibrium mole fraction of

\mathrm{NH_3}

in terms of the reaction quotient

Q

and the equilibrium constant

K

Intuitive explanation

When you compress the vessel, the total pressure rises instantly, which increases $Q$ (the ratio of product to reactant partial pressures). Since $Q$ becomes less than the fixed equilibrium constant $K$ at that temperature, the system must shift right to re-establish equilibrium—and shifting right produces more $\mathrm{NH_3}$ because the forward reaction consumes 4 moles of gas to make 2 moles, favoring the side with fewer particles at higher pressure.

Intuitive explanation

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