Equilibrium Composition Calculations

Learn and apply Equilibrium Composition Calculations for H2 Chemistry 9476.

GCE A-Level H2 Chemistry 9476-2027

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Equilibrium Composition Calculations: Orientation

Equilibrium composition questions are “ICE tables under pressure”: set up the changes with coefficients, substitute into K_c or Kₚ, solve for x, then sanity-check the result against the size of K.

Anchor this lesson with Equilibrium Constants (Kc, Kp) and the Chemical Equilibria hub so you can move between concept and calculation questions.

What this page is really testing

Can you set up an ICE table with correct coefficient-based changes?
Can you translate that table into the correct K_c/Kₚ expression?
Can you defend your final value with a quick physical sanity check?

Definitions (Must Know)

A. Equilibrium composition

Equilibrium composition means the amounts/concentrations/partial pressures of species at equilibrium.

B. ICE table

An ICE table tracks:

Initial values
Change (using x and coefficients)
Equilibrium values

Key Ideas (What Earns Marks)

Use an ICE (Initial–Change–Equilibrium) table.
Only make an approximation if you can justify it (and check it afterwards).
Sanity checks: concentrations can’t be negative; results should fit the size of K.
This syllabus does not require solving quadratic equations in equilibrium calculations (your setup should lead to linear algebra or a valid approximation).

If you assume a-x ≈ a, check x/a < 0.05 afterwards.

Example A ⇌ B with Kc = 0.200: initial [A] = 0.500, [B] = 0.000; equilibrium [A] ≈ 0.417 and [B] ≈ 0.0833.

Data table

Species	Initial	Equilibrium
A	0.5	0.4
B	0	0.1

Detailed Explanations

A. ICE table method (workflow)

Write the balanced equation.
Decide whether you are using concentrations (K_c) or partial pressures (Kₚ).
Fill in the ICE table using x and coefficients (e.g. -2x if the coefficient is 2).
Write the K_c or Kₚ expression and substitute the equilibrium row.
Solve for x, then calculate the equilibrium composition asked for.

B. Approximation sanity check

If you assume a-x ≈ a, check:

x/a < 0.05

If the check fails, do not force an approximation. In this syllabus, questions are set so that:

either the approximation is valid, or
the algebra simplifies without needing a quadratic (e.g. 1:1 stoichiometry, or one equilibrium value is given).

C. Worked method (full-mark layout)

Use this layout in your script:

Balanced equation.
ICE table with symbols and units.
K expression with powers from coefficients.
Substitution line (equilibrium row only).
Solve for x.
Convert x into the exact quantity asked (concentration/partial pressure/amount).
Sanity check: sign, magnitude, and compatibility with K.

Worked Examples

Problem

For

A(g) ⇌ B(g), initially

[A] = 0.50 mol dm⁻³ and

[B] = 0. At equilibrium,

[B] = x. Write

K_c in terms of

Study the worked solution

Build the change row
Method
Use the 1:1 coefficients to pair a decrease of x in A with an increase of x in B.
Reason
Reaction progress changes species in their stoichiometric ratio.
Working
Δ[A] = -x; Δ[B] = +x.
Write the equilibrium row
Method
Add each change to its initial concentration.
Reason
Only equilibrium values may be inserted into K_c.
Working
[A]_eq = 0.50-x; [B]_eq = x.
Substitute symbolically
Method
Insert the equilibrium row into products over reactants.
Reason
Both coefficients are 1.
Working
K_c = [B]/[A] = x/(0.50-x).

Problem

For

A(g) ⇌ B(g) at 298 K,

K_c = 0.200. Initially,

[A] = 0.500 mol dm⁻³ and

[B] = 0. Calculate the equilibrium concentration of B.

Try this before viewing the solution

ICE rows, Kc substitution and result

Hints

Hint 1: write equilibrium values

For the 1:1 equation, use

[A]_eq = 0.500-x and

[B]_eq = x.

Hint 2: form one equation

Set

x/(0.500-x) equal to 0.200 and solve without rounding early.

View solution step by step

Set up ICE
Method
Apply equal and opposite changes to A and B.
Reason
The balanced ratio is 1:1.
Working
[A]_eq = 0.500-x; [B]_eq = x.
Substitute into Kc
Method
Use the equilibrium concentrations in the expression.
Reason
K_c = [B]/[A] for this equation.
Working
x/(0.500-x) = 0.200.
Solve and report
Method
Rearrange the linear equation and identify x as B’s equilibrium concentration.
Reason
B starts at zero and increases by x.
Working
1.200x = 0.100, so [B]_eq = x = 0.0833 mol dm⁻³.

Proposed method

For

A(g) ⇌ B(g),

K_c = 0.0100, initially

[A] = 1.00 mol dm⁻³ and

[B] = 0. A learner uses

1.00-x ≈ 1.00. Estimate

[B]_eq and decide whether the approximation is justified.

Test the approximation

x divided by initial A

View solution step by step

Estimate x
Method
Replace 1.00-x by 1.00 in the denominator.
Reason
The proposed approximation treats the reactant decrease as small.
Working
0.0100 ≈ x/1.00, so x ≈ 0.0100 mol dm⁻³.
Test the assumption
Method
Compare the estimated change with the initial concentration.
Reason
The stated small-change check is x/a < 0.05.
Working
x/1.00 = 0.0100 < 0.05, so the approximation is valid and [B]_eq ≈ 0.0100 mol dm⁻³.

Problem

For

H₂(g) + I₂(g) ⇌ 2HI(g), equilibrium concentrations are

[H₂] = 0.20,

[I₂] = 0.20 and

[HI] = 0.60 mol dm⁻³. Calculate

K_c. [3 marks]

Try this before viewing the solution

Expression, substitution and Kc

View solution step by step

Write the expression
1 mark
Method
Use the coefficient 2 as the power on HI.
Reason
Coefficient powers are part of the Kc definition.
Working
K_c = [HI]²/[H₂][I₂].
Substitute equilibrium data
1 mark
Method
Insert the three measured equilibrium concentrations.
Reason
No ICE deduction is needed when every required equilibrium value is supplied.
Working
K_c = (0.60)²/(0.20)(0.20) = 0.36/0.040.
Evaluate
1 mark
Method
Complete the ratio.
Reason
The overall concentration powers cancel for this equation.
Working
K_c = 9.0.

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 powered expression, correct substitution and final value.

Mark points

1 mark: Writes the correct Kc expression including [HI] squared.
1 mark: Substitutes equilibrium concentrations correctly.
1 mark: Obtains Kc = 9.0.

Problem

For

A(g) ⇌ B(g) at 298 K,

Kₚ = 0.500. Initially,

p_A = 80.0 kPa and

p_B = 0. Calculate

p_B at equilibrium.

Try this before viewing the solution

Pressure ICE setup, Kp equation and result

Hints

Hint 1: change the representation

Use the same 1:1 ICE logic, but write partial pressures rather than concentrations.

Hint 2: solve for the product pressure

Set

x/(80.0-x) = 0.500 and remember that

x is

p_B.

View solution step by step

Build the pressure ICE row
Method
Decrease A by x kPa and increase B by x kPa.
Reason
The gaseous stoichiometric ratio is 1:1.
Working
p_(A,eq) = 80.0-x; p_(B,eq) = x.
Apply Kp
Method
Substitute equilibrium partial pressures into Kₚ = p_B/p_A.
Reason
Kp uses partial pressures of the gaseous species.
Working
x/(80.0-x) = 0.500.
Solve and identify the target
Method
Solve the linear equation and report the product pressure.
Reason
B starts at zero, so its equilibrium partial pressure equals x.
Working
1.50x = 40.0, so p_B = 26.7 kPa.

Next step

Try the impossible-root Mind Stretcher independently.

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

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.

Practise

Work through questions with marking and feedback as you learn.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

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

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

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

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

Connect This To

Equilibrium Constants (Kc, Kp) for writing expressions correctly before substitution.
Dynamic Equilibrium and Le Chatelier for predicting direction before calculating numbers.
Haber Process (Case Study) for applying the same equilibrium logic to industrial conditions.

Mind stretcher 1Extension

Explain why a “sanity check” is needed even if your algebra is correct.

Show Hint

Build the change row from one variable and the balanced coefficients before inserting equilibrium values into K.

Show Answer

Mark scheme:

Approximations can give numerically plausible but physically impossible results.
Algebra mistakes can yield negative concentrations or values inconsistent with the size of K.
A sanity check confirms the result is physically meaningful and matches the expected equilibrium position.

Mind stretcher 2: Rejecting an impossible rootExtension

Question. For an ICE table with initial [A] = 0.40 mol dm⁻³ and change -x, algebra gives x = 0.52 or x = 0.18 mol dm⁻³. Select the physical root and justify it.

Show Hint

No equilibrium concentration may be negative.

Show Answer

x = 0.52 would give [A]_eq = -0.12 mol dm⁻³ and is impossible. The physical root is x = 0.18, giving [A]_eq = 0.22 mol dm⁻³.

Continue where you stopped

Equilibrium Composition Calculations: Orientation

What this page is really testing

Definitions (Must Know)

A. Equilibrium composition

B. ICE table

Key Ideas (What Earns Marks)

Detailed Explanations

A. ICE table method (workflow)

B. Approximation sanity check

C. Worked method (full-mark layout)

Worked Examples

Starting check complete

Your answers

Practice complete

Your answers

Practice complete

Your answers

Progress check complete

Your answers

Progress check complete

Your answers

Review complete

Your answers

Mind Stretchers

Connect This To

Mind stretcher 1Extension

Mind stretcher 2: Rejecting an impossible rootExtension