$K{}^p$ and K_c are equilibrium constants of a reversible reaction when concentrations are expressed in terms of partial pressure and mole litre^-1 respectively. Which one of the following reactions will have $K{}^p$= Kc?

CONCEPT:

Relationship between K_p and K_c

• The equilibrium constants K_p and K_c are related by the equation:

  K_p = K_c(RT)^Δn

• Here:
  • K_p = equilibrium constant in terms of partial pressures
  • K_c = equilibrium constant in terms of concentrations
  • R = universal gas constant
  • T = temperature in kelvin
  • Δn = change in the number of moles of gas between products and reactants
• If Δn = 0, then K_p = K_c because (RT)^Δn = (RT)⁰ = 1.

EXPLANATION:

• 1) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
  • Δn = (1 + 1) - 1 = 1 (K_p ≠ K_c)
• 2) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
  • Δn = 2 - (1 + 3) = -2 (K_p ≠ K_c)
• 3) COCl₂(g) ⇌ CO(g) + Cl₂(g)
  • Δn = (1 + 1) - 1 = 1 (K_p ≠ K_c)
• 4) H₂(g) + I₂(g) ⇌ 2HI(g)
  • Δn = 2 - (1 + 1) = 0 (K_p = K_c)

From the calculations, only reaction 4) H₂(g) + I₂(g) ⇌ 2HI(g) has Δn = 0, which means K_p = K_c.

Therefore, the correct answer is \(\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g} )\)