Questions: Equilibrium Constants: Kc and Kp

Question 1

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what is the value of Δn used in Kp = Kc(RT)^Δn?

• A. -2
• B. +2
• C. -4
• D. +4

Answer: (A)

Δn = moles of gaseous products − moles of gaseous reactants = 2 − (1 + 3) = −2. This negative Δn means Kp < Kc for this reaction at any temperature above 0 K, because (RT)^−2 is less than 1.

Question 2

When writing Kc for a reaction that involves a solid reactant (e.g., CaCO₃(s) ⇌ CaO(s) + CO₂(g)), the concentration of CaCO₃ should appear in the denominator of the equilibrium expression.

• T. True
• F. False

Answer: False

Pure solids and pure liquids are omitted from equilibrium expressions because their concentrations are constant (their 'concentration' is absorbed into the value of K). Only gases and dissolved solutes appear. For this reaction, Kc = [CO₂] and Kp = P(CO₂).

Question 3

Why are Kc and Kp numerically different for most reactions, even though they describe the same equilibrium?

Model answer: Kc measures concentrations in mol/L while Kp measures partial pressures in atm (or Pa). They are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. Unless Δn = 0, the (RT)^Δn factor shifts the numerical value, so the two constants differ.

The conversion factor (RT)^Δn bridges the concentration and pressure scales. When Δn = 0 (equal moles of gas on both sides), Kp = Kc. For all other reactions the two constants diverge in proportion to temperature and the imbalance in gas moles.