Consider the equilibrium CO(g) + 3H2(g) <= CH4(g) + H2O(g) If the pressure applied over the system increases by two…

Step 1 — Analyse the reaction

$\ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)}$

Moles of gas: Reactants = 1 + 3 = 4; Products = 1 + 1 = 2

$\Delta n_g = 2 - 4 = -2$ (fewer moles on product side)

Step 2 — Effect of doubling pressure

When pressure is doubled at constant temperature (by compressing volume to half):

(A) Concentrations of all species increase — YES ✓ (volume halved → all concentrations double)

(B) Equilibrium shifts forward — YES ✓ (Le Chatelier: increasing pressure shifts equilibrium toward fewer moles of gas, i.e., forward direction since $\Delta n_g = -2$)

(C) Equilibrium constant increases — NO ✗ ($K_c$ and $K_p$ depend only on temperature, not pressure)

(D) Equilibrium constant remains unchanged — Partially true, but the statement "concentration of reactants and products remain same" is FALSE (concentrations do change initially before re-equilibration)

Answer: (A) and (B) only → Option (a)

Key Points to Remember:

• $K_c$ and $K_p$ are temperature-dependent only — pressure does NOT change $K$
• Increasing pressure: shifts equilibrium toward side with fewer moles of gas
• $\Delta n_g = -2$ → forward shift on pressure increase
• All concentrations increase when volume is compressed