General Chemistry II · Chemical Equilibrium

Relating Kp and Kc

5 min read

On this page 8 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Study tools
Sources & references

In 30 seconds

For gas-phase equilibria we can express the constant two ways: in concentrations (Kc) or in partial pressures (Kp). The two are related through the ideal gas law. The conversion depends on Δn, the change in the number of moles of gas from reactants to products:

$K p = K c (RT) Δ n$

When a reaction produces the same number of gas molecules it consumes (Δn = 0), the two constants are numerically equal.

Why this matters

Real gas-phase data are often reported as pressures (Kp) while calculations and ICE tables use concentrations (Kc). Being able to convert cleanly — and knowing that Δn counts only gas-phase species — prevents sign errors and lets you move between tabulated data and worked problems. It also reinforces the ideal-gas-law bridge between pressure and concentration.

The college version

Core Concept

$K p = K c (RT) Δ n$

When a reaction produces the same number of gas molecules it consumes (Δn = 0), the two constants are numerically equal.

Key Ideas

Δn = (moles of gaseous products) − (moles of gaseous reactants), using only gas-phase species.
R must be chosen to match the units of Kp: 0.08206 L·atm·mol⁻¹·K⁻¹ when pressures are in atm.
Temperature must be in kelvin.
Solids and liquids contribute nothing to Δn.

Equations and Variables

$K p = K c (RT) Δ n$

Kp = pressure-based equilibrium constant (units from partial pressures)
Kc = concentration-based equilibrium constant
R = 0.08206 L·atm·mol⁻¹·K⁻¹ (ideal gas constant)
T = absolute temperature (K)
Δn = (moles gaseous products) − (moles gaseous reactants)

How It Works

The connection comes from the ideal gas law, P = (n/V)RT = [X]RT. Substituting P_X = [X]RT into the Kp expression and factoring out (RT) for each gas species leaves (RT) raised to the difference in gas moles:

$K p = K c (RT) (gas products - gas reactants) = K c (RT) Δ n$

Three cases matter:

Δn > 0 (more gas produced): Kp > Kc at any T above absolute zero.
Δn < 0 (gas consumed): Kp < Kc.
Δn = 0: Kp = Kc exactly, and the two constants carry the same value.

Worked Example

(a) N₂O₄(g) ⇌ 2 NO₂(g) at 25 °C (298 K), Kc = 4.63 × 10⁻³.

Δn = 2 − 1 = +1 (gas only):

$K p = K c (RT) 1 = (4.63 \times 10 -3)(0.08206 \times 298) = 4.63 \times 10 -3 \times 24.45 = 0.113 atm$

(b) H₂(g) + I₂(g) ⇌ 2 HI(g) at 445 °C, Kc = 50.5.

Δn = 2 − (1 + 1) = 0:

$K p = K c (RT) 0 = K c = 50.5$

(c) N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), Δn = 2 − 4 = −2.

$K p = K c (RT) -2 = K c (RT) 2$

Kp is smaller than Kc because the reaction reduces the number of gas molecules.

How it works

$K p = K c (RT) (gas products - gas reactants) = K c (RT) Δ n$

Three cases matter:

Δn > 0 (more gas produced): Kp > Kc at any T above absolute zero.
Δn < 0 (gas consumed): Kp < Kc.
Δn = 0: Kp = Kc exactly, and the two constants carry the same value.

Common confusions

Counting solids/liquids in Δn. Only gas-phase species contribute.
Using the wrong R. For atm pressures, use 0.08206 L·atm·mol⁻¹·K⁻¹; using 8.314 gives wrong units.
Wrong sign on Δn. It is products minus reactants, not the reverse.
Using °C. Temperature must be in kelvin.

Quick review

Write the equation relating Kp and Kc.
Define Δn precisely.
For which condition is Kp = Kc?
For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), is Kp larger or smaller than Kc?
What value of R should you use when pressures are in atm?

Eli explains

The same idea, in plain words

Explain it like I’m 10

Kc and Kp are just two ways to describe the same balance — one counts molecules by concentration, the other by pressure. Converting between them is like converting between "people per room" and "air pressure from all those people." The conversion depends on whether the reaction makes more gas or less gas. If it makes more gas (Δn positive), the pressure-based K is bigger because you've got more "push." If it makes the same amount of gas (Δn = 0), the two numbers are identical — no conversion needed at all.

Worked example

Worked Example

(a) N₂O₄(g) ⇌ 2 NO₂(g) at 25 °C (298 K), Kc = 4.63 × 10⁻³.

Δn = 2 − 1 = +1 (gas only):

$K p = K c (RT) 1 = (4.63 \times 10 -3)(0.08206 \times 298) = 4.63 \times 10 -3 \times 24.45 = 0.113 atm$

(b) H₂(g) + I₂(g) ⇌ 2 HI(g) at 445 °C, Kc = 50.5.

Δn = 2 − (1 + 1) = 0:

$K p = K c (RT) 0 = K c = 50.5$

(c) N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), Δn = 2 − 4 = −2.

$K p = K c (RT) -2 = K c (RT) 2$

Kp is smaller than Kc because the reaction reduces the number of gas molecules.

Key takeaways

### High-Yield Facts
Kp = Kc(RT)^Δn, with Δn from gas species only.
Δn = 0 ⇒ Kp = Kc.
Δn > 0 ⇒ Kp > Kc; Δn < 0 ⇒ Kp < Kc.
Use R = 0.08206 L·atm·mol⁻¹·K⁻¹ with atm, and T in kelvin.
Solids and pure liquids are excluded from Δn.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice General Chemistry II

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

Convert between Kp and Kc using Kp = Kc(RT)^Δn.
Calculate Δn correctly as the change in moles of gas (not all species).
Recognize the special case Δn = 0, where Kp = Kc.
Use the appropriate value of the gas constant R.

Sources & references

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.

In 30 seconds

Why this matters

The college version

Core Concept

Key Ideas

Equations and Variables

How It Works

Worked Example

How it works

Common confusions

Quick review

The same idea, in plain words

Worked example

Worked Example

Key takeaways

Keep learning

You’ll learn to

Related lessons

Sources & references