Equilibrium constants

What you'll learn

How to write and use expressions for $K_c$ and $K_p$ .
How to calculate equilibrium constants and unknown equilibrium quantities.
What the size of an equilibrium constant tells you about equilibrium position.
How temperature affects $K_c$ and $K_p$ , and how an esterification practical can determine $K_c$ .

1. Starting point: dynamic equilibrium

A reversible reaction is a reaction that can proceed in both directions, shown using ⇌. In a closed system, the forward and reverse reactions can reach dynamic equilibrium.

Definition

Dynamic equilibrium

A dynamic equilibrium is reached when the forward and reverse reactions are still happening, but at equal rates, so the concentrations of reactants and products stay constant.

The diagram summarises the key ideas: equilibrium is dynamic, the equilibrium constant has products over reactants, and the size of $K$ shows which side is favoured.

Diagram showing dynamic equilibrium, equilibrium constant expressions, magnitude of K and temperature effects

Common Mistake

Equal rates, not equal amounts

At equilibrium, the rates of the forward and reverse reactions are equal. The concentrations of reactants and products are usually not equal.

2. $K_c$ : using equilibrium concentrations

$K_c$ is the equilibrium constant written using equilibrium concentrations, usually in mol dm⁻³.

For a general homogeneous reaction:

\rightleftharpoons cC + dD

the expression is:

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Square brackets mean equilibrium concentration, not initial concentration.

Definition

Equilibrium constant Kc

$K_c$ is the value of the products’ equilibrium concentrations divided by the reactants’ equilibrium concentrations, each raised to the power of its balancing number.

Pure solids are not included in equilibrium expressions. Pure liquids are usually omitted in heterogeneous equilibria, but in the esterification practical, the reacting liquid substances are treated as variable concentrations and are included.

Example

Writing a Kc expression

For the Haber equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Identify the products and reactants from the balanced equation: NH₃ is the product; N₂ and H₂ are reactants.
Put products over reactants, using square brackets for equilibrium concentrations:
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
Work out the units from the powers: top power is 2, bottom total power is 4, so the overall unit is $dm−3)−2\left(\text{mol dm}^{-3}\right)^{-2}$ , often written as dm⁶ mol⁻².

Common Mistake

Using initial concentrations

Equilibrium constants must be calculated from equilibrium concentrations or partial pressures. Initial amounts are only a starting point for working out the equilibrium amounts.

3. Calculating $K_c$ from equilibrium data

You often use the amount–concentration relationship:

n = c V

so:

\frac{n}{V}

where volume must be in dm³ if concentration is in mol dm⁻³.

Example

Calculating Kc from equilibrium amounts

At a fixed temperature, hydrogen and iodine reach equilibrium:

H₂(g) + I₂(g) ⇌ 2HI(g)

In a 1.00 dm³ vessel at equilibrium, there are 0.100 mol H₂, 0.100 mol I₂ and 0.780 mol HI.

Convert each amount into concentration using $\frac{n}{V}$ :
$[H2]=0.100,[I2]=0.100,[HI]=0.780[\text{H}_2] = 0.100,\quad [\text{I}_2] = 0.100,\quad [\text{HI}] = 0.780$
all in mol dm⁻³.
Write the expression:
$Kc=[HI]2[H2][I2]K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$
Substitute the equilibrium concentrations:
$Kc=0.78020.100×0.100=60.8K_c = \frac{0.780^2}{0.100 \times 0.100} = 60.8$
Check the units: the total concentration power on top and bottom is 2, so the units cancel. Here, $K_c = 60.8$ with no units.

4. Finding equilibrium quantities

When the equilibrium amount is unknown, use a change variable such as $x$ . This is sometimes called an ICE method: initial, change, equilibrium.

Example

Finding equilibrium concentrations from Kc

For H₂(g) + I₂(g) ⇌ 2HI(g), $K_c = 49.0$ at a particular temperature. Initially, $[H2]=0.500[\text{H}_2] = 0.500$ mol dm⁻³ and $[I2]=0.500[\text{I}_2] = 0.500$ mol dm⁻³, with no HI present.

Let $x$ be the concentration of H₂ and I₂ that reacts. The equilibrium concentrations are:
$[H2]=0.500−x,[I2]=0.500−x,[HI]=2x[\text{H}_2] = 0.500 - x,\quad [\text{I}_2] = 0.500 - x,\quad [\text{HI}] = 2x$
Substitute into the $K_c$ expression:
$\frac{(2x)^2}{(0.500 - x)(0.500 - x)}$
Take the square root to simplify:
$\frac{2x}{0.500 - x}$
Solve for $x$ :
$7.00(0.500−x)=2x3.50−7.00x=2x3.50=9.00xx=0.389\begin{aligned} 7.00(0.500 - x) &= 2x \\ 3.50 - 7.00x &= 2x \\ 3.50 &= 9.00x \\ x &= 0.389 \end{aligned}$
Calculate the equilibrium concentrations:
$dm−3[\text{H}_2] = [\text{I}_2] = 0.111\ \text{mol dm}^{-3},\quad [\text{HI}] = 0.778\ \text{mol dm}^{-3}$

Tip

Sanity check

Your value of $x$ cannot be larger than the starting concentration of a limiting reactant. If it is, something has gone wrong in the algebra.

5. $K_p$ : using partial pressures

For gaseous equilibria, you may use $K_p$ , which is written using partial pressures.

Definition

Partial pressure

The partial pressure of a gas is the pressure it would exert if it alone occupied the container. In a gas mixture, $pi=χiptotalp_i = \chi_i p_{\text{total}}$ , where $χi\chi_i$ is the mole fraction of gas $i$ .

For:

\rightleftharpoons 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}

Only gaseous species appear in a $K_p$ expression.

Example

Calculating Kp from mole fractions

For N₂O₄(g) ⇌ 2NO₂(g), an equilibrium mixture contains 0.400 mol N₂O₄ and 0.600 mol NO₂. The total pressure is 200 kPa.

Calculate the total number of moles and mole fractions:
$ntotal=0.400+0.600=1.000n_{\text{total}} = 0.400 + 0.600 = 1.000$ $χN2O4=0.400,χNO2=0.600\chi_{\text{N}_2\text{O}_4} = 0.400,\quad \chi_{\text{NO}_2} = 0.600$
Calculate partial pressures using $pi=χiptotalp_i = \chi_i p_{\text{total}}$ :
$kPap_{\text{N}_2\text{O}_4} = 0.400 \times 200 = 80.0\ \text{kPa}$ $kPap_{\text{NO}_2} = 0.600 \times 200 = 120\ \text{kPa}$
Substitute into the expression:
$kPaK_p = \frac{(p_{\text{NO}_2})^2}{p_{\text{N}_2\text{O}_4}} = \frac{120^2}{80.0} = 180\ \text{kPa}$

Common Mistake

Using total pressure in Kp

$K_p$ uses partial pressures, not the total pressure directly. Use mole fractions first if the mixture composition is given.

6. What the size of $K$ means

The position of equilibrium describes whether the equilibrium mixture contains mostly reactants or mostly products.

Key Idea

Magnitude of K

If $\gg 1$ , products are favoured and equilibrium lies to the right. If $\ll 1$ , reactants are favoured and equilibrium lies to the left. If $K$ is close to 1, appreciable amounts of both reactants and products are present.

A large $K$ does not mean the reaction is fast. Rate and equilibrium position are different ideas.

Example

Interpreting the size of K

Two reversible reactions have these equilibrium constants at the same temperature: reaction A has $Kc=2.0×10−4K_c = 2.0 \times 10^{-4}$ and reaction B has $Kc=1.5×103K_c = 1.5 \times 10^3$ .

Compare each value with 1: reaction A has $Kc≪1K_c \ll 1$ , while reaction B has $Kc≫1K_c \gg 1$ .
For reaction A, the denominator of the $K_c$ expression is relatively large, so reactants are favoured.
For reaction B, the numerator is relatively large, so products are favoured.

7. Temperature changes $K_c$ and $K_p$

At a fixed temperature, $K_c$ and $K_p$ are constant for a given balanced equation. Changing concentration or pressure can change the equilibrium composition, but it does not change the value of $K$ if temperature stays the same.

Temperature is different because it changes the relative favourability of the forward and reverse reactions.

For an exothermic forward reaction, increasing temperature decreases $K$ .
For an endothermic forward reaction, increasing temperature increases $K$ .

Example

Predicting the temperature effect on Kp

For the Haber process:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), $mol−1\Delta H = -92\ \text{kJ mol}^{-1}$

The forward reaction is exothermic because $ΔH\Delta H$ is negative.
Increasing temperature favours the endothermic direction, which is the reverse reaction.
The equilibrium mixture contains a smaller proportion of NH₃, so the numerator of the $K_p$ expression becomes smaller relative to the denominator.
Therefore $K_p$ decreases when temperature increases.

Tip

Catalysts and K

A catalyst helps equilibrium be reached faster, but it increases the rates of the forward and reverse reactions equally. It does not change $K_c$ , $K_p$ or the equilibrium position.

8. Practical: determining $K_c$ for esterification

A common practical is the equilibrium between ethanol and ethanoic acid:

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

This is catalysed by acid, often concentrated sulfuric acid.

A typical method:

Measure known initial amounts of ethanoic acid and ethanol into a stoppered flask.
Add a small amount of acid catalyst and keep the mixture at constant temperature until equilibrium is reached.
Take a measured sample and titrate the remaining ethanoic acid with standard sodium hydroxide.
Correct for any acid catalyst that is also neutralised.
Use the remaining ethanoic acid to calculate the equilibrium amounts of all species, then calculate $K_c$ .

Example

Calculating Kc from esterification titration data

An equilibrium mixture was prepared from 0.100 mol ethanoic acid and 0.100 mol ethanol, with no ester or water initially. After equilibrium, the mixture was analysed. A corrected titration showed that 0.0330 mol ethanoic acid remained in the whole mixture.

Calculate the amount of ethanoic acid that reacted:
$moln_{\text{reacted}} = 0.100 - 0.0330 = 0.0670\ \text{mol}$
Use the 1:1:1:1 stoichiometry to find equilibrium amounts:
$nCH3COOH=0.0330,nC2H5OH=0.0330n_{\text{CH}_3\text{COOH}} = 0.0330,\quad n_{\text{C}_2\text{H}_5\text{OH}} = 0.0330$ $nester=0.0670,nwater=0.0670n_{\text{ester}} = 0.0670,\quad n_{\text{water}} = 0.0670$
Write the expression:
$Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$
Because the powers on top and bottom are equal, the same volume factor cancels, so moles can be used directly:
$Kc=0.0670×0.06700.0330×0.0330=4.12K_c = \frac{0.0670 \times 0.0670}{0.0330 \times 0.0330} = 4.12$

Common Mistake

Practical accuracy

The mixture must be at constant temperature and genuinely at equilibrium. Ethanol and ethyl ethanoate are flammable, and concentrated sulfuric acid is corrosive, so use eye protection, avoid naked flames and handle acids carefully.

Exam technique

In the exam

Start with the balanced equation, then write the equilibrium expression with products over reactants.
Use equilibrium concentrations for $K_c$ and partial pressures for $K_p$ ; convert cm³ to dm³ where needed.
If an equilibrium amount is unknown, define $x$ and build the equilibrium quantities from the stoichiometry.
For temperature questions, identify whether the forward reaction is exothermic or endothermic, then link the shift to how $K$ changes.
Remember: pressure, concentration and catalysts do not change $K$ at constant temperature.

Self review

Check yourself

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what is the expression for $K_c$ and what are its units?
How do you calculate partial pressures from mole amounts and total pressure?
For an exothermic forward reaction, what happens to $K$ when temperature is increased?

What you'll learn

How to write and use expressions for $K_c$ and $K_p$ .
How to calculate equilibrium constants and unknown equilibrium quantities.
What the size of an equilibrium constant tells you about equilibrium position.
How temperature affects $K_c$ and $K_p$ , and how an esterification practical can determine $K_c$ .

1. Starting point: dynamic equilibrium

A reversible reaction is a reaction that can proceed in both directions, shown using ⇌. In a closed system, the forward and reverse reactions can reach dynamic equilibrium.

Definition

Dynamic equilibrium

A dynamic equilibrium is reached when the forward and reverse reactions are still happening, but at equal rates, so the concentrations of reactants and products stay constant.

The diagram summarises the key ideas: equilibrium is dynamic, the equilibrium constant has products over reactants, and the size of $K$ shows which side is favoured.

Diagram showing dynamic equilibrium, equilibrium constant expressions, magnitude of K and temperature effects

Common Mistake

Equal rates, not equal amounts

At equilibrium, the rates of the forward and reverse reactions are equal. The concentrations of reactants and products are usually not equal.

2. $K_c$ : using equilibrium concentrations

$K_c$ is the equilibrium constant written using equilibrium concentrations, usually in mol dm⁻³.

For a general homogeneous reaction:

\rightleftharpoons cC + dD

the expression is:

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Square brackets mean equilibrium concentration, not initial concentration.

Definition

Equilibrium constant Kc

$K_c$ is the value of the products’ equilibrium concentrations divided by the reactants’ equilibrium concentrations, each raised to the power of its balancing number.

Example

Writing a Kc expression

For the Haber equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Identify the products and reactants from the balanced equation: NH₃ is the product; N₂ and H₂ are reactants.
Put products over reactants, using square brackets for equilibrium concentrations:
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
Work out the units from the powers: top power is 2, bottom total power is 4, so the overall unit is $dm−3)−2\left(\text{mol dm}^{-3}\right)^{-2}$ , often written as dm⁶ mol⁻².

Common Mistake

Using initial concentrations

Equilibrium constants must be calculated from equilibrium concentrations or partial pressures. Initial amounts are only a starting point for working out the equilibrium amounts.

3. Calculating $K_c$ from equilibrium data

You often use the amount–concentration relationship:

n = c V

so:

\frac{n}{V}

where volume must be in dm³ if concentration is in mol dm⁻³.

Example

Calculating Kc from equilibrium amounts

At a fixed temperature, hydrogen and iodine reach equilibrium:

H₂(g) + I₂(g) ⇌ 2HI(g)

In a 1.00 dm³ vessel at equilibrium, there are 0.100 mol H₂, 0.100 mol I₂ and 0.780 mol HI.

Convert each amount into concentration using $\frac{n}{V}$ :
$[H2]=0.100,[I2]=0.100,[HI]=0.780[\text{H}_2] = 0.100,\quad [\text{I}_2] = 0.100,\quad [\text{HI}] = 0.780$
all in mol dm⁻³.
Write the expression:
$Kc=[HI]2[H2][I2]K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$
Substitute the equilibrium concentrations:
$Kc=0.78020.100×0.100=60.8K_c = \frac{0.780^2}{0.100 \times 0.100} = 60.8$
Check the units: the total concentration power on top and bottom is 2, so the units cancel. Here, $K_c = 60.8$ with no units.

4. Finding equilibrium quantities

When the equilibrium amount is unknown, use a change variable such as $x$ . This is sometimes called an ICE method: initial, change, equilibrium.

Example

Finding equilibrium concentrations from Kc

Let $x$ be the concentration of H₂ and I₂ that reacts. The equilibrium concentrations are:
$[H2]=0.500−x,[I2]=0.500−x,[HI]=2x[\text{H}_2] = 0.500 - x,\quad [\text{I}_2] = 0.500 - x,\quad [\text{HI}] = 2x$
Substitute into the $K_c$ expression:
$\frac{(2x)^2}{(0.500 - x)(0.500 - x)}$
Take the square root to simplify:
$\frac{2x}{0.500 - x}$
Solve for $x$ :
$7.00(0.500−x)=2x3.50−7.00x=2x3.50=9.00xx=0.389\begin{aligned} 7.00(0.500 - x) &= 2x \\ 3.50 - 7.00x &= 2x \\ 3.50 &= 9.00x \\ x &= 0.389 \end{aligned}$
Calculate the equilibrium concentrations:
$dm−3[\text{H}_2] = [\text{I}_2] = 0.111\ \text{mol dm}^{-3},\quad [\text{HI}] = 0.778\ \text{mol dm}^{-3}$

Tip

Sanity check

Your value of $x$ cannot be larger than the starting concentration of a limiting reactant. If it is, something has gone wrong in the algebra.

5. $K_p$ : using partial pressures

For gaseous equilibria, you may use $K_p$ , which is written using partial pressures.

Definition

Partial pressure

For:

\rightleftharpoons 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}

Only gaseous species appear in a $K_p$ expression.

Example

Calculating Kp from mole fractions

For N₂O₄(g) ⇌ 2NO₂(g), an equilibrium mixture contains 0.400 mol N₂O₄ and 0.600 mol NO₂. The total pressure is 200 kPa.

Calculate the total number of moles and mole fractions:
$ntotal=0.400+0.600=1.000n_{\text{total}} = 0.400 + 0.600 = 1.000$ $χN2O4=0.400,χNO2=0.600\chi_{\text{N}_2\text{O}_4} = 0.400,\quad \chi_{\text{NO}_2} = 0.600$
Calculate partial pressures using $pi=χiptotalp_i = \chi_i p_{\text{total}}$ :
$kPap_{\text{N}_2\text{O}_4} = 0.400 \times 200 = 80.0\ \text{kPa}$ $kPap_{\text{NO}_2} = 0.600 \times 200 = 120\ \text{kPa}$
Substitute into the expression:
$kPaK_p = \frac{(p_{\text{NO}_2})^2}{p_{\text{N}_2\text{O}_4}} = \frac{120^2}{80.0} = 180\ \text{kPa}$

Common Mistake

Using total pressure in Kp

$K_p$ uses partial pressures, not the total pressure directly. Use mole fractions first if the mixture composition is given.

6. What the size of $K$ means

The position of equilibrium describes whether the equilibrium mixture contains mostly reactants or mostly products.

Key Idea

Magnitude of K

A large $K$ does not mean the reaction is fast. Rate and equilibrium position are different ideas.

Example

Interpreting the size of K

Two reversible reactions have these equilibrium constants at the same temperature: reaction A has $Kc=2.0×10−4K_c = 2.0 \times 10^{-4}$ and reaction B has $Kc=1.5×103K_c = 1.5 \times 10^3$ .

Compare each value with 1: reaction A has $Kc≪1K_c \ll 1$ , while reaction B has $Kc≫1K_c \gg 1$ .
For reaction A, the denominator of the $K_c$ expression is relatively large, so reactants are favoured.
For reaction B, the numerator is relatively large, so products are favoured.

7. Temperature changes $K_c$ and $K_p$

Temperature is different because it changes the relative favourability of the forward and reverse reactions.

For an exothermic forward reaction, increasing temperature decreases $K$ .
For an endothermic forward reaction, increasing temperature increases $K$ .

Example

Predicting the temperature effect on Kp

For the Haber process:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), $mol−1\Delta H = -92\ \text{kJ mol}^{-1}$

The forward reaction is exothermic because $ΔH\Delta H$ is negative.
Increasing temperature favours the endothermic direction, which is the reverse reaction.
The equilibrium mixture contains a smaller proportion of NH₃, so the numerator of the $K_p$ expression becomes smaller relative to the denominator.
Therefore $K_p$ decreases when temperature increases.

Tip

Catalysts and K

A catalyst helps equilibrium be reached faster, but it increases the rates of the forward and reverse reactions equally. It does not change $K_c$ , $K_p$ or the equilibrium position.

8. Practical: determining $K_c$ for esterification

A common practical is the equilibrium between ethanol and ethanoic acid:

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

This is catalysed by acid, often concentrated sulfuric acid.

A typical method:

Measure known initial amounts of ethanoic acid and ethanol into a stoppered flask.
Add a small amount of acid catalyst and keep the mixture at constant temperature until equilibrium is reached.
Take a measured sample and titrate the remaining ethanoic acid with standard sodium hydroxide.
Correct for any acid catalyst that is also neutralised.
Use the remaining ethanoic acid to calculate the equilibrium amounts of all species, then calculate $K_c$ .

Example

Calculating Kc from esterification titration data

Calculate the amount of ethanoic acid that reacted:
$moln_{\text{reacted}} = 0.100 - 0.0330 = 0.0670\ \text{mol}$
Use the 1:1:1:1 stoichiometry to find equilibrium amounts:
$nCH3COOH=0.0330,nC2H5OH=0.0330n_{\text{CH}_3\text{COOH}} = 0.0330,\quad n_{\text{C}_2\text{H}_5\text{OH}} = 0.0330$ $nester=0.0670,nwater=0.0670n_{\text{ester}} = 0.0670,\quad n_{\text{water}} = 0.0670$
Write the expression:
$Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]K_c = \frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}$
Because the powers on top and bottom are equal, the same volume factor cancels, so moles can be used directly:
$Kc=0.0670×0.06700.0330×0.0330=4.12K_c = \frac{0.0670 \times 0.0670}{0.0330 \times 0.0330} = 4.12$

Common Mistake

Practical accuracy

Exam technique

In the exam

Start with the balanced equation, then write the equilibrium expression with products over reactants.
Use equilibrium concentrations for $K_c$ and partial pressures for $K_p$ ; convert cm³ to dm³ where needed.
If an equilibrium amount is unknown, define $x$ and build the equilibrium quantities from the stoichiometry.
For temperature questions, identify whether the forward reaction is exothermic or endothermic, then link the shift to how $K$ changes.
Remember: pressure, concentration and catalysts do not change $K$ at constant temperature.

Self review

Check yourself

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what is the expression for $K_c$ and what are its units?
How do you calculate partial pressures from mole amounts and total pressure?
For an exothermic forward reaction, what happens to $K$ when temperature is increased?

What you'll learn

1. Starting point: dynamic equilibrium

2. $K_c$ : using equilibrium concentrations

3. Calculating $K_c$ from equilibrium data

4. Finding equilibrium quantities

5. $K_p$ : using partial pressures

6. What the size of $K$ means

7. Temperature changes $K_c$ and $K_p$

8. Practical: determining $K_c$ for esterification

Review Equilibrium constants by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constants

What you'll learn

1. Starting point: dynamic equilibrium

2. $K_c$ : using equilibrium concentrations

3. Calculating $K_c$ from equilibrium data

4. Finding equilibrium quantities

5. $K_p$ : using partial pressures

6. What the size of $K$ means

7. Temperature changes $K_c$ and $K_p$

8. Practical: determining $K_c$ for esterification

Review Equilibrium constants by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constants

What you'll learn

1. Starting point: dynamic equilibrium

2. KcK_cKc​: using equilibrium concentrations

3. Calculating KcK_cKc​ from equilibrium data

4. Finding equilibrium quantities

5. KpK_pKp​: using partial pressures

6. What the size of KKK means

7. Temperature changes KcK_cKc​ and KpK_pKp​

8. Practical: determining KcK_cKc​ for esterification

Review Equilibrium constants by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constants Revision Guide

Equilibrium constants

What you'll learn

1. Starting point: dynamic equilibrium

2. KcK_cKc​: using equilibrium concentrations

3. Calculating KcK_cKc​ from equilibrium data

4. Finding equilibrium quantities

5. KpK_pKp​: using partial pressures

6. What the size of KKK means

7. Temperature changes KcK_cKc​ and KpK_pKp​

8. Practical: determining KcK_cKc​ for esterification

Review Equilibrium constants by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constants Revision Guide

2. $K_c$ : using equilibrium concentrations

3. Calculating $K_c$ from equilibrium data

5. $K_p$ : using partial pressures

6. What the size of $K$ means

7. Temperature changes $K_c$ and $K_p$

8. Practical: determining $K_c$ for esterification

2. $K_c$ : using equilibrium concentrations

3. Calculating $K_c$ from equilibrium data

5. $K_p$ : using partial pressures

6. What the size of $K$ means

7. Temperature changes $K_c$ and $K_p$

8. Practical: determining $K_c$ for esterification