Equilibrium constant $K_p$ for homogeneous systems (A-level only)

What you'll learn

How to calculate partial pressures from mole fractions and total pressure.
How to write a $K_p$ expression from a balanced gas-phase equation.
How to calculate $K_p$ using equilibrium data.
How pressure, temperature and catalysts affect equilibrium yield and the value of $K_p$ .

This topic extends the AS idea of equilibrium constants into gas-phase systems. It is A-Level only, so it is assessed in the full A-level papers, not AS.

The equilibrium background

A reversible reaction is one that can proceed in both directions: reactants form products, and products can reform reactants. In a closed system, it may reach dynamic equilibrium.

Definition

Dynamic equilibrium

A dynamic equilibrium is reached when the forward and reverse reactions occur at the same rate, so the amounts of reactants and products remain constant, provided temperature stays constant.

For $K_p$ , we are dealing with gas-phase equilibria.

Definition

Homogeneous gas equilibrium

A homogeneous equilibrium is one in which all reactants and products are in the same physical phase. In this topic, every species in the equilibrium expression is a gas, shown by the state symbol (g).

Partial pressures

In a mixture of gases, each gas contributes to the total pressure. That contribution is called its partial pressure.

Definition

Mole fraction and partial pressure

The mole fraction of gas $i$ is $xi=nintotalx_i = \frac{n_i}{n_{\text{total}}}$ . Its partial pressure is $pi=xiptotalp_i = x_i p_{\text{total}}$ , where $ptotalp_{\text{total}}$ is the total pressure of the gas mixture.

The mole fraction has no units because it is a fraction of amounts in mol. Partial pressure has the same unit as total pressure, usually kPa in A-level calculations.

The diagram summarises the route from equilibrium moles to the $K_p$ expression.

Gas-phase equilibrium partial pressures and Kp workflow

Example

Calculating partial pressures

A gas mixture at equilibrium contains 0.200 mol of N₂, 0.600 mol of H₂ and 0.400 mol of NH₃. The total pressure is 200 kPa. Calculate the partial pressure of each gas.

Find the total amount of gas: $moln_{\text{total}} = 0.200 + 0.600 + 0.400 = 1.200\ \text{mol}$ .
Calculate each mole fraction: $xN2=0.2001.200=0.167x_{\text{N}_2} = \frac{0.200}{1.200} = 0.167$ , $xH2=0.6001.200=0.500x_{\text{H}_2} = \frac{0.600}{1.200} = 0.500$ , and $xNH3=0.4001.200=0.333x_{\text{NH}_3} = \frac{0.400}{1.200} = 0.333$ .
Multiply each mole fraction by the total pressure: $kPap(\text{N}_2) = 0.167 \times 200 = 33.3\ \text{kPa}$ , $kPap(\text{H}_2) = 0.500 \times 200 = 100\ \text{kPa}$ , and $kPap(\text{NH}_3) = 0.333 \times 200 = 66.7\ \text{kPa}$ .
Check the partial pressures add to the total pressure: $200\ \text{kPa}$ .

Constructing a $K_p$ expression

$K_p$ is the equilibrium constant calculated using equilibrium partial pressures.

Definition

$K_p$ is the equilibrium constant for a gas-phase system at constant temperature, calculated from the partial pressures of gases at equilibrium.

For the general reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

the expression is:

Kp=(pC)c(pD)d(pA)a(pB)bK_p = \frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}

The powers come from the stoichiometric coefficients in the balanced equation.

Key Idea

Products over reactants

For $K_p$ , put equilibrium partial pressures of gaseous products on the top, gaseous reactants on the bottom, and raise each partial pressure to its balancing number.

Example

Writing a Kp expression and units

For the Haber equilibrium, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), write the $K_p$ expression and state its units if pressures are measured in kPa.

Put the product partial pressure on the top and reactant partial pressures on the bottom: ammonia is the product, while nitrogen and hydrogen are reactants.
Apply the balancing numbers as powers: $Kp=p(NH3)2p(N2)p(H2)3K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)p(\text{H}_2)^3}$ .
Work out the units from the expression: $kPa2kPa⋅kPa3=kPa−2\frac{\text{kPa}^2}{\text{kPa}\cdot\text{kPa}^3} = \text{kPa}^{-2}$ .

Common Mistake

Forgetting that the equation matters

$K_p$ belongs to the exact balanced equation written. If you reverse the equation, the new $K_p$ is the reciprocal; if you multiply the whole equation by 2, the powers in the expression double.

Calculating $K_p$ from equilibrium data

A typical calculation follows the same sequence every time:

Use the balanced equation to find equilibrium moles.
Add them to get total moles of gas.
Convert each amount into a mole fraction.
Convert mole fractions into partial pressures.
Substitute into the $K_p$ expression, including units.

Common Mistake

Using initial amounts

The $K_p$ expression uses equilibrium partial pressures, not initial amounts or initial pressures. If the question gives starting amounts, you must first use the reaction stoichiometry to find the amounts present at equilibrium.

Example

Calculating Kp from a degree of dissociation

Dinitrogen tetroxide dissociates according to N₂O₄(g) ⇌ 2NO₂(g). Initially, 1.00 mol of N₂O₄ is placed in a vessel. At equilibrium, 40.0% has dissociated. The total pressure at equilibrium is 125 kPa. Calculate $K_p$ .

Convert the percentage dissociation into reacting moles: 40.0% of 1.00 mol is 0.400 mol, so equilibrium moles are 0.600 mol N₂O₄ and $\times 0.400 = 0.800\ \text{mol}$ NO₂.
Find the total moles at equilibrium: $moln_{\text{total}} = 0.600 + 0.800 = 1.400\ \text{mol}$ .
Calculate mole fractions: $xN2O4=0.6001.400=0.429x_{\text{N}_2\text{O}_4} = \frac{0.600}{1.400} = 0.429$ and $xNO2=0.8001.400=0.571x_{\text{NO}_2} = \frac{0.800}{1.400} = 0.571$ .
Calculate partial pressures: $kPap(\text{N}_2\text{O}_4) = 0.429 \times 125 = 53.6\ \text{kPa}$ and $kPap(\text{NO}_2) = 0.571 \times 125 = 71.4\ \text{kPa}$ .
Substitute into the expression: $kPaK_p = \frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \frac{71.4^2}{53.6} = 95.2\ \text{kPa}$ .

What $K_p$ tells you about yield

The position of equilibrium describes the relative amounts of reactants and products at equilibrium.

A large value of $K_p$ usually means products are favoured. A small value of $K_p$ usually means reactants are favoured. Be careful: because $K_p$ units depend on the equation, do not compare unrelated reactions just by their numerical values.

Effect of pressure on equilibrium position

For gas equilibria, changing pressure can change the equilibrium mixture.

Increasing pressure by decreasing volume favours the side with fewer gas molecules.
Decreasing pressure favours the side with more gas molecules.
If both sides have the same number of gas molecules, pressure has no effect on the equilibrium position.

At constant temperature, changing pressure does not change the value of $K_p$ . Instead, the mixture shifts until the partial pressures fit the same $K_p$ value again.

Common Mistake

Pressure changes

The usual pressure rule assumes the pressure is changed by changing the volume of the container. Adding an inert gas at constant volume does not change the partial pressures of the reacting gases.

Effect of temperature on equilibrium position and $K_p$

Temperature is different: it changes both the equilibrium position and the value of $K_p$ .

For an exothermic forward reaction, increasing temperature shifts equilibrium to the reactants and decreases $K_p$ .

For an endothermic forward reaction, increasing temperature shifts equilibrium to the products and increases $K_p$ .

Key Idea

Only temperature changes Kp

For a given reaction, $K_p$ is constant only at a fixed temperature. Pressure and concentration changes can alter the equilibrium position, but temperature changes alter the value of $K_p$ itself.

Effect of a catalyst

A catalyst increases the rate of reaction by providing an alternative route with lower activation energy. In a reversible reaction, it speeds up both the forward and reverse reactions.

So a catalyst:

makes equilibrium reached faster
does not change the position of equilibrium
does not change $K_p$
does not increase equilibrium yield

Example

Predicting changes for the Haber equilibrium

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the forward reaction is exothermic. Predict the effects of increasing pressure, increasing temperature and adding a catalyst.

Compare gas moles for pressure: the left side has 4 mol of gas and the right side has 2 mol of gas, so increasing pressure shifts equilibrium to the right and increases the equilibrium yield of NH₃.
Use the enthalpy change for temperature: the forward reaction is exothermic, so increasing temperature favours the reverse endothermic direction, decreasing the NH₃ yield and decreasing $K_p$ .
Apply the catalyst rule: a catalyst helps the system reach equilibrium faster, but the equilibrium composition and $K_p$ are unchanged.

Tip

Sanity check

If a question asks whether $K_p$ changes, your first thought should be temperature. If the temperature is unchanged, $K_p$ is unchanged.

Exam technique

In the exam

Always write the balanced equation first, then build the $K_p$ expression directly from it.
Use equilibrium amounts to calculate mole fractions and partial pressures; do not substitute initial amounts into $K_p$ .
Keep pressure units consistent throughout, then derive the units of $K_p$ from the expression and round to a suitable number of significant figures.

Self review

Check yourself

Can you calculate partial pressures from equilibrium moles and total pressure?
Can you write the $K_p$ expression, including units, for any balanced gas-phase equation?
Can you explain why temperature can change $K_p$ , but a catalyst cannot?

What you'll learn

How to calculate partial pressures from mole fractions and total pressure.
How to write a $K_p$ expression from a balanced gas-phase equation.
How to calculate $K_p$ using equilibrium data.
How pressure, temperature and catalysts affect equilibrium yield and the value of $K_p$ .

This topic extends the AS idea of equilibrium constants into gas-phase systems. It is A-Level only, so it is assessed in the full A-level papers, not AS.

The equilibrium background

A reversible reaction is one that can proceed in both directions: reactants form products, and products can reform reactants. In a closed system, it may reach dynamic equilibrium.

Definition

Dynamic equilibrium

A dynamic equilibrium is reached when the forward and reverse reactions occur at the same rate, so the amounts of reactants and products remain constant, provided temperature stays constant.

For $K_p$ , we are dealing with gas-phase equilibria.

Definition

Homogeneous gas equilibrium

Partial pressures

In a mixture of gases, each gas contributes to the total pressure. That contribution is called its partial pressure.

Definition

Mole fraction and partial pressure

The mole fraction has no units because it is a fraction of amounts in mol. Partial pressure has the same unit as total pressure, usually kPa in A-level calculations.

The diagram summarises the route from equilibrium moles to the $K_p$ expression.

Gas-phase equilibrium partial pressures and Kp workflow

Example

Calculating partial pressures

A gas mixture at equilibrium contains 0.200 mol of N₂, 0.600 mol of H₂ and 0.400 mol of NH₃. The total pressure is 200 kPa. Calculate the partial pressure of each gas.

Find the total amount of gas: $moln_{\text{total}} = 0.200 + 0.600 + 0.400 = 1.200\ \text{mol}$ .
Calculate each mole fraction: $xN2=0.2001.200=0.167x_{\text{N}_2} = \frac{0.200}{1.200} = 0.167$ , $xH2=0.6001.200=0.500x_{\text{H}_2} = \frac{0.600}{1.200} = 0.500$ , and $xNH3=0.4001.200=0.333x_{\text{NH}_3} = \frac{0.400}{1.200} = 0.333$ .
Multiply each mole fraction by the total pressure: $kPap(\text{N}_2) = 0.167 \times 200 = 33.3\ \text{kPa}$ , $kPap(\text{H}_2) = 0.500 \times 200 = 100\ \text{kPa}$ , and $kPap(\text{NH}_3) = 0.333 \times 200 = 66.7\ \text{kPa}$ .
Check the partial pressures add to the total pressure: $200\ \text{kPa}$ .

Constructing a $K_p$ expression

$K_p$ is the equilibrium constant calculated using equilibrium partial pressures.

Definition

$K_p$ is the equilibrium constant for a gas-phase system at constant temperature, calculated from the partial pressures of gases at equilibrium.

For the general reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

the expression is:

Kp=(pC)c(pD)d(pA)a(pB)bK_p = \frac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}

The powers come from the stoichiometric coefficients in the balanced equation.

Key Idea

Products over reactants

For $K_p$ , put equilibrium partial pressures of gaseous products on the top, gaseous reactants on the bottom, and raise each partial pressure to its balancing number.

Example

Writing a Kp expression and units

For the Haber equilibrium, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), write the $K_p$ expression and state its units if pressures are measured in kPa.

Put the product partial pressure on the top and reactant partial pressures on the bottom: ammonia is the product, while nitrogen and hydrogen are reactants.
Apply the balancing numbers as powers: $Kp=p(NH3)2p(N2)p(H2)3K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)p(\text{H}_2)^3}$ .
Work out the units from the expression: $kPa2kPa⋅kPa3=kPa−2\frac{\text{kPa}^2}{\text{kPa}\cdot\text{kPa}^3} = \text{kPa}^{-2}$ .

Common Mistake

Forgetting that the equation matters

$K_p$ belongs to the exact balanced equation written. If you reverse the equation, the new $K_p$ is the reciprocal; if you multiply the whole equation by 2, the powers in the expression double.

Calculating $K_p$ from equilibrium data

A typical calculation follows the same sequence every time:

Use the balanced equation to find equilibrium moles.
Add them to get total moles of gas.
Convert each amount into a mole fraction.
Convert mole fractions into partial pressures.
Substitute into the $K_p$ expression, including units.

Common Mistake

Using initial amounts

Example

Calculating Kp from a degree of dissociation

Convert the percentage dissociation into reacting moles: 40.0% of 1.00 mol is 0.400 mol, so equilibrium moles are 0.600 mol N₂O₄ and $\times 0.400 = 0.800\ \text{mol}$ NO₂.
Find the total moles at equilibrium: $moln_{\text{total}} = 0.600 + 0.800 = 1.400\ \text{mol}$ .
Calculate mole fractions: $xN2O4=0.6001.400=0.429x_{\text{N}_2\text{O}_4} = \frac{0.600}{1.400} = 0.429$ and $xNO2=0.8001.400=0.571x_{\text{NO}_2} = \frac{0.800}{1.400} = 0.571$ .
Calculate partial pressures: $kPap(\text{N}_2\text{O}_4) = 0.429 \times 125 = 53.6\ \text{kPa}$ and $kPap(\text{NO}_2) = 0.571 \times 125 = 71.4\ \text{kPa}$ .
Substitute into the expression: $kPaK_p = \frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \frac{71.4^2}{53.6} = 95.2\ \text{kPa}$ .

What $K_p$ tells you about yield

The position of equilibrium describes the relative amounts of reactants and products at equilibrium.

Effect of pressure on equilibrium position

For gas equilibria, changing pressure can change the equilibrium mixture.

Increasing pressure by decreasing volume favours the side with fewer gas molecules.
Decreasing pressure favours the side with more gas molecules.
If both sides have the same number of gas molecules, pressure has no effect on the equilibrium position.

At constant temperature, changing pressure does not change the value of $K_p$ . Instead, the mixture shifts until the partial pressures fit the same $K_p$ value again.

Common Mistake

Pressure changes

The usual pressure rule assumes the pressure is changed by changing the volume of the container. Adding an inert gas at constant volume does not change the partial pressures of the reacting gases.

Effect of temperature on equilibrium position and $K_p$

Temperature is different: it changes both the equilibrium position and the value of $K_p$ .

For an exothermic forward reaction, increasing temperature shifts equilibrium to the reactants and decreases $K_p$ .

For an endothermic forward reaction, increasing temperature shifts equilibrium to the products and increases $K_p$ .

Key Idea

Only temperature changes Kp

For a given reaction, $K_p$ is constant only at a fixed temperature. Pressure and concentration changes can alter the equilibrium position, but temperature changes alter the value of $K_p$ itself.

Effect of a catalyst

A catalyst increases the rate of reaction by providing an alternative route with lower activation energy. In a reversible reaction, it speeds up both the forward and reverse reactions.

So a catalyst:

makes equilibrium reached faster
does not change the position of equilibrium
does not change $K_p$
does not increase equilibrium yield

Example

Predicting changes for the Haber equilibrium

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the forward reaction is exothermic. Predict the effects of increasing pressure, increasing temperature and adding a catalyst.

Compare gas moles for pressure: the left side has 4 mol of gas and the right side has 2 mol of gas, so increasing pressure shifts equilibrium to the right and increases the equilibrium yield of NH₃.
Use the enthalpy change for temperature: the forward reaction is exothermic, so increasing temperature favours the reverse endothermic direction, decreasing the NH₃ yield and decreasing $K_p$ .
Apply the catalyst rule: a catalyst helps the system reach equilibrium faster, but the equilibrium composition and $K_p$ are unchanged.

Tip

Sanity check

If a question asks whether $K_p$ changes, your first thought should be temperature. If the temperature is unchanged, $K_p$ is unchanged.

Exam technique

In the exam

Always write the balanced equation first, then build the $K_p$ expression directly from it.
Use equilibrium amounts to calculate mole fractions and partial pressures; do not substitute initial amounts into $K_p$ .
Keep pressure units consistent throughout, then derive the units of $K_p$ from the expression and round to a suitable number of significant figures.

Self review

Check yourself

Can you calculate partial pressures from equilibrium moles and total pressure?
Can you write the $K_p$ expression, including units, for any balanced gas-phase equation?
Can you explain why temperature can change $K_p$ , but a catalyst cannot?

What you'll learn

The equilibrium background

Partial pressures

Constructing a $K_p$ expression

Calculating $K_p$ from equilibrium data

What $K_p$ tells you about yield

Effect of pressure on equilibrium position

Effect of temperature on equilibrium position and $K_p$

Effect of a catalyst

Recap questions

Review Equilibrium constant $K_p$ for homogeneous systems (A-level only) by teaching Genie

Remember key concepts with flashcards

Equilibrium constant $K_p$ for homogeneous systems (A-level only)

What you'll learn

The equilibrium background

Partial pressures

Constructing a $K_p$ expression

Calculating $K_p$ from equilibrium data

What $K_p$ tells you about yield

Effect of pressure on equilibrium position

Effect of temperature on equilibrium position and $K_p$

Effect of a catalyst

Recap questions

Review Equilibrium constant $K_p$ for homogeneous systems (A-level only) by teaching Genie

Remember key concepts with flashcards

Equilibrium constant $K_p$ for homogeneous systems (A-level only)

What you'll learn

The equilibrium background

Partial pressures

Constructing a KpK_pKp​ expression

Calculating KpK_pKp​ from equilibrium data

What KpK_pKp​ tells you about yield

Effect of pressure on equilibrium position

Effect of temperature on equilibrium position and KpK_pKp​

Effect of a catalyst

Recap questions

Review Equilibrium constant $K_p$ for homogeneous systems (A-level only) by teaching Genie

Remember key concepts with flashcards

Equilibrium constant $K_p$ for homogeneous systems (A-level only) Revision Guide

Equilibrium constant $K_p$ for homogeneous systems (A-level only)

What you'll learn

The equilibrium background

Partial pressures

Constructing a KpK_pKp​ expression

Calculating KpK_pKp​ from equilibrium data

What KpK_pKp​ tells you about yield

Effect of pressure on equilibrium position

Effect of temperature on equilibrium position and KpK_pKp​

Effect of a catalyst

Recap questions

Review Equilibrium constant $K_p$ for homogeneous systems (A-level only) by teaching Genie

Remember key concepts with flashcards

Equilibrium constant $K_p$ for homogeneous systems (A-level only) Revision Guide

Constructing a $K_p$ expression

Calculating $K_p$ from equilibrium data

What $K_p$ tells you about yield

Effect of temperature on equilibrium position and $K_p$

Constructing a $K_p$ expression

Calculating $K_p$ from equilibrium data

What $K_p$ tells you about yield

Effect of temperature on equilibrium position and $K_p$