Equilibrium constant problems — Kp, Kc, Ksp selection and calculation

Question

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ , $K_c = 0.5$ at 400°C. Calculate $K_p$ at the same temperature. ( $R = 0.0821$ L atm mol $^{-1}$ K $^{-1}$ )

(JEE Main / CBSE 11 — Equilibrium)

Equilibrium Constant Selection

flowchart TD
    A["Equilibrium Problem"] --> B{Phase of reactants?}
    B -->|All gases| C{Given data in?}
    B -->|Solution (aqueous)| D["Use Kc"]
    B -->|Sparingly soluble salt| E["Use Ksp"]
    C -->|Concentrations| F["Use Kc"]
    C -->|Partial pressures| G["Use Kp"]
    F --> H["Kp = Kc(RT)^delta_n"]
    G --> H
    E --> I["Ksp = product of ion concentrations"]

Solution — Step by Step

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$

$\Delta n_g$ = moles of gaseous products - moles of gaseous reactants

$\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$

K_p = K_c(RT)^{\Delta n_g}

$T = 400°C = 673$ K, $R = 0.0821$ L atm mol $^{-1}$ K $^{-1}$

K_p = 0.5 \times (0.0821 \times 673)^{-2}

K_p = 0.5 \times (55.25)^{-2}

K_p = 0.5 \times \frac{1}{3052.6}

\boxed{K_p = 1.64 \times 10^{-4} \text{ atm}^{-2}}

$K_p$ is much smaller than $K_c$ because $\Delta n_g < 0$ . When the reaction decreases the total moles of gas, the pressure-based constant is smaller than the concentration-based constant. This makes physical sense — the forward reaction reduces pressure, so at equilibrium, the pressure contribution from products is proportionally smaller.

Why This Works

$K_p$ and $K_c$ differ because pressure and concentration are related by the ideal gas law ( $P = cRT$ ). When $\Delta n_g = 0$ , $K_p = K_c$ because the RT factors cancel. When $\Delta n_g \neq 0$ , the $(RT)^{\Delta n_g}$ factor converts between the two.

The key insight: $K_c$ uses mol/L, $K_p$ uses atm (or Pa). Since $P = (n/V)RT = cRT$ , each concentration term in $K_c$ gets multiplied by $RT$ when converting to pressure, and the net effect depends on the difference in total gas moles.

Alternative Method — When to Use Ksp

For sparingly soluble salts like AgCl:

\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)

K_{sp} = [\text{Ag}^+][\text{Cl}^-]

The solid does not appear in the expression. If $K_{sp} = 1.8 \times 10^{-10}$ , the solubility $s = \sqrt{K_{sp}} = 1.34 \times 10^{-5}$ mol/L.

For JEE, always check the units. $K_c$ is unitless when written as activities, but in practice has units of (mol/L) $^{\Delta n}$ . $K_p$ has units of (atm) $^{\Delta n}$ . If a problem gives concentrations, use $K_c$ . If it gives partial pressures, use $K_p$ . Converting between them requires the temperature in Kelvin — never Celsius.

Common Mistake

The most common error: using temperature in Celsius instead of Kelvin in the $K_p = K_c(RT)^{\Delta n_g}$ formula. At 400°C, $T = 673$ K, not 400. Using 400 gives $(0.0821 \times 400)^{-2} = (32.84)^{-2}$ instead of $(55.25)^{-2}$ — a completely different answer. Always convert to Kelvin first.

Another mistake: getting the sign of $\Delta n_g$ wrong. It is (gaseous products) minus (gaseous reactants). Including solids or liquids in the count is wrong — they do not appear in $K_p$ or $K_c$ expressions.