2H_2(g) + O_2(g) ⇋ 2H₂O_(g) ΔH = - ve. What happens to the equilibrium constant of the reaction above if the temperature is increased?

Correct answer: it decrease.

Explanation: The reaction is exothermic (ΔH < 0). By Le Châtelier/van ’t Hoff, increasing temperature favors the endothermic direction (the reverse here), so fewer products (H₂O) are present at equilibrium. Therefore the equilibrium constant K (products/reactants) decreases as temperature increases.

• It is unaffected — Only true if ΔH = 0.
• It becomes zero — Impossible for a finite reversible system.
• It increases — Would be true for an endothermic forward reaction (ΔH > 0), not here.