Equilibrium Constants Kc and Kp

Learn and apply Equilibrium Constants Kc and Kp for H2 Chemistry 9476.

GCE A-Level H2 Chemistry 9476-2027

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Equilibrium Constants Kc and Kp: Orientation

This lesson is about writing exam-safe K expressions and avoiding the classic traps: wrong powers, including solids/liquids, and mixing up “equilibrium position” with “K value”.

Anchor this lesson with Dynamic Equilibrium and Le Chatelier and the Chemical Equilibria hub so you can move between concept and calculation questions.

Definitions (Must Know)

A. Equilibrium constant, K

An equilibrium constant, K, is a constant value (at a fixed temperature) that relates the equilibrium amounts of products to reactants for a given equilibrium equation.

B. K_c

K_c is an equilibrium constant written in terms of equilibrium concentrations (typically mol dm⁻³).

C. Kₚ

Kₚ is an equilibrium constant written in terms of equilibrium partial pressures (gases only).

D. Reaction quotient style (brackets / p)

[X] means the equilibrium concentration of X.
p_X means the equilibrium partial pressure of X.

Key Ideas (What Earns Marks)

Use equilibrium values only (not initial).
Include only species that appear in the equilibrium (gases for Kₚ).
Do not include pure solids and pure liquids in K_c or Kₚ expressions.
K changes only with temperature.
If the equation is changed:
- reversed → K becomes 1/K
- multiplied by k → K becomes K^k

Products on top, reactants on bottom.
Powers come from coefficients.
Exclude pure solids and liquids.

Detailed Explanations

A. Writing a K_c expression (workflow)

Write the balanced equilibrium equation.
Put equilibrium concentrations of products in the numerator and reactants in the denominator.
Raise each concentration to the power of its coefficient.
Remove any pure solids and pure liquids.

For: aA + bB ⇌ cC + dD K_c = [C]^c[D]^d/[A]^a[B]^b

Mini example: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) K_c = [SO₃]²/[SO₂]²[O₂]

B. Writing a Kₚ expression (workflow)

Use partial pressures (gases only): Kₚ = (p_C)^c(p_D)^d/(p_A)^a(p_B)^b

Mini example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Kₚ = (p_NH₃)²/p_N₂(p_H₂)³

C. Interpreting magnitude

Large K means equilibrium lies to the right (mostly products).
Small K means equilibrium lies to the left (mostly reactants).

Because K is a ratio of “products over reactants” at equilibrium, a large K means you need a lot of products (relative to reactants) to satisfy the equilibrium expression.

Example with A ⇌ B and a total concentration of 1.00 (A + B): larger K gives a larger fraction of B at equilibrium.

Data table

K value	B (products)	A (reactants)
0.01	0	1
0.1	0.1	0.9
1	0.5	0.5
10	0.9	0.1
100	1	0

D. How K changes when you change the equation

If you reverse the equilibrium equation, the numerator and denominator swap, so: K_reversed = 1/K

If you multiply every coefficient by k, every concentration/pressure term is raised to a power k, so: K_new = K^k

Plotted as log10(K) to show the effect clearly: reversing gives 1/K and multiplying coefficients by 2 gives K^2 (example K = 10).

Data table

Equation change	log10(K)
Original	1
Reversed	-1
Multiplied by 2	2

Worked Examples

Problem

Write the

K_c expression for

2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

Study the worked solution

Place product over reactants
Method
Put the equilibrium concentration of SO₃ in the numerator and those of SO₂ and O₂ in the denominator.
Reason
The equilibrium expression follows products over reactants for the equation as written.
Working
K_c = [SO₃]^?/[SO₂]^?[O₂]^?.
Apply stoichiometric powers
Method
Use each balanced coefficient as the power of its concentration term.
Reason
The coefficients define how the reaction quotient changes with composition.
Working
K_c = [SO₃]²/[SO₂]²[O₂].

Problem

Write the

Kₚ expression for

N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

Try this before viewing the solution

Kp expression with coefficient powers

Hints

Hint 1: choose the quantities

Use equilibrium partial pressures of gaseous species, with products over reactants.

Hint 2: apply coefficients

The coefficients 2, 1 and 3 become the powers on the ammonia, nitrogen and hydrogen terms.

View solution step by step

Arrange the pressure terms
Method
Place ammonia pressure above nitrogen and hydrogen pressures.
Reason
Ammonia is the product for the equation as written.
Working
Kₚ = (p_NH₃)^?/(p_N₂)^?(p_H₂)^?.
Apply the powers
Method
Use the balanced coefficients as exponents.
Reason
Two ammonia, one nitrogen and three hydrogen appear in the balanced equation.
Working
Kₚ = (p_NH₃)²/p_N₂(p_H₂)³.

Learner claim

At 298 K, a reaction has

K_c = 1.0 × 10⁻⁵. A learner concludes, “The forward reaction must be very slow.” Identify what the value does establish and what it cannot establish.

Interpret Kc

A very small Kc directly describes

Equilibrium compositionReaction speed

View solution step by step

Interpret the magnitude
Method
State that the equilibrium lies far to the left.
Reason
A very small products-over-reactants ratio requires reactants to predominate at equilibrium.
Working
The equilibrium mixture contains mostly reactants.
Set the boundary
Method
Reject the conclusion about speed.
Reason
K_c describes equilibrium composition; reaction rate depends on kinetic factors such as activation energy.
Working
No forward-reaction timescale can be deduced from K_c alone.

Problem

For

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium partial pressures are

p_N₂ = 20.0 kPa,

p_H₂ = 40.0 kPa and

p_NH₃ = 10.0 kPa. Calculate

Kₚ. [3 marks]

Try this before viewing the solution

Expression, substitution and result

View solution step by step

Write the expression
1 mark
Method
Construct Kp before substituting values.
Reason
The balanced coefficients determine the pressure powers.
Working
Kₚ = (p_NH₃)²/p_N₂(p_H₂)³.
Substitute consistently
1 mark
Method
Insert all three equilibrium pressures in kPa.
Reason
A single pressure convention must be used throughout the expression.
Working
Kₚ = (10.0)²/(20.0)(40.0)³ = 100/(1.28 × 10⁶).
Evaluate
1 mark
Method
Report the value to three significant figures.
Reason
The supplied pressures each have three significant figures.
Working
Kₚ = 7.81 × 10⁻⁵ kPa⁻².

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Credit the correct expression, consistent substitution and evaluated result.

Mark points

1 mark: Uses the correct Kp expression and powers.
1 mark: Substitutes all equilibrium partial pressures consistently.
1 mark: Obtains 7.81 × 10^-5 kPa^-2 to three significant figures.

Problem

An equilibrium equation has

K_c = 10. Every coefficient in that equation is multiplied by 2. Determine the equilibrium constant for the scaled equation and explain the transformation.

Try this before viewing the solution

Transformation rule and new Kc

Hints

Hint 1: track expression powers

Doubling every coefficient doubles every power in the equilibrium expression.

Hint 2: transform the full ratio

Raising every term to power 2 raises the original products-over-reactants ratio to power 2.

View solution step by step

Translate the equation change
Method
Raise the original equilibrium expression to power 2.
Reason
Every stoichiometric coefficient, and therefore every exponent, has been doubled.
Working
K_new = K_c².
Evaluate
Method
Square the original constant.
Reason
The equation has been scaled rather than merely rewritten in the same stoichiometric form.
Working
K_new = 10² = 100.

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Hydrogen and iodine only are sealed in a flask at 700 K and react: H₂(g) + I₂(g) ⇌ 2HI(g). Which statement describes the rates as the mixture approaches equilibrium?

The forward rate falls and the reverse rate rises from zero until the two rates are equal; both reactions then continue.
The forward rate falls to zero at equilibrium, and the reverse rate stays at zero.
Both rates rise until the concentrations of H₂, I₂ and HI are equal.
The reverse reaction begins only after all the H₂ and I₂ has reacted.

N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. The equilibrium mixture is compressed to a smaller volume at constant temperature. How does the position of equilibrium change?

It shifts to the left, towards the side with fewer gas molecules, which partly opposes the pressure increase.
It shifts to the right, because the higher pressure makes collisions more frequent.
It does not change, because the temperature is constant.
It shifts to the right, because the forward reaction is endothermic.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹. Which change alters the value of Kc?

Raising the temperature, which decreases Kc.
Raising the pressure, which increases Kc.
Adding a catalyst, which increases Kc.
Removing SO₃ as it forms, which decreases Kc.

2NO(g) + O₂(g) ⇌ 2NO₂(g). At equilibrium a 1.00 dm³ vessel contains 0.20 mol NO, 0.10 mol O₂ and 0.40 mol NO₂. What is the value of Kc?

40 mol⁻¹ dm³
20 mol⁻¹ dm³
0.025 mol dm⁻³
40 mol dm⁻³

1.00 mol CO and 3.00 mol H₂ are sealed in a vessel: CO(g) + 2H₂(g) ⇌ CH₃OH(g). At equilibrium the vessel contains 0.40 mol CH₃OH. What amounts of CO and H₂ are present at equilibrium?

0.60 mol CO and 2.20 mol H₂
0.60 mol CO and 2.60 mol H₂
0.60 mol CO and 2.80 mol H₂
0.40 mol CO and 0.80 mol H₂

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. Why is the Haber process operated at about 450 °C rather than a much lower temperature?

A lower temperature would give a higher equilibrium yield, but equilibrium would be reached too slowly even with the iron catalyst; 450 °C is a compromise between yield and rate.
A higher temperature gives a higher equilibrium yield of ammonia.
The iron catalyst increases the equilibrium yield only when it is hot.
Below 450 °C the forward reaction becomes endothermic.

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A little radioactive iodine is added to an equilibrium mixture of H₂(g), I₂(g) and HI(g) at constant temperature. Later, radioactivity is found in both I₂ and HI, but the concentrations of all three substances are unchanged. What does this show?

The forward and reverse reactions are both still happening, at equal rates.
The position of equilibrium has shifted to the right.
The mixture had not yet reached equilibrium.
The radioactive iodine acts as a catalyst for the reaction.

CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. The temperature of the equilibrium mixture is raised at constant pressure. What happens to the equilibrium amount of CH₃OH?

It decreases, because the position shifts in the endothermic (reverse) direction to absorb the added heat.
It increases, because both reactions become faster.
It is unchanged, because temperature affects only the rate.
It increases, because the position shifts to the side with fewer gas molecules.

N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. Which change increases the value of Kc?

Raising the temperature.
Lowering the pressure at constant temperature.
Removing NO₂ as it forms at constant temperature.
Adding a catalyst.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol dm⁻³. What is the value of Kc?

90 mol⁻¹ dm³
30 mol⁻¹ dm³
90 mol dm⁻³
0.011 mol dm⁻³

0.80 mol H₂S is heated in a sealed vessel: 2H₂S(g) ⇌ 2H₂(g) + S₂(g). At equilibrium 25% of the H₂S has dissociated. What amount of S₂ is present at equilibrium?

0.10 mol
0.20 mol
0.40 mol
0.60 mol

In the Haber process a pressure of about 200 atm is used. A higher pressure would give a higher equilibrium yield of ammonia. Why is a much higher pressure not used?

The extra yield does not justify the higher cost of stronger plant and of the energy for compression, and the greater safety risk.
Higher pressure would decrease Kc, lowering the yield.
Higher pressure would shift the position to the left, towards more gas molecules.
Higher pressure would poison the iron catalyst.

CaCO₃(s) ⇌ CaO(s) + CO₂(g). At 1100 K the equilibrium partial pressure of carbon dioxide is 0.45 atm. What is Kp at this temperature?

0.45 atm
2.2 atm⁻¹
It cannot be found without the amounts of CaCO₃ and CaO.
0.45, with no units

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

The forward and reverse reactions are both still happening, at equal rates.
The position of equilibrium has shifted to the right.
The mixture had not yet reached equilibrium.
The radioactive iodine acts as a catalyst for the reaction.

CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹. The temperature of the equilibrium mixture is raised at constant pressure. What happens to the equilibrium amount of CH₃OH?

It decreases, because the position shifts in the endothermic (reverse) direction to absorb the added heat.
It increases, because both reactions become faster.
It is unchanged, because temperature affects only the rate.
It increases, because the position shifts to the side with fewer gas molecules.

N₂O₄(g) ⇌ 2NO₂(g), ΔH = +58 kJ mol⁻¹. Which change increases the value of Kc?

Raising the temperature.
Lowering the pressure at constant temperature.
Removing NO₂ as it forms at constant temperature.
Adding a catalyst.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol dm⁻³. What is the value of Kc?

90 mol⁻¹ dm³
30 mol⁻¹ dm³
90 mol dm⁻³
0.011 mol dm⁻³

0.80 mol H₂S is heated in a sealed vessel: 2H₂S(g) ⇌ 2H₂(g) + S₂(g). At equilibrium 25% of the H₂S has dissociated. What amount of S₂ is present at equilibrium?

0.10 mol
0.20 mol
0.40 mol
0.60 mol

In the Haber process a pressure of about 200 atm is used. A higher pressure would give a higher equilibrium yield of ammonia. Why is a much higher pressure not used?

The extra yield does not justify the higher cost of stronger plant and of the energy for compression, and the greater safety risk.
Higher pressure would decrease Kc, lowering the yield.
Higher pressure would shift the position to the left, towards more gas molecules.
Higher pressure would poison the iron catalyst.

CaCO₃(s) ⇌ CaO(s) + CO₂(g). At 1100 K the equilibrium partial pressure of carbon dioxide is 0.45 atm. What is Kp at this temperature?

0.45 atm
2.2 atm⁻¹
It cannot be found without the amounts of CaCO₃ and CaO.
0.45, with no units

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Mind Stretchers

Mind stretcher 1Extension

For a reaction with K_c = 10, find K_c for: (a) the reversed equation, and (b) the equation multiplied by 3.

Show Hint

Write the balanced equation first; powers come from coefficients and pure solids or liquids are omitted.

Show Answer

Mark scheme:

Reversed: Kᵣₑᵥ = 1/K = 1/10 = 0.1
Multiplied by 3: K_new = K³ = 10³ = 1000

Mind stretcher 2: Reversing and scaling an equilibriumExtension

Question. An equilibrium equation has K_c = 16. State the new constant when the equation is reversed, then when that reversed equation is divided by two.

Show Hint

Reversal takes the reciprocal; multiplying all coefficients by a factor raises K to that factor.

Show Answer

Reversal gives K_c' = 1/16. Halving every coefficient in the reversed equation gives K_c'' = (1/16)^(1/2) = 1/4.

Continue where you stopped

Equilibrium Constants Kc and Kp: Orientation

Definitions (Must Know)

A. Equilibrium constant, K

B. K_c

C. Kₚ

D. Reaction quotient style (brackets / p)

Key Ideas (What Earns Marks)

Detailed Explanations

A. Writing a K_c expression (workflow)

B. Writing a Kₚ expression (workflow)

C. Interpreting magnitude

D. How K changes when you change the equation

Worked Examples

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Mind Stretchers

Mind stretcher 1Extension

Mind stretcher 2: Reversing and scaling an equilibriumExtension