Kp vs Kc relation — derive and calculate for N₂ + 3H₂ ⇌ 2NH₃

Question

Derive the relationship between $K_p$ and $K_c$ . For the reaction $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , if $K_c = 0.5\text{ mol}^{-2}\text{L}^2$ at 400°C, calculate $K_p$ .

(Given: $R = 0.082\text{ L atm mol}^{-1}\text{K}^{-1}$ , $T = 400°C = 673\text{ K}$ )

Solution — Step by Step

For a gaseous reaction, we have two equilibrium constants:

$K_c$ : equilibrium constant in terms of molar concentrations (mol/L)
$K_p$ : equilibrium constant in terms of partial pressures (atm)

Both describe the same equilibrium state — they differ only in how concentrations are expressed.

For an ideal gas: $PV = nRT$ , so partial pressure $p = (n/V)RT = [C]RT$

For a general reaction $aA + bB \rightleftharpoons cC + dD$ :

K_p = \frac{p_C^c \cdot p_D^d}{p_A^a \cdot p_B^b} = \frac{[C]^c(RT)^c \cdot [D]^d(RT)^d}{[A]^a(RT)^a \cdot [B]^b(RT)^b}

K_p = \frac{[C]^c[D]^d}{[A]^a[B]^b} \times (RT)^{(c+d)-(a+b)} = K_c \cdot (RT)^{\Delta n_g}

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

Where $\Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}$

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

Moles of gaseous products = 2 (NH₃)

Moles of gaseous reactants = 1 + 3 = 4 (N₂ + H₂)

\Delta n_g = 2 - 4 = -2

K_p = K_c(RT)^{\Delta n_g} = 0.5 \times (0.082 \times 673)^{-2}

First compute $RT = 0.082 \times 673 = 55.186$ L·atm/mol

$(RT)^{-2} = \frac{1}{(55.186)^2} = \frac{1}{3045.5} = 3.284 \times 10^{-4}$

K_p = 0.5 \times 3.284 \times 10^{-4} = 1.64 \times 10^{-4} \text{ atm}^{-2}

Why This Works

The relationship $K_p = K_c(RT)^{\Delta n_g}$ arises directly from the ideal gas law. Since concentration $[C] = n/V = p/RT$ , replacing concentrations with pressures introduces factors of $RT$ for each mole of gas.

Notice that if $\Delta n_g = 0$ (equal moles of gas on both sides), then $K_p = K_c$ — the two constants are numerically equal. This happens in reactions like $H_2 + I_2 \rightleftharpoons 2HI$ .

For the Haber process (NH₃ synthesis), $\Delta n_g = -2$ means $K_p < K_c$ . Physically, this means using pressure units (atm) instead of concentration units makes the equilibrium constant smaller — consistent with the fact that high pressure favors the forward reaction (fewer moles of gas on product side).

Alternative Method

Units check: $K_c$ has units of $\text{mol}^{-2}\text{L}^2$ (from 2 − 4 = −2 power of mol/L). $K_p$ should have units of atm⁻². We can verify: $(RT)^{-2}$ has units of $(\text{L atm mol}^{-1})^{-2} = \text{mol}^2\text{L}^{-2}\text{atm}^{-2}$ .

So $K_p = K_c \times (RT)^{-2}$ has units: $(\text{mol}^{-2}\text{L}^2) \times (\text{mol}^2\text{L}^{-2}\text{atm}^{-2}) = \text{atm}^{-2}$ ✓

In JEE Main and CBSE Class 11, $\Delta n_g$ calculation is the most error-prone step. Always count only gaseous moles — ignore solids and liquids (their “concentrations” are incorporated into $K$ ). For the reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ : $\Delta n_g = 1 - 0 = 1$ (only CO₂ is a gas; CaCO₃ and CaO are solids and don’t count). When the problem gives $R$ with units L·atm, always use $T$ in Kelvin.

Common Mistake

Students frequently forget to convert Celsius to Kelvin before substituting $T$ . Using $T = 400$ instead of $T = 673\text{ K}$ gives a completely wrong answer. Also, when $\Delta n_g$ is negative (as here), $(RT)^{\Delta n_g}$ is less than 1, so $K_p < K_c$ . Students sometimes try to “make it positive” by inverting — don’t. A negative $\Delta n_g$ means fewer gas molecules on the product side, and $K_p < K_c$ is physically correct for this case.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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