The equilibrium constant for the reaction 2 HF (g) ⇌ H₂ (g) + F₂ (g) is 0.210 at a particular temperature. What is the equilibrium constant for the equation ½ H₂ (g) + ½ F₂ (g) ⇌ HF (g)?

Answer 1

First, the equation in question is the reverse of the one given, so the Keq will be the inverse of the original. Next, all coefficients are 1/2 of the original, so the Keq will be the square root of the original. Putting these together, the Keq for the reaction in question will be √1/Keq = √1/0.210 = 2.18 (See below)

2HF <==> H₂ + F₂ Keq = 0.210

H₂ + F₂ <==> 2HF Keq = 1/0.210 = 4.76

1/2 H₂ + 1/2 F₂ <==> HF Keq = √4.76 = 2.18