JAM Chemical Equilibrium — Kp and Kc Conversions

By Aniket Bhardwaj · 2 September 2026 · IIT-JAM Chemistry

Converting between K_p, K_c and K_x looks like a one-line formula, and it is — but IIT-JAM questions build entire multi-step problems on top of it: degree of dissociation, total pressure, ICE tables, and the effect of adding an inert gas. Every one of those depends on getting Δn right and on choosing the value of R that matches the pressure unit. This guide works through the whole family with the arithmetic shown.

The conversion, and where Δn comes from

K_p = K_c (RT)^Δn where Δn = (moles of gaseous product) − (moles of gaseous reactant)

K_p = K_x P^Δn so K_x = K_p P^−Δn

Only gases count in Δn. Pure solids and pure liquids have unit activity and never appear in K at all, so they contribute nothing to the exponent either. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_p = p(CO₂) and Δn = +1, even though the balanced equation has three species.

Choosing R correctly is half the battle.

Pressure unit for K_p	R to use	Other units
atm	0.08206 L atm K⁻¹ mol⁻¹	c in mol L⁻¹, T in K
bar	0.08314 L bar K⁻¹ mol⁻¹	c in mol L⁻¹, T in K
Pa	8.314 J K⁻¹ mol⁻¹	c in mol m⁻³, T in K

Using 8.314 with concentrations in mol L⁻¹ and pressures in atm is by far the most common wrong answer in this topic, and it is off by a factor of about 100 per unit of Δn.

Worked example 1 — ammonia synthesis (negative Δn)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), K_c = 0.500 at 700 K. Find K_p in atm.

Δn = 2 − (1 + 3) = −2
RT = 0.08206 × 700 = 57.44 L atm mol⁻¹
(RT)⁻² = 1 / 57.44² = 1 / 3 299 = 3.031 × 10⁻⁴

K_p = 0.500 × 3.031 × 10⁻⁴ = 1.52 × 10⁻⁴ (in atm⁻²)

Sanity check on the direction: Δn is negative, so K_p should come out much smaller than K_c. It does.

Worked example 2 — phosphorus pentachloride (positive Δn)

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), K_p = 1.78 atm at 523 K. Find K_c.

Δn = 2 − 1 = +1
RT = 0.08206 × 523 = 42.92 L atm mol⁻¹
K_c = K_p / (RT)¹ = 1.78 / 42.92 = 4.15 × 10⁻² mol L⁻¹

Worked example 3 — the full ICE-table problem

2.00 mol of PCl₅ is placed in a 5.00 L vessel at 523 K and 40.0% dissociates at equilibrium. Find K_c, K_p, the total pressure and K_x.

Initial [PCl₅] = 2.00 / 5.00 = 0.400 M. Amount dissociated x = 0.400 × 0.400 = 0.160 M.

	PCl₅	PCl₃	Cl₂
Initial (M)	0.400	0	0
Change	−0.160	+0.160	+0.160
Equilibrium	0.240	0.160	0.160

K_c = (0.160)(0.160) / 0.240 = 0.02560 / 0.240 = 0.1067 mol L⁻¹

K_p = K_c(RT)⁺¹ = 0.1067 × 42.92 = 4.58 atm

Total moles at equilibrium = 2.00(1 + α) = 2.00 × 1.400 = 2.80 mol, so
P = nRT/V = 2.80 × 42.92 / 5.00 = 24.0 atm

K_x = K_p P⁻¹ = 4.58 / 24.0 = 0.191

The degree-of-dissociation shortcut, and how to check it

For a one-into-two dissociation A ⇌ B + C starting from pure A at total pressure P, mole fractions give

K_p = α² P / (1 − α²)

Testing it against the numbers just obtained: α = 0.400, P = 24.0 atm.
K_p = (0.400)² × 24.0 / (1 − 0.160) = 3.84 / 0.840 = 4.57 atm — agreeing with the ICE-table value of 4.58 atm to rounding.

Two independent routes landing on the same number is the strongest check available in an exam. Build the habit.

The formula also shows immediately that α falls as P rises, which is Le Chatelier for a Δn > 0 reaction. When α is very small, 1 − α² ≈ 1 and α ≈ √(K_p/P) — a useful approximation, but state that you have made it.

Cases where Δn = 0

For H₂(g) + I₂(g) ⇌ 2HI(g), Δn = 2 − 2 = 0, so K_p = K_c = K_x, all numerically equal and independent of pressure. Adding an inert gas at constant volume changes nothing for any reaction (partial pressures are unaltered); adding one at constant total pressure shifts the equilibrium towards the side with more moles of gas — but a Δn = 0 reaction is unaffected even then.

Is K dimensionless?

Strictly, yes. The thermodynamic equilibrium constant is built from activities, each of which is a ratio to a standard state (c° = 1 mol L⁻¹ for solutes, p° = 1 bar for gases), so every term is a pure number and ΔG° = −RT ln K is well defined. The "units" carried through in exam working — atm⁻², mol L⁻¹ and so on — are a bookkeeping convenience. Quote them if the question quotes them, but never write ln of a quantity with units in a thermodynamics answer.

Common mistakes

Counting solids and liquids in Δn. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Δn = +1, not 0.
Using R = 8.314 with atm and mol L⁻¹. Match R to the units of K_p.
Getting the sign of Δn backwards. Products minus reactants, always. For NH₃ synthesis Δn = −2, so K_p < K_c.
Forgetting that α changes the total mole count. In the PCl₅ example, 2.00 mol becomes 2.80 mol, and the total pressure follows from that, not from the starting amount.
Applying K_p = α²P/(1 − α²) to the wrong stoichiometry. It is derived for one mole giving two; 2NH₃ ⇌ N₂ + 3H₂ needs its own derivation.
Assuming K changes with pressure. K_p and K_c depend only on temperature. K_x does depend on P whenever Δn ≠ 0 — that is the one genuine exception, and it is a favourite trick.

Quick reference

Reaction	Δn	Relation
N₂ + 3H₂ ⇌ 2NH₃	−2	K_p = K_c(RT)⁻²
PCl₅ ⇌ PCl₃ + Cl₂	+1	K_p = K_c(RT)
H₂ + I₂ ⇌ 2HI	0	K_p = K_c
2SO₂ + O₂ ⇌ 2SO₃	−1	K_p = K_c(RT)⁻¹
CaCO₃(s) ⇌ CaO(s) + CO₂(g)	+1	K_p = p(CO₂)
N₂O₄ ⇌ 2NO₂	+1	K_p = 4α²P/(1 − α²)

The total pressure step is where marks vanish. Once you have the equilibrium mole count, P = nRT/V decides K_x and every pressure-dependence question that follows. Run that calculation through the ideal gas tool and confirm it before building the rest of the answer on top of it.

Open the Ideal Gas Law Calculator (PV = nRT) →

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