Equilibrium constant $K_c$ for homogeneous systems

What you'll learn

What $K_c$ means for a reversible reaction at equilibrium.
How to construct a $K_c$ expression from a balanced equation.
How to calculate $K_c$ using equilibrium concentrations in mol dm⁻³.
How concentration, catalysts and temperature affect — or do not affect — $K_c$ .

Starting point: reversible reactions and equilibrium

A reversible reaction is one that can go in both directions. We show this using the reversible arrow, $⇌\rightleftharpoons$ .

At dynamic equilibrium, the forward and reverse reactions are still happening, but at the same rate. This means the concentrations of reactants and products stay constant.

Definition

Dynamic equilibrium

A dynamic equilibrium exists in a closed system when the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products remain constant.

A closed system means no substances can enter or leave. Equilibrium can only be established properly if matter is not escaping.

Homogeneous systems

A species is any chemical particle involved in a reaction, such as a molecule, ion or atom.

Definition

Homogeneous system

A homogeneous system is a reaction mixture in which all the reacting species are in the same physical phase, for example all gases or all dissolved in solution.

For this topic, all the species in the equilibrium expression are in the same phase. Common examples include gas equilibria and solution equilibria.

Concentration notation: square brackets

The concentration of a species X is written as $[X]$ .

Definition

Square bracket notation

$[X]$ means the concentration of species X, measured in mol dm⁻³. In a $K_c$ expression, square brackets always mean equilibrium concentration, not initial concentration.

For example, $[H2][\text{H}_2]$ means the equilibrium concentration of hydrogen.

Common Mistake

Using initial concentrations

The most common $K_c$ error is substituting the starting concentrations into the expression. $K_c$ uses concentrations at equilibrium only.

Building the $K_c$ expression

For a general reversible reaction:

\rightleftharpoons cC + dD

the equilibrium constant is:

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

The products go on the top, reactants go on the bottom, and the powers come from the balanced equation.

Schematic showing a homogeneous equilibrium, the general Kc expression, and an ICE table workflow

Key Idea

The structure of Kc

$K_c$ is built directly from the balanced reversible equation: products over reactants, with each concentration raised to the power of its coefficient.

Example

Writing a Kc expression

For the equilibrium:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

write the expression for $K_c$ .

Identify the products and reactants. The product is $NH3\text{NH}_3$ ; the reactants are $N2\text{N}_2$ and $H2\text{H}_2$ .
Put products over reactants:
$Kc=[NH3][N2][H2]K_c = \frac{[\text{NH}_3]}{[\text{N}_2][\text{H}_2]}$
Apply the powers from the balanced equation. The coefficient of $NH3\text{NH}_3$ is two, and the coefficient of $H2\text{H}_2$ is three:
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

Common Mistake

Forgetting the powers

If the equation contains $2NH32\text{NH}_3$ , the expression contains $[NH3]2[\text{NH}_3]^2$ , not $2[NH3]2[\text{NH}_3]$ .

What the value of Kc tells you

The size of $K_c$ tells you about the position of equilibrium.

If $K_c$ is large, the numerator is relatively large, so the equilibrium mixture contains a high proportion of products.
If $K_c$ is small, the denominator is relatively large, so the equilibrium mixture contains a high proportion of reactants.
If $K_c$ is close to one, there are significant amounts of both reactants and products.

This does not mean the reaction has stopped. It only describes the equilibrium composition at a particular temperature.

Units of Kc

The units of $K_c$ depend on the expression. You work them out by substituting the unit mol dm⁻³ into the $K_c$ expression and simplifying.

Sometimes the concentration units cancel completely. In that case, $K_c$ has no units, often described as dimensionless.

Example

Finding the units of Kc

For:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

the expression is:

Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Replace each concentration with mol dm⁻³:
$dm−3)3\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3}$
Combine the powers in the denominator:
$dm−3)4\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4}$
Subtract the powers:
$dm−3)−2(\text{mol dm}^{-3})^{-2}$
Rewrite with positive powers:
$mol−2\text{dm}^6\ \text{mol}^{-2}$

Tip

Unit shortcut

Compare the total powers on the top and bottom of the $K_c$ expression. If they are the same, the units cancel.

Calculating Kc from equilibrium concentrations

If you are given equilibrium concentrations, the calculation is usually direct:

Write the balanced equation.
Write the $K_c$ expression.
Substitute equilibrium concentrations.
Calculate the value and units.

Example

Calculating Kc from equilibrium concentrations

For:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

at equilibrium:

$dm−3[\text{N}_2] = 0.250\ \text{mol dm}^{-3}$
$dm−3[\text{H}_2] = 0.150\ \text{mol dm}^{-3}$
$dm−3[\text{NH}_3] = 0.300\ \text{mol dm}^{-3}$

Calculate $K_c$ .

Write the expression from the balanced equation:
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
Substitute the equilibrium concentrations:
$Kc=(0.300)2(0.250)(0.150)3K_c = \frac{(0.300)^2}{(0.250)(0.150)^3}$
Calculate the numerator and denominator:
$Kc=0.09000.00084375K_c = \frac{0.0900}{0.00084375}$
Evaluate and give appropriate units:
$mol−2K_c = 107\ \text{dm}^6\ \text{mol}^{-2}$

Tip

Significant figures

Give your final answer to a sensible number of significant figures, usually matching the least precise data given in the question.

When you are given amounts instead of concentrations

Often, you are given amounts in mol and a volume in dm³. Convert amount to concentration using:

dm3\text{concentration} = \frac{\text{amount in mol}}{\text{volume in dm}^3}

You may need to work out the equilibrium amounts first using the stoichiometry of the balanced equation. This is often called an ICE method: Initial, Change, Equilibrium.

Example

Using equilibrium amounts to calculate Kc

Dinitrogen tetroxide decomposes in a sealed flask:

N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

Initially, 0.200 mol of $N2O4\text{N}_2\text{O}_4$ is placed in a 2.00 dm³ flask. At equilibrium, 0.0800 mol of $N2O4\text{N}_2\text{O}_4$ remains. Calculate $K_c$ .

Work out the amount of $N2O4\text{N}_2\text{O}_4$ that reacted:
$0.120\ \text{mol}$
Use the mole ratio. For every one mol of $N2O4\text{N}_2\text{O}_4$ that reacts, two mol of $NO2\text{NO}_2$ form:
$mol\text{amount of NO}_2 = 2 \times 0.120 = 0.240\ \text{mol}$
Convert equilibrium amounts to concentrations using the 2.00 dm³ volume:
$dm−3[\text{N}_2\text{O}_4] = \frac{0.0800}{2.00} = 0.0400\ \text{mol dm}^{-3}$ $dm−3[\text{NO}_2] = \frac{0.240}{2.00} = 0.120\ \text{mol dm}^{-3}$
Substitute into the expression:
$Kc=[NO2]2[N2O4]K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}$ $dm−3K_c = \frac{(0.120)^2}{0.0400} = 0.360\ \text{mol dm}^{-3}$

Common Mistake

Same volume does not always mean ignore volume

For some $K_c$ expressions, volume cancels out, but for others it does not. The safest method is to convert equilibrium amounts into concentrations before substituting.

Changes in concentration and catalysts

At a constant temperature, changing concentration does not change the value of $K_c$ .

If you add more reactant or product, the equilibrium position shifts to reduce the change, according to Le Chatelier’s principle. However, once equilibrium is re-established at the same temperature, the ratio in the $K_c$ expression returns to the same value.

A catalyst also does not change $K_c$ . It speeds up both the forward and reverse reactions, so equilibrium is reached faster, but the equilibrium composition is unchanged.

Key Idea

What can change Kc?

For a given reaction, only temperature changes the value of $K_c$ . Concentration changes and catalysts do not change $K_c$ .

Example

Predicting the effect of adding a reactant

For:

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

extra hydrogen is added at constant temperature. Explain what happens to $K_c$ .

Adding $H2\text{H}_2$ increases the denominator of the reaction quotient temporarily, so the mixture is no longer at equilibrium.
The equilibrium shifts to the right, using up some $H2\text{H}_2$ and $I2\text{I}_2$ and forming more HI.
Because the temperature has not changed, the system settles at a new equilibrium mixture with the same value of $K_c$ .

Temperature and Kc

Temperature is different: it changes the value of $K_c$ .

Use the enthalpy change of the forward reaction to decide what happens.

If the forward reaction is exothermic, increasing temperature favours the reverse reaction, so $K_c$ decreases.
If the forward reaction is endothermic, increasing temperature favours the forward reaction, so $K_c$ increases.

The opposite happens when temperature is decreased.

Example

Predicting the effect of temperature on Kc

For the Haber equilibrium:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

the forward reaction is exothermic. Predict the effect on $K_c$ when temperature is increased.

Since the forward reaction is exothermic, the reverse reaction is endothermic.
Increasing temperature favours the endothermic direction, so equilibrium shifts to the left.
This decreases $[NH3][\text{NH}_3]$ and increases $[N2][\text{N}_2]$ and $[H2][\text{H}_2]$ .
In the expression
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
the numerator decreases and the denominator increases, so $K_c$ decreases.

Practical link: esterification

A common equilibrium used to determine $K_c$ is the reaction between ethanol and ethanoic acid to form ethyl ethanoate and water:

CH3COOH+C2H5OH⇌CH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}

The expression is:

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}]}

A strong acid catalyst may be used to help the mixture reach equilibrium faster. It does not change the value of $K_c$ at the same temperature.

Exam technique

In the exam

Always start by writing the balanced equation and the correct $K_c$ expression before substituting numbers.
Check that you are using equilibrium concentrations in mol dm⁻³, not initial amounts in mol.
For temperature questions, identify whether the forward reaction is exothermic or endothermic, then decide whether products or reactants are favoured.

Self review

Check yourself

For $2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ , can you write the expression for $K_c$ ?
If all concentrations in a $K_c$ expression are doubled at the same temperature, does $K_c$ change?
For an endothermic forward reaction, what happens to $K_c$ when temperature is increased?

What you'll learn

What $K_c$ means for a reversible reaction at equilibrium.
How to construct a $K_c$ expression from a balanced equation.
How to calculate $K_c$ using equilibrium concentrations in mol dm⁻³.
How concentration, catalysts and temperature affect — or do not affect — $K_c$ .

Starting point: reversible reactions and equilibrium

A reversible reaction is one that can go in both directions. We show this using the reversible arrow, $⇌\rightleftharpoons$ .

At dynamic equilibrium, the forward and reverse reactions are still happening, but at the same rate. This means the concentrations of reactants and products stay constant.

Definition

Dynamic equilibrium

A dynamic equilibrium exists in a closed system when the forward and reverse reactions occur at equal rates, so the concentrations of reactants and products remain constant.

A closed system means no substances can enter or leave. Equilibrium can only be established properly if matter is not escaping.

Homogeneous systems

A species is any chemical particle involved in a reaction, such as a molecule, ion or atom.

Definition

Homogeneous system

A homogeneous system is a reaction mixture in which all the reacting species are in the same physical phase, for example all gases or all dissolved in solution.

For this topic, all the species in the equilibrium expression are in the same phase. Common examples include gas equilibria and solution equilibria.

Concentration notation: square brackets

The concentration of a species X is written as $[X]$ .

Definition

Square bracket notation

$[X]$ means the concentration of species X, measured in mol dm⁻³. In a $K_c$ expression, square brackets always mean equilibrium concentration, not initial concentration.

For example, $[H2][\text{H}_2]$ means the equilibrium concentration of hydrogen.

Common Mistake

Using initial concentrations

The most common $K_c$ error is substituting the starting concentrations into the expression. $K_c$ uses concentrations at equilibrium only.

Building the $K_c$ expression

For a general reversible reaction:

\rightleftharpoons cC + dD

the equilibrium constant is:

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

The products go on the top, reactants go on the bottom, and the powers come from the balanced equation.

Schematic showing a homogeneous equilibrium, the general Kc expression, and an ICE table workflow

Key Idea

The structure of Kc

$K_c$ is built directly from the balanced reversible equation: products over reactants, with each concentration raised to the power of its coefficient.

Example

Writing a Kc expression

For the equilibrium:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

write the expression for $K_c$ .

Identify the products and reactants. The product is $NH3\text{NH}_3$ ; the reactants are $N2\text{N}_2$ and $H2\text{H}_2$ .
Put products over reactants:
$Kc=[NH3][N2][H2]K_c = \frac{[\text{NH}_3]}{[\text{N}_2][\text{H}_2]}$
Apply the powers from the balanced equation. The coefficient of $NH3\text{NH}_3$ is two, and the coefficient of $H2\text{H}_2$ is three:
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

Common Mistake

Forgetting the powers

If the equation contains $2NH32\text{NH}_3$ , the expression contains $[NH3]2[\text{NH}_3]^2$ , not $2[NH3]2[\text{NH}_3]$ .

What the value of Kc tells you

The size of $K_c$ tells you about the position of equilibrium.

If $K_c$ is large, the numerator is relatively large, so the equilibrium mixture contains a high proportion of products.
If $K_c$ is small, the denominator is relatively large, so the equilibrium mixture contains a high proportion of reactants.
If $K_c$ is close to one, there are significant amounts of both reactants and products.

This does not mean the reaction has stopped. It only describes the equilibrium composition at a particular temperature.

Units of Kc

The units of $K_c$ depend on the expression. You work them out by substituting the unit mol dm⁻³ into the $K_c$ expression and simplifying.

Sometimes the concentration units cancel completely. In that case, $K_c$ has no units, often described as dimensionless.

Example

Finding the units of Kc

For:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

the expression is:

Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}

Replace each concentration with mol dm⁻³:
$dm−3)3\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3}$
Combine the powers in the denominator:
$dm−3)4\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4}$
Subtract the powers:
$dm−3)−2(\text{mol dm}^{-3})^{-2}$
Rewrite with positive powers:
$mol−2\text{dm}^6\ \text{mol}^{-2}$

Tip

Unit shortcut

Compare the total powers on the top and bottom of the $K_c$ expression. If they are the same, the units cancel.

Calculating Kc from equilibrium concentrations

If you are given equilibrium concentrations, the calculation is usually direct:

Write the balanced equation.
Write the $K_c$ expression.
Substitute equilibrium concentrations.
Calculate the value and units.

Example

Calculating Kc from equilibrium concentrations

For:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

at equilibrium:

$dm−3[\text{N}_2] = 0.250\ \text{mol dm}^{-3}$
$dm−3[\text{H}_2] = 0.150\ \text{mol dm}^{-3}$
$dm−3[\text{NH}_3] = 0.300\ \text{mol dm}^{-3}$

Calculate $K_c$ .

Write the expression from the balanced equation:
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
Substitute the equilibrium concentrations:
$Kc=(0.300)2(0.250)(0.150)3K_c = \frac{(0.300)^2}{(0.250)(0.150)^3}$
Calculate the numerator and denominator:
$Kc=0.09000.00084375K_c = \frac{0.0900}{0.00084375}$
Evaluate and give appropriate units:
$mol−2K_c = 107\ \text{dm}^6\ \text{mol}^{-2}$

Tip

Significant figures

Give your final answer to a sensible number of significant figures, usually matching the least precise data given in the question.

When you are given amounts instead of concentrations

Often, you are given amounts in mol and a volume in dm³. Convert amount to concentration using:

dm3\text{concentration} = \frac{\text{amount in mol}}{\text{volume in dm}^3}

You may need to work out the equilibrium amounts first using the stoichiometry of the balanced equation. This is often called an ICE method: Initial, Change, Equilibrium.

Example

Using equilibrium amounts to calculate Kc

Dinitrogen tetroxide decomposes in a sealed flask:

N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

Initially, 0.200 mol of $N2O4\text{N}_2\text{O}_4$ is placed in a 2.00 dm³ flask. At equilibrium, 0.0800 mol of $N2O4\text{N}_2\text{O}_4$ remains. Calculate $K_c$ .

Work out the amount of $N2O4\text{N}_2\text{O}_4$ that reacted:
$0.120\ \text{mol}$
Use the mole ratio. For every one mol of $N2O4\text{N}_2\text{O}_4$ that reacts, two mol of $NO2\text{NO}_2$ form:
$mol\text{amount of NO}_2 = 2 \times 0.120 = 0.240\ \text{mol}$
Convert equilibrium amounts to concentrations using the 2.00 dm³ volume:
$dm−3[\text{N}_2\text{O}_4] = \frac{0.0800}{2.00} = 0.0400\ \text{mol dm}^{-3}$ $dm−3[\text{NO}_2] = \frac{0.240}{2.00} = 0.120\ \text{mol dm}^{-3}$
Substitute into the expression:
$Kc=[NO2]2[N2O4]K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}$ $dm−3K_c = \frac{(0.120)^2}{0.0400} = 0.360\ \text{mol dm}^{-3}$

Common Mistake

Same volume does not always mean ignore volume

For some $K_c$ expressions, volume cancels out, but for others it does not. The safest method is to convert equilibrium amounts into concentrations before substituting.

Changes in concentration and catalysts

At a constant temperature, changing concentration does not change the value of $K_c$ .

A catalyst also does not change $K_c$ . It speeds up both the forward and reverse reactions, so equilibrium is reached faster, but the equilibrium composition is unchanged.

Key Idea

What can change Kc?

For a given reaction, only temperature changes the value of $K_c$ . Concentration changes and catalysts do not change $K_c$ .

Example

Predicting the effect of adding a reactant

For:

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

extra hydrogen is added at constant temperature. Explain what happens to $K_c$ .

Adding $H2\text{H}_2$ increases the denominator of the reaction quotient temporarily, so the mixture is no longer at equilibrium.
The equilibrium shifts to the right, using up some $H2\text{H}_2$ and $I2\text{I}_2$ and forming more HI.
Because the temperature has not changed, the system settles at a new equilibrium mixture with the same value of $K_c$ .

Temperature and Kc

Temperature is different: it changes the value of $K_c$ .

Use the enthalpy change of the forward reaction to decide what happens.

If the forward reaction is exothermic, increasing temperature favours the reverse reaction, so $K_c$ decreases.
If the forward reaction is endothermic, increasing temperature favours the forward reaction, so $K_c$ increases.

The opposite happens when temperature is decreased.

Example

Predicting the effect of temperature on Kc

For the Haber equilibrium:

N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

the forward reaction is exothermic. Predict the effect on $K_c$ when temperature is increased.

Since the forward reaction is exothermic, the reverse reaction is endothermic.
Increasing temperature favours the endothermic direction, so equilibrium shifts to the left.
This decreases $[NH3][\text{NH}_3]$ and increases $[N2][\text{N}_2]$ and $[H2][\text{H}_2]$ .
In the expression
$Kc=[NH3]2[N2][H2]3K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
the numerator decreases and the denominator increases, so $K_c$ decreases.

Practical link: esterification

A common equilibrium used to determine $K_c$ is the reaction between ethanol and ethanoic acid to form ethyl ethanoate and water:

CH3COOH+C2H5OH⇌CH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}

The expression is:

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}]}

A strong acid catalyst may be used to help the mixture reach equilibrium faster. It does not change the value of $K_c$ at the same temperature.

Exam technique

In the exam

Always start by writing the balanced equation and the correct $K_c$ expression before substituting numbers.
Check that you are using equilibrium concentrations in mol dm⁻³, not initial amounts in mol.
For temperature questions, identify whether the forward reaction is exothermic or endothermic, then decide whether products or reactants are favoured.

Self review

Check yourself

For $2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ , can you write the expression for $K_c$ ?
If all concentrations in a $K_c$ expression are doubled at the same temperature, does $K_c$ change?
For an endothermic forward reaction, what happens to $K_c$ when temperature is increased?

What you'll learn

Starting point: reversible reactions and equilibrium

Homogeneous systems

Concentration notation: square brackets

Building the $K_c$ expression

What the value of Kc tells you

Units of Kc

Calculating Kc from equilibrium concentrations

When you are given amounts instead of concentrations

Changes in concentration and catalysts

Temperature and Kc

Practical link: esterification

Review Equilibrium constant $K_c$ for homogeneous systems by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constant $K_c$ for homogeneous systems

What you'll learn

Starting point: reversible reactions and equilibrium

Homogeneous systems

Concentration notation: square brackets

Building the $K_c$ expression

What the value of Kc tells you

Units of Kc

Calculating Kc from equilibrium concentrations

When you are given amounts instead of concentrations

Changes in concentration and catalysts

Temperature and Kc

Practical link: esterification

Review Equilibrium constant $K_c$ for homogeneous systems by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constant $K_c$ for homogeneous systems

What you'll learn

Starting point: reversible reactions and equilibrium

Homogeneous systems

Concentration notation: square brackets

Building the KcK_cKc​ expression

What the value of Kc tells you

Units of Kc

Calculating Kc from equilibrium concentrations

When you are given amounts instead of concentrations

Changes in concentration and catalysts

Temperature and Kc

Practical link: esterification

Review Equilibrium constant $K_c$ for homogeneous systems by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constant $K_c$ for homogeneous systems Revision Guide

Equilibrium constant $K_c$ for homogeneous systems

What you'll learn

Starting point: reversible reactions and equilibrium

Homogeneous systems

Concentration notation: square brackets

Building the KcK_cKc​ expression

What the value of Kc tells you

Units of Kc

Calculating Kc from equilibrium concentrations

When you are given amounts instead of concentrations

Changes in concentration and catalysts

Temperature and Kc

Practical link: esterification

Review Equilibrium constant $K_c$ for homogeneous systems by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Equilibrium constant $K_c$ for homogeneous systems Revision Guide

Building the $K_c$ expression

Building the $K_c$ expression