Given below are certain reactions. Identify the reaction for which K_p < K_c:

We are asked to find which of the given gas‑phase reactions has an equilibrium constant K_p that is smaller than its concentration equilibrium constant K_c.

For any gaseous reaction, the two constants are related by the equation

$Kp\; =\; Kc · (RT)^\{\Delta n\}$

where

• R is the gas constant,
• T is the absolute temperature,
• Δn = (sum of stoichiometric coefficients of gaseous products) – (sum of stoichiometric coefficients of gaseous reactants).

Because R > 0 and T > 0, the factor (RT)^{Δn} is greater than 1 when Δn > 0, equal to 1 when Δn = 0, and less than 1 when Δn < 0. Consequently:

• If Δn > 0, then Kp > Kc.
• If Δn = 0, then Kp = Kc.
• If Δn < 0, then Kp < Kc.

Thus we must examine each reaction and compute Δn.

1. $H2(g)\; +\; I2(g)\; \leftrightharpoons \; 2 HI(g)$

Δn = (2) – (1 + 1) = 2 – 2 = 0 → Kp = Kc.

2. $N2(g)\; +\; O2(g)\; \leftrightharpoons \; 2 NO(g)$

Δn = (2) – (1 + 1) = 0 → Kp = Kc.

3. $N2(g)\; +\; 3 H2(g)\; \leftrightharpoons \; 2 NH3(g)$

Reactant moles = 1 + 3 = 4
Product moles = 2
Δn = 2 – 4 = –2.

Since Δn is negative, (RT)^{Δn} = 1/(RT)^{2} < 1, so Kp = Kc · 1/(RT)^{2} < Kc.

4. $H2O(g)\; +\; CO(g)\; \leftrightharpoons \; H2(g)\; +\; CO2(g)$

Δn = (1 + 1) – (1 + 1) = 2 – 2 = 0 → Kp = Kc.

Only the third reaction has Δn < 0, giving Kp < Kc.

Therefore, the reaction for which K_p < K_c is

$N2(g)\; +\; 3 H2(g)\; \leftrightharpoons \; 2 NH3(g)$