Homogeneous equilibria: Kp and Kc

What is a homogeneous equilibrium?

In plain words: when everything is a gas, it is often easier to measure pressures than concentrations. For an ideal gas at fixed temperature, partial pressure is concentration times RT, so a constant written with pressures is the same idea as one written with concentrations; the two differ only by powers of RT, one for each mole of gas gained or lost in the reaction.

Equilibrium constant in gaseous systems

From the ideal gas equation, pV = nRT, so p = (n/V)RT. With concentration c in mol L⁻¹ and pressure in bar (NCERT Section 6.4.1, pages 177–178):

In the ideal-gas approximation at constant temperature, partial pressure is proportional to concentration. For H₂(g) + I₂(g) ⇌ 2HI(g):

NCERT’s Eq. 6.12 labels the pressure ratio K_c; it should read K_p.

The RT factors cancel because there are 2 moles of gas on each side. That is not always so. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), 2 moles of gas on the right but 4 on the left leave (RT)² on top and (RT)⁴ underneath, so K_p = K_c(RT)⁻² (NCERT Eq. 6.14).

The relation between K_p and K_c

For an ideal-gas reaction a A + b B ⇌ c C + d D, the same steps give (NCERT Eq. 6.15, page 178):

So:

• Δn = 0: K_p = K_c (H₂ + I₂ ⇌ 2HI).
• Δn > 0 (more gas made): the numerical value of K_p in bar-based units exceeds that of K_c in mol L⁻¹-based units when the numerical value of RT exceeds 1 in the matching units. This is a unit-dependent comparison, not an inequality between quantities with different dimensions.
• Δn < 0 (gas used up): the numerical pressure-based value is smaller in those same units and ordinary temperature range.

For a dimensional pressure product, any consistent pressure unit can be used with the matching R. For a dimensionless thermodynamic constant, divide each pressure by the chosen standard pressure, usually p° = 1 bar = 10^⁵ Pa. The ratio is unit-independent: 2 bar/1 bar equals 200 000 Pa/100 000 Pa. If both K_p and K_c are normalised, their relation contains [RTc°/p°]^Δn, not the dimensionful RT alone.

Some K_p values (NCERT Table 6.5, page 178): N₂ + 3H₂ ⇌ 2NH₃, 6.8 × 10⁵ at 298 K, 41 at 400 K and 3.6 × 10⁻² at 500 K; 2SO₂ + O₂ ⇌ 2SO₃, 4.0 × 10²⁴ at 298 K, 2.5 × 10¹⁰ at 500 K and 3.0 × 10⁴ at 700 K; N₂O₄ ⇌ 2NO₂, 47.9 at 400 K and 1700 at 500 K. For N₂O₄ ⇌ 2NO₂ at 298 K the table prints 0.98. Δ_fG^⊖ values (NO₂ 51.31, N₂O₄ 97.89 kJ mol⁻¹) give ΔG^⊖ = +4.73 kJ mol⁻¹ and K_p ≈ 0.15, and only a value of that size fits the table’s own 400 K and 500 K entries. NCERT itself, on page 182, gives K_c = 4.64 × 10⁻³ at 25 °C for this reaction, which corresponds to K_p ≈ 0.11.

Units of the equilibrium constant

With concentrations in mol L⁻¹ and pressures in bar, K has units unless the powers on top and bottom are equal (NCERT box, page 180):

• H₂(g) + I₂(g) ⇌ 2HI(g): K_c and K_p have no unit;
• N₂O₄(g) ⇌ 2NO₂(g): K_c is in mol L⁻¹ and K_p in bar.

For these all-gas expressions, the unit of K_c is (mol L⁻¹)^Δn and that of K_p is bar^Δn. If each concentration is divided by the standard 1 mol L⁻¹ and each pressure by the standard 1 bar, both constants become pure numbers; they still differ in value (unless Δn = 0), because the standard states differ (page 180). That is how K can appear inside a logarithm, as in ΔG^⊖ = −RT ln K (see Gibbs energy and equilibrium and K, Q and Gibbs energy); for ideal gases with the usual 1 bar gas standard state, use pressure ratios p/p°; choosing a different standard state also requires the matching standard Gibbs energy.

Solved examples

Common mistakes to avoid

Quick revision summary

• Homogeneous equilibrium: all reactants and products in one phase.
• For gases p = [gas]RT (R = 0.0831 bar L mol⁻¹ K⁻¹); K_p uses partial pressures in bar.
• K_p = K_c(RT)^Δn; Δn = gaseous moles of products minus reactants; Δn = 0 gives K_p = K_c.
• Units: (mol L⁻¹)^Δn for K_c, bar^Δn for K_p; dimensionless when referred to standard states.

Questions students ask

What is the difference between K_p and K_c?

K_c is the equilibrium constant written with molar concentrations in mol L^⁻¹, while K_p is written with partial pressures in bar, so only gases enter it; pure solids and liquids are left out of both. For the same ideal-gas equation at the same temperature the two are linked by K_p = K_c(RT)^Δn, so Δn = 0 makes them equal independently of temperature and unit conversion. Both depend only on temperature and say the same thing about the extent of reaction; K_p is simply more convenient when pressures are what is measured.

How are K_p, K_c and K_x related?

K_x is the equilibrium constant written with mole fractions, and for ideal gases each partial pressure is the mole fraction times the total pressure, p_i = x_iP, the three are linked by K_p = K_c(RT)^Δn = K_xP^Δn, with P in bar. K_p and K_c depend only on temperature, but K_x also changes with the total pressure unless Δn = 0: when Δn is negative, raising P raises K_x and the mole fractions shift towards the products, which is Le Chatelier’s principle in numbers. K_x is not in NCERT.

How is the degree of dissociation of PCl₅ related to K_p?

For an ideal gas starting with pure PCl₅ and no other gases, PCl₅ ⇌ PCl₃ + Cl₂ gives K_p = α^²P/(1 − α^²), where α is the degree of dissociation and P the total pressure: one mole of PCl₅ leaves (1 − α) mol of PCl₅ and α mol each of PCl₃ and Cl₂, (1 + α) mol in all, and each partial pressure is its mole fraction times P. For example, α = 0.40 at P = 2.0 bar gives K_p = 0.16 × 2.0/0.84 ≈ 0.38 bar. When α is small, α ≈ √(K_p/P), so the dissociation falls as the pressure rises, as Le Chatelier’s principle predicts for a reaction that makes more moles of gas.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K_p = 3.0 × 10⁴ at 700 K (NCERT Table 6.5). Find K_c.

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   Δn = 2 − 3 = −1, so K_p = K_c(RT)⁻¹ and K_c = K_p × RT = 3.0 × 10⁴ × 0.0831 × 700, about 1.7 × 10⁶ (L mol⁻¹).

2. Q2. Give the units of K_c and K_p for SO₂(g) + ½O₂(g) ⇌ SO₃(g).

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   Δn = 1 − 1½ = −½, so K_c is in (mol L⁻¹)^−1/2 = L^1/2 mol^−1/2, and K_p in bar^−1/2.

3. Q3. Using K_c = 1.79 at 500 K for PCl₅ ⇌ PCl₃ + Cl₂ (Example 1), find K_p.

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   Δn = +1, so K_p = 1.79 × 0.0831 × 500 = 1.79 × 41.6, about 74 bar.