Kc and Kp equilibria

What you'll learn

How equilibrium constants describe the position of a reversible reaction at a fixed temperature.
How to write expressions for $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ from balanced equations.
How to calculate $KcK_\mathrm{c}$ from equilibrium concentrations and $KpK_\mathrm{p}$ from partial pressures.
How changes in temperature, pressure and concentration affect equilibrium constants.

Starting point: dynamic equilibrium

A reversible reaction is one that can go in both directions. We write it using the equilibrium arrow, ⇌.

A reaction can only reach equilibrium in a closed system, where substances cannot escape. For gases, this usually means a sealed container.

Definition

Dynamic equilibrium

A dynamic equilibrium is reached when the forward and reverse reactions are both still happening, but at the same rate, so the concentrations of reactants and products remain constant.

“Constant” does not mean “equal”. At equilibrium, there might be far more product than reactant, or far more reactant than product.

Key Idea

Equilibrium constants need equilibrium values

$KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ are calculated using the composition of the mixture at equilibrium, not the starting amounts.

Why use an equilibrium constant?

For a reversible reaction, the equilibrium mixture depends on how far the reaction goes towards products or reactants. An equilibrium constant gives a numerical measure of this position at a particular temperature.

For the general equilibrium:

aA+bB⇌cC+dDa\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}

the coefficients in the balanced equation become powers in the equilibrium expression.

Diagram comparing Kc and Kp equilibrium expressions

$KcK_\mathrm{c}$ : equilibrium constant using concentrations

Definition

$KcK_\mathrm{c}$ is the equilibrium constant written using the equilibrium concentrations of substances, usually in mol dm⁻³.

For:

aA+bB⇌cC+dDa\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}

the expression is:

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

Square brackets mean concentration, so $[A][\mathrm{A}]$ means the equilibrium concentration of A.

What goes into a $KcK_\mathrm{c}$ expression?

Include species whose concentration can change significantly in the equilibrium mixture. At A-Level, you usually:

include gases, aqueous substances and solutes
omit pure solids
omit pure liquids

For example, in:

CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3(s)}\rightleftharpoons \mathrm{CaO(s)}+\mathrm{CO_2(g)}

the solids are omitted, so:

Kc=[CO2]K_\mathrm{c}=[\mathrm{CO_2}]

Common Mistake

Including solids and pure liquids

Do not put pure solids or pure liquids into $KcK_\mathrm{c}$ or $KpK_\mathrm{p}$ expressions. Their concentration is effectively constant, so it is absorbed into the value of the equilibrium constant.

Example

Writing a Kc expression and units

For the equilibrium:

2SO2(g)+O2(g)⇌2SO3(g)2\mathrm{SO_2(g)}+\mathrm{O_2(g)}\rightleftharpoons2\mathrm{SO_3(g)}

All species are gases, so all three appear in the $KcK_\mathrm{c}$ expression.
Put products over reactants, with powers from the balanced equation:

Kc=[SO3]2[SO2]2[O2]K_\mathrm{c}=\frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]}

Work out the units by replacing each concentration with mol dm⁻³:

dm3\frac{(\mathrm{mol\,dm^{-3}})^2}{(\mathrm{mol\,dm^{-3}})^3} =(\mathrm{mol\,dm^{-3}})^{-1} =\mathrm{mol}^{-1}\,\mathrm{dm}^{3}

Calculating $KcK_\mathrm{c}$ from equilibrium amounts

Exam questions often give amounts in mol and a volume in dm³. You must convert amounts into concentrations before substituting into $KcK_\mathrm{c}$ :

c=nVc=\frac{n}{V}

where $c$ is concentration in mol dm⁻³, $n$ is amount in mol, and $V$ is volume in dm³.

If the question gives starting amounts and one equilibrium amount, use the balanced equation to build an amount-change-equilibrium calculation.

Example

Calculating Kc from equilibrium amounts

Hydrogen and iodine react in a sealed 2.00 dm³ container:

H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g)}+\mathrm{I_2(g)}\rightleftharpoons2\mathrm{HI(g)}

Initially, 1.00 mol of hydrogen and 1.00 mol of iodine are present. At equilibrium, there are 1.20 mol of hydrogen iodide. Calculate $KcK_\mathrm{c}$ .

Use the stoichiometry to find how much reactant has been used. Forming 1.20 mol of HI uses half as much H₂ and I₂ because the ratio is 1 : 1 : 2:

moln(\mathrm{H_2\ used})=n(\mathrm{I_2\ used})=\frac{1.20}{2}=0.600\ \mathrm{mol}

Find the equilibrium amounts:

mol\begin{aligned} n(\mathrm{H_2})&=1.00-0.600=0.400\ \mathrm{mol}\\ n(\mathrm{I_2})&=1.00-0.600=0.400\ \mathrm{mol}\\ n(\mathrm{HI})&=1.20\ \mathrm{mol} \end{aligned}

Convert amounts into concentrations by dividing by 2.00 dm³:

dm−3\begin{aligned} [\mathrm{H_2}]&=\frac{0.400}{2.00}=0.200\ \mathrm{mol\,dm^{-3}}\\ [\mathrm{I_2}]&=\frac{0.400}{2.00}=0.200\ \mathrm{mol\,dm^{-3}}\\ [\mathrm{HI}]&=\frac{1.20}{2.00}=0.600\ \mathrm{mol\,dm^{-3}} \end{aligned}

Substitute into the expression:

Kc=[HI]2[H2][I2]=0.60020.200×0.200=9.00K_\mathrm{c}=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} =\frac{0.600^2}{0.200\times0.200}=9.00

Check the units. The total powers on the top and bottom are both 2, so the units cancel. $Kc=9.00K_\mathrm{c}=9.00$ with no units.

Tip

Use a clear amount table

For harder questions, set up rows for initial amount, change, and equilibrium amount. Do stoichiometry in mol first, then convert to concentrations at the end.

$KpK_\mathrm{p}$ : equilibrium constant using partial pressures

$KpK_\mathrm{p}$ is used for equilibria involving gases. Instead of concentrations, it uses partial pressures.

Definition

Partial pressure

The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the container at the same temperature. For gas $i$ , $pi=xiPtotalp_i=x_iP_\text{total}$ , where $xi=nintotalx_i=\frac{n_i}{n_\text{total}}$ is its mole fraction.

For:

aA(g)+bB(g)⇌cC(g)+dD(g)a\mathrm{A(g)}+b\mathrm{B(g)}\rightleftharpoons c\mathrm{C(g)}+d\mathrm{D(g)}

the expression is:

Kp=pCcpDdpAapBbK_\mathrm{p}=\frac{p_{\mathrm{C}}^c p_{\mathrm{D}}^d}{p_{\mathrm{A}}^a p_{\mathrm{B}}^b}

Use one pressure unit consistently, usually kPa or Pa.

Example

Calculating Kp from mole fractions

For the equilibrium:

N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g)}\rightleftharpoons2\mathrm{NO_2(g)}

an equilibrium mixture contains 0.300 mol of N₂O₄ and 0.700 mol of NO₂ at a total pressure of 200 kPa. Calculate $KpK_\mathrm{p}$ .

Calculate the total amount and mole fractions:

molxN2O4=0.3001.000=0.300xNO2=0.7001.000=0.700\begin{aligned} n_\text{total}&=0.300+0.700=1.000\ \mathrm{mol}\\ x_{\mathrm{N_2O_4}}&=\frac{0.300}{1.000}=0.300\\ x_{\mathrm{NO_2}}&=\frac{0.700}{1.000}=0.700 \end{aligned}

Convert mole fractions into partial pressures:

kPa\begin{aligned} p_{\mathrm{N_2O_4}}&=0.300\times200=60.0\ \mathrm{kPa}\\ p_{\mathrm{NO_2}}&=0.700\times200=140\ \mathrm{kPa} \end{aligned}

Substitute into the $KpK_\mathrm{p}$ expression:

kPaK_\mathrm{p}=\frac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}} =\frac{140^2}{60.0} =3.27\times10^2\ \mathrm{kPa}

Check the units from the pressure powers: pressure squared divided by pressure leaves kPa.

Common Mistake

Mixing pressure units

If you use kPa for one partial pressure, use kPa for all of them. Do not mix Pa and kPa in the same $KpK_\mathrm{p}$ calculation.

What does the size of $K$ tell you?

A large equilibrium constant means products are favoured at equilibrium. A small equilibrium constant means reactants are favoured.

If $K$ is much greater than 1, the equilibrium lies mainly to the right.
If $K$ is much less than 1, the equilibrium lies mainly to the left.
If $K$ is close to 1, appreciable amounts of both reactants and products are present.

Be careful: this is a qualitative interpretation. The exact amounts still depend on the balanced equation and starting conditions.

What changes $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ ?

At a fixed temperature, the value of $KcK_\mathrm{c}$ or $KpK_\mathrm{p}$ is constant for a particular reaction.

Changing concentration or pressure can change the equilibrium composition, but the mixture shifts until the same equilibrium constant is restored. A catalyst does not change the equilibrium constant; it only helps the system reach equilibrium faster.

Key Idea

Temperature is the only condition that changes K

For a given reaction, $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ change only when temperature changes.

For temperature changes, use the enthalpy change of the forward reaction:

If the forward reaction is exothermic, increasing temperature decreases $K$ .
If the forward reaction is endothermic, increasing temperature increases $K$ .

Example

Predicting how temperature changes Kp

For the Haber equilibrium:

N2(g)+3H2(g)⇌2NH3(g)ΔH<0\mathrm{N_2(g)}+3\mathrm{H_2(g)}\rightleftharpoons2\mathrm{NH_3(g)} \qquad \Delta H<0

predict what happens to $KpK_\mathrm{p}$ when temperature is increased.

The forward reaction is exothermic, so the reverse reaction is endothermic.
Increasing temperature favours the endothermic direction, so equilibrium shifts left towards N₂ and H₂.
In the expression

Kp=pNH32pN2pH23K_\mathrm{p}=\frac{p_{\mathrm{NH_3}}^2}{p_{\mathrm{N_2}}p_{\mathrm{H_2}}^3}

the numerator becomes smaller relative to the denominator, so $KpK_\mathrm{p}$ decreases.

Common Mistake

Converting between Kc and Kp

Only use $Kp=Kc(RT)ΔnK_\mathrm{p}=K_\mathrm{c}(RT)^{\Delta n}$ if the question asks for it and the units are consistent: pressure in Pa, concentration in mol m⁻³, temperature in K, and $Δn\Delta n$ is gaseous product moles minus gaseous reactant moles.

Exam technique

In the exam

Start from the balanced equation, then write the equilibrium expression before substituting numbers.
Use equilibrium concentrations for $KcK_\mathrm{c}$ and equilibrium partial pressures for $KpK_\mathrm{p}$ .
Convert mol to mol dm⁻³ for $KcK_\mathrm{c}$ , and convert mole fractions to partial pressures for $KpK_\mathrm{p}$ .
Check units from the powers in the expression, and state “no units” only when they cancel.
For changes in conditions, remember: temperature changes $K$ ; pressure, concentration and catalysts do not.

Self review

Check yourself

Can you write the $KcK_\mathrm{c}$ expression for $2NO2(g)⇌N2O4(g)2\mathrm{NO_2(g)}\rightleftharpoons\mathrm{N_2O_4(g)}$ and work out its units?
If a gas mixture has mole fraction 0.25 of CO₂ at 400 kPa total pressure, what is the partial pressure of CO₂?
For an endothermic forward reaction, what happens to the value of $K$ when temperature is increased?

What you'll learn

How equilibrium constants describe the position of a reversible reaction at a fixed temperature.
How to write expressions for $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ from balanced equations.
How to calculate $KcK_\mathrm{c}$ from equilibrium concentrations and $KpK_\mathrm{p}$ from partial pressures.
How changes in temperature, pressure and concentration affect equilibrium constants.

Starting point: dynamic equilibrium

A reversible reaction is one that can go in both directions. We write it using the equilibrium arrow, ⇌.

A reaction can only reach equilibrium in a closed system, where substances cannot escape. For gases, this usually means a sealed container.

Definition

Dynamic equilibrium

A dynamic equilibrium is reached when the forward and reverse reactions are both still happening, but at the same rate, so the concentrations of reactants and products remain constant.

“Constant” does not mean “equal”. At equilibrium, there might be far more product than reactant, or far more reactant than product.

Key Idea

Equilibrium constants need equilibrium values

$KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ are calculated using the composition of the mixture at equilibrium, not the starting amounts.

Why use an equilibrium constant?

For the general equilibrium:

aA+bB⇌cC+dDa\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}

the coefficients in the balanced equation become powers in the equilibrium expression.

Diagram comparing Kc and Kp equilibrium expressions

$KcK_\mathrm{c}$ : equilibrium constant using concentrations

Definition

$KcK_\mathrm{c}$ is the equilibrium constant written using the equilibrium concentrations of substances, usually in mol dm⁻³.

For:

aA+bB⇌cC+dDa\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}

the expression is:

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

Square brackets mean concentration, so $[A][\mathrm{A}]$ means the equilibrium concentration of A.

What goes into a $KcK_\mathrm{c}$ expression?

Include species whose concentration can change significantly in the equilibrium mixture. At A-Level, you usually:

include gases, aqueous substances and solutes
omit pure solids
omit pure liquids

For example, in:

CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3(s)}\rightleftharpoons \mathrm{CaO(s)}+\mathrm{CO_2(g)}

the solids are omitted, so:

Kc=[CO2]K_\mathrm{c}=[\mathrm{CO_2}]

Common Mistake

Including solids and pure liquids

Example

Writing a Kc expression and units

For the equilibrium:

2SO2(g)+O2(g)⇌2SO3(g)2\mathrm{SO_2(g)}+\mathrm{O_2(g)}\rightleftharpoons2\mathrm{SO_3(g)}

All species are gases, so all three appear in the $KcK_\mathrm{c}$ expression.
Put products over reactants, with powers from the balanced equation:

Kc=[SO3]2[SO2]2[O2]K_\mathrm{c}=\frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]}

Work out the units by replacing each concentration with mol dm⁻³:

dm3\frac{(\mathrm{mol\,dm^{-3}})^2}{(\mathrm{mol\,dm^{-3}})^3} =(\mathrm{mol\,dm^{-3}})^{-1} =\mathrm{mol}^{-1}\,\mathrm{dm}^{3}

Calculating $KcK_\mathrm{c}$ from equilibrium amounts

Exam questions often give amounts in mol and a volume in dm³. You must convert amounts into concentrations before substituting into $KcK_\mathrm{c}$ :

c=nVc=\frac{n}{V}

where $c$ is concentration in mol dm⁻³, $n$ is amount in mol, and $V$ is volume in dm³.

If the question gives starting amounts and one equilibrium amount, use the balanced equation to build an amount-change-equilibrium calculation.

Example

Calculating Kc from equilibrium amounts

Hydrogen and iodine react in a sealed 2.00 dm³ container:

H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g)}+\mathrm{I_2(g)}\rightleftharpoons2\mathrm{HI(g)}

Initially, 1.00 mol of hydrogen and 1.00 mol of iodine are present. At equilibrium, there are 1.20 mol of hydrogen iodide. Calculate $KcK_\mathrm{c}$ .

Use the stoichiometry to find how much reactant has been used. Forming 1.20 mol of HI uses half as much H₂ and I₂ because the ratio is 1 : 1 : 2:

moln(\mathrm{H_2\ used})=n(\mathrm{I_2\ used})=\frac{1.20}{2}=0.600\ \mathrm{mol}

Find the equilibrium amounts:

mol\begin{aligned} n(\mathrm{H_2})&=1.00-0.600=0.400\ \mathrm{mol}\\ n(\mathrm{I_2})&=1.00-0.600=0.400\ \mathrm{mol}\\ n(\mathrm{HI})&=1.20\ \mathrm{mol} \end{aligned}

Convert amounts into concentrations by dividing by 2.00 dm³:

dm−3\begin{aligned} [\mathrm{H_2}]&=\frac{0.400}{2.00}=0.200\ \mathrm{mol\,dm^{-3}}\\ [\mathrm{I_2}]&=\frac{0.400}{2.00}=0.200\ \mathrm{mol\,dm^{-3}}\\ [\mathrm{HI}]&=\frac{1.20}{2.00}=0.600\ \mathrm{mol\,dm^{-3}} \end{aligned}

Substitute into the expression:

Kc=[HI]2[H2][I2]=0.60020.200×0.200=9.00K_\mathrm{c}=\frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} =\frac{0.600^2}{0.200\times0.200}=9.00

Check the units. The total powers on the top and bottom are both 2, so the units cancel. $Kc=9.00K_\mathrm{c}=9.00$ with no units.

Tip

Use a clear amount table

For harder questions, set up rows for initial amount, change, and equilibrium amount. Do stoichiometry in mol first, then convert to concentrations at the end.

$KpK_\mathrm{p}$ : equilibrium constant using partial pressures

$KpK_\mathrm{p}$ is used for equilibria involving gases. Instead of concentrations, it uses partial pressures.

Definition

Partial pressure

For:

aA(g)+bB(g)⇌cC(g)+dD(g)a\mathrm{A(g)}+b\mathrm{B(g)}\rightleftharpoons c\mathrm{C(g)}+d\mathrm{D(g)}

the expression is:

Kp=pCcpDdpAapBbK_\mathrm{p}=\frac{p_{\mathrm{C}}^c p_{\mathrm{D}}^d}{p_{\mathrm{A}}^a p_{\mathrm{B}}^b}

Use one pressure unit consistently, usually kPa or Pa.

Example

Calculating Kp from mole fractions

For the equilibrium:

N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g)}\rightleftharpoons2\mathrm{NO_2(g)}

an equilibrium mixture contains 0.300 mol of N₂O₄ and 0.700 mol of NO₂ at a total pressure of 200 kPa. Calculate $KpK_\mathrm{p}$ .

Calculate the total amount and mole fractions:

molxN2O4=0.3001.000=0.300xNO2=0.7001.000=0.700\begin{aligned} n_\text{total}&=0.300+0.700=1.000\ \mathrm{mol}\\ x_{\mathrm{N_2O_4}}&=\frac{0.300}{1.000}=0.300\\ x_{\mathrm{NO_2}}&=\frac{0.700}{1.000}=0.700 \end{aligned}

Convert mole fractions into partial pressures:

kPa\begin{aligned} p_{\mathrm{N_2O_4}}&=0.300\times200=60.0\ \mathrm{kPa}\\ p_{\mathrm{NO_2}}&=0.700\times200=140\ \mathrm{kPa} \end{aligned}

Substitute into the $KpK_\mathrm{p}$ expression:

kPaK_\mathrm{p}=\frac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}} =\frac{140^2}{60.0} =3.27\times10^2\ \mathrm{kPa}

Check the units from the pressure powers: pressure squared divided by pressure leaves kPa.

Common Mistake

Mixing pressure units

If you use kPa for one partial pressure, use kPa for all of them. Do not mix Pa and kPa in the same $KpK_\mathrm{p}$ calculation.

What does the size of $K$ tell you?

A large equilibrium constant means products are favoured at equilibrium. A small equilibrium constant means reactants are favoured.

If $K$ is much greater than 1, the equilibrium lies mainly to the right.
If $K$ is much less than 1, the equilibrium lies mainly to the left.
If $K$ is close to 1, appreciable amounts of both reactants and products are present.

Be careful: this is a qualitative interpretation. The exact amounts still depend on the balanced equation and starting conditions.

What changes $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ ?

At a fixed temperature, the value of $KcK_\mathrm{c}$ or $KpK_\mathrm{p}$ is constant for a particular reaction.

Key Idea

Temperature is the only condition that changes K

For a given reaction, $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ change only when temperature changes.

For temperature changes, use the enthalpy change of the forward reaction:

If the forward reaction is exothermic, increasing temperature decreases $K$ .
If the forward reaction is endothermic, increasing temperature increases $K$ .

Example

Predicting how temperature changes Kp

For the Haber equilibrium:

N2(g)+3H2(g)⇌2NH3(g)ΔH<0\mathrm{N_2(g)}+3\mathrm{H_2(g)}\rightleftharpoons2\mathrm{NH_3(g)} \qquad \Delta H<0

predict what happens to $KpK_\mathrm{p}$ when temperature is increased.

The forward reaction is exothermic, so the reverse reaction is endothermic.
Increasing temperature favours the endothermic direction, so equilibrium shifts left towards N₂ and H₂.
In the expression

Kp=pNH32pN2pH23K_\mathrm{p}=\frac{p_{\mathrm{NH_3}}^2}{p_{\mathrm{N_2}}p_{\mathrm{H_2}}^3}

the numerator becomes smaller relative to the denominator, so $KpK_\mathrm{p}$ decreases.

Common Mistake

Converting between Kc and Kp

Exam technique

In the exam

Start from the balanced equation, then write the equilibrium expression before substituting numbers.
Use equilibrium concentrations for $KcK_\mathrm{c}$ and equilibrium partial pressures for $KpK_\mathrm{p}$ .
Convert mol to mol dm⁻³ for $KcK_\mathrm{c}$ , and convert mole fractions to partial pressures for $KpK_\mathrm{p}$ .
Check units from the powers in the expression, and state “no units” only when they cancel.
For changes in conditions, remember: temperature changes $K$ ; pressure, concentration and catalysts do not.

Self review

Check yourself

Can you write the $KcK_\mathrm{c}$ expression for $2NO2(g)⇌N2O4(g)2\mathrm{NO_2(g)}\rightleftharpoons\mathrm{N_2O_4(g)}$ and work out its units?
If a gas mixture has mole fraction 0.25 of CO₂ at 400 kPa total pressure, what is the partial pressure of CO₂?
For an endothermic forward reaction, what happens to the value of $K$ when temperature is increased?

What you'll learn

Starting point: dynamic equilibrium

Why use an equilibrium constant?

$KcK_\mathrm{c}$ : equilibrium constant using concentrations

What goes into a $KcK_\mathrm{c}$ expression?

Calculating $KcK_\mathrm{c}$ from equilibrium amounts

$KpK_\mathrm{p}$ : equilibrium constant using partial pressures

What does the size of $K$ tell you?

What changes $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ ?

Review Kc and Kp equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kc and Kp equilibria

What you'll learn

Starting point: dynamic equilibrium

Why use an equilibrium constant?

$KcK_\mathrm{c}$ : equilibrium constant using concentrations

What goes into a $KcK_\mathrm{c}$ expression?

Calculating $KcK_\mathrm{c}$ from equilibrium amounts

$KpK_\mathrm{p}$ : equilibrium constant using partial pressures

What does the size of $K$ tell you?

What changes $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ ?

Review Kc and Kp equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kc and Kp equilibria

What you'll learn

Starting point: dynamic equilibrium

Why use an equilibrium constant?

KcK_\mathrm{c}Kc​: equilibrium constant using concentrations

What goes into a KcK_\mathrm{c}Kc​ expression?

Calculating KcK_\mathrm{c}Kc​ from equilibrium amounts

KpK_\mathrm{p}Kp​: equilibrium constant using partial pressures

What does the size of KKK tell you?

What changes KcK_\mathrm{c}Kc​ and KpK_\mathrm{p}Kp​?

Review Kc and Kp equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kc and Kp equilibria Revision Guide

Kc and Kp equilibria

What you'll learn

Starting point: dynamic equilibrium

Why use an equilibrium constant?

KcK_\mathrm{c}Kc​: equilibrium constant using concentrations

What goes into a KcK_\mathrm{c}Kc​ expression?

Calculating KcK_\mathrm{c}Kc​ from equilibrium amounts

KpK_\mathrm{p}Kp​: equilibrium constant using partial pressures

What does the size of KKK tell you?

What changes KcK_\mathrm{c}Kc​ and KpK_\mathrm{p}Kp​?

Review Kc and Kp equilibria by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kc and Kp equilibria Revision Guide

$KcK_\mathrm{c}$ : equilibrium constant using concentrations

What goes into a $KcK_\mathrm{c}$ expression?

Calculating $KcK_\mathrm{c}$ from equilibrium amounts

$KpK_\mathrm{p}$ : equilibrium constant using partial pressures

What does the size of $K$ tell you?

What changes $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ ?

$KcK_\mathrm{c}$ : equilibrium constant using concentrations

What goes into a $KcK_\mathrm{c}$ expression?

Calculating $KcK_\mathrm{c}$ from equilibrium amounts

$KpK_\mathrm{p}$ : equilibrium constant using partial pressures

What does the size of $K$ tell you?

What changes $KcK_\mathrm{c}$ and $KpK_\mathrm{p}$ ?