Equilibrium Constants Kc and Kp

AS · 10 min

Le Chatelier's principle tells you which way an equilibrium shifts; the equilibrium constant tells you how far. For any reaction at a given temperature, a particular ratio of product to reactant concentrations always has the same value at equilibrium, whatever the starting amounts. This note shows how to write the expressions for $K_c$ (in concentrations) and $K_p$ (in partial pressures), work out their units, calculate equilibrium amounts from starting amounts, and decide which changes alter the value of $K$ . Expect a multi-step calculation in Paper 2; the syllabus guarantees none will need a quadratic equation.

The equilibrium constant Kc

For a general reaction at equilibrium:

a\ce{A} + b\ce{B} \ce{<=>} c\ce{C} + d\ce{D}

Key result

K_c = \frac{[\ce{C}]^c\,[\ce{D}]^d}{[\ce{A}]^a\,[\ce{B}]^b}

Square brackets mean equilibrium concentration in $\text{mol dm}^{-3}$ .
Products on top, reactants on the bottom.
Each concentration is raised to the power of its coefficient in the balanced equation.

For example:

\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad K_c = \frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}

\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)} \qquad K_c = \frac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{C2H5OH}]}

In the esterification, every substance is in the same liquid mixture, so all four appear, including water.

Tip

In heterogeneous equilibria (more than one phase), the concentrations of pure solids are constant and are left out of the expression. For $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ , $K_c = [\ce{CO2}]$ . Water is left out only when it is the solvent in large excess; in the esterification above it is a product in a mixture, so it stays in.

Units of Kc

The units depend on the expression. Substitute $\text{mol dm}^{-3}$ for every concentration and cancel.

reaction	expression	units
$\ce{H2 + I2 <=> 2HI}$	$\dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$	$\dfrac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2}$ : no units
$\ce{N2O4 <=> 2NO2}$	$\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$	$\text{mol dm}^{-3}$
$\ce{2SO2 + O2 <=> 2SO3}$	$\dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$	$\dfrac{1}{\text{mol dm}^{-3}} = \text{mol}^{-1}\ \text{dm}^3$
$\ce{N2 + 3H2 <=> 2NH3}$	$\dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}$	$(\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\ \text{dm}^6$

What the size of K tells you

$K$ much greater than 1: the position of equilibrium lies far to the right; the equilibrium mixture is mostly products.
$K$ much less than 1: the position lies far to the left; mostly reactants.

$K$ says nothing about how fast equilibrium is reached.

Partial pressures and Kp

For reactions between gases, it is often more convenient to use pressures than concentrations. In a mixture of gases, each gas contributes part of the total pressure.

Definition

The mole fraction of a gas A in a mixture is the amount of A divided by the total amount of gas:

x_{\ce{A}} = \frac{n_{\ce{A}}}{n_{\text{total}}}

The partial pressure of a gas A is the pressure that gas would exert if it alone occupied the whole volume of the mixture:

p_{\ce{A}} = x_{\ce{A}} \times p_{\text{total}}

Mole fractions add up to 1, and partial pressures add up to the total pressure.

For the general gaseous reaction above:

Key result

K_p = \frac{p_{\ce{C}}^{\,c}\,p_{\ce{D}}^{\,d}}{p_{\ce{A}}^{\,a}\,p_{\ce{B}}^{\,b}}

where each $p$ is an equilibrium partial pressure (in Pa, kPa or atm). Units are found in the same way as for $K_c$ , with pressure units in place of $\text{mol dm}^{-3}$ . Only gases appear in $K_p$ .

\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad K_p = \frac{p_{\ce{NH3}}^{\,2}}{p_{\ce{N2}}\,p_{\ce{H2}}^{\,3}} \quad (\text{units: kPa}^{-2})

(You do not need the relationship between $K_p$ and $K_c$ .)

Calculating equilibrium amounts

Most questions give you starting amounts and one equilibrium amount and ask for $K$ . The tool is an ICE table: Initial, Change, Equilibrium.

Method

Equilibrium calculations

Write the balanced equation and the expression for $K_c$ or $K_p$ .
Draw an ICE table with a column for each species. Fill in the initial moles.
Use the one known equilibrium amount to find the change for that species, then use the equation's mole ratio to find the changes for all others (reactants decrease, products increase).
Add to get the equilibrium moles.
For $K_c$ : divide each by the volume (in $\text{dm}^3$ ) to get concentrations. For $K_p$ : divide each by the total moles to get mole fractions, then multiply by total pressure.
Substitute into the expression, calculate, and give units.

Tip

If the total moles on each side of the equation are equal (as for $\ce{H2 + I2 <=> 2HI}$ ), the volume cancels and you can put moles straight into $K_c$ . Otherwise, you must convert to concentrations.

What changes the value of K

Key result

change	position of equilibrium	value of $K$
concentration	may move	unchanged
pressure	may move	unchanged
catalyst	unchanged	unchanged
temperature	moves	changes

Why does a concentration change not alter $K$ ? If you add more reactant, the expression momentarily no longer equals $K$ ; the system reacts (the position shifts right) until the ratio returns to the same value of $K$ . The position moves; the constant does not.

Temperature is different: $K$ itself depends on temperature.

For an exothermic forward reaction, increasing temperature moves the position to the left and decreases $K$ .
For an endothermic forward reaction, increasing temperature moves the position to the right and increases $K$ .

Worked examples

Routine: Kc with no units

$1.00\ \text{mol}$ of hydrogen and $1.00\ \text{mol}$ of iodine were heated in a sealed $1.00\ \text{dm}^3$ flask until equilibrium was reached. The equilibrium mixture contained $1.56\ \text{mol}$ of hydrogen iodide. Calculate $K_c$ for $\ce{H2(g) + I2(g) <=> 2HI(g)}$ .

Solution

	$\ce{H2}$	$\ce{I2}$	$\ce{HI}$
initial / mol	1.00	1.00	0
change / mol	$-0.78$	$-0.78$	$+1.56$
equilibrium / mol	0.22	0.22	1.56

Forming $1.56\ \text{mol}$ HI uses $\tfrac{1.56}{2} = 0.78\ \text{mol}$ of each reactant.

Volume $= 1.00\ \text{dm}^3$ , so concentrations equal moles.

K_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} = \frac{1.56^2}{0.22 \times 0.22} = 50.3 \quad (\text{no units})

Kc where volume matters

$0.500\ \text{mol}$ of phosphorus pentachloride was heated in a sealed $2.00\ \text{dm}^3$ container. At equilibrium, $0.200\ \text{mol}$ of chlorine was present. Calculate $K_c$ for $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ .

Solution

	$\ce{PCl5}$	$\ce{PCl3}$	$\ce{Cl2}$
initial / mol	0.500	0	0
change / mol	$-0.200$	$+0.200$	$+0.200$
equilibrium / mol	0.300	0.200	0.200
concentration / mol dm⁻³	0.150	0.100	0.100

K_c = \frac{[\ce{PCl3}][\ce{Cl2}]}{[\ce{PCl5}]} = \frac{0.100 \times 0.100}{0.150} = 0.0667\ \text{mol dm}^{-3}

Esterification

$1.00\ \text{mol}$ of ethanoic acid and $1.00\ \text{mol}$ of ethanol were mixed with a little acid catalyst and left to reach equilibrium. Titration showed that $0.333\ \text{mol}$ of ethanoic acid remained. (a) Calculate $K_c$ . (b) In another equilibrium mixture at the same temperature there are $0.50\ \text{mol}$ of ester, $0.50\ \text{mol}$ of water and $0.25\ \text{mol}$ of ethanoic acid. Calculate the amount of ethanol present.

Solution

(a) $0.667\ \text{mol}$ of acid reacted, so $0.667\ \text{mol}$ each of ester and water formed, and $0.333\ \text{mol}$ of ethanol remains. The volume $V$ cancels (2 moles on each side):

K_c = \frac{(0.667/V)(0.667/V)}{(0.333/V)(0.333/V)} = \frac{0.667^2}{0.333^2} = 4.0 \quad (\text{no units})

(b) Let the amount of ethanol be $x$ :

4.0 = \frac{0.50 \times 0.50}{0.25 \times x} \quad\Longrightarrow\quad x = \frac{0.25}{4.0 \times 0.25} = 0.25\ \text{mol}

Kp from a degree of dissociation

$1.00\ \text{mol}$ of dinitrogen tetraoxide was allowed to reach equilibrium at a total pressure of $100\ \text{kPa}$ ; $20.0\%$ of it dissociated: $\ce{N2O4(g) <=> 2NO2(g)}$ . Calculate $K_p$ .

Solution

$20.0\%$ dissociated: $0.200\ \text{mol}$ $\ce{N2O4}$ reacts, forming $0.400\ \text{mol}$ $\ce{NO2}$ .

	$\ce{N2O4}$	$\ce{NO2}$
equilibrium / mol	0.800	0.400
mole fraction	$\tfrac{0.800}{1.200} = 0.667$	$\tfrac{0.400}{1.200} = 0.333$
partial pressure / kPa	66.7	33.3

K_p = \frac{p_{\ce{NO2}}^{\,2}}{p_{\ce{N2O4}}} = \frac{33.3^2}{66.7} = 16.7\ \text{kPa}

Exam-hard: Kp for the Haber process

At $500\ ^\circ\text{C}$ , an equilibrium mixture for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ contains $0.60\ \text{mol}$ $\ce{N2}$ , $1.80\ \text{mol}$ $\ce{H2}$ and $0.80\ \text{mol}$ $\ce{NH3}$ at a total pressure of $2.00 \times 10^{4}\ \text{kPa}$ .

(a) Calculate $K_p$ , with units. (b) The mixture was made from nitrogen and hydrogen only. Calculate the initial amounts. (c) State and explain the effect on $K_p$ of raising the temperature to $600\ ^\circ\text{C}$ .

Solution

(a) Total moles $= 0.60 + 1.80 + 0.80 = 3.20\ \text{mol}$ .

p_{\ce{N2}} = \frac{0.60}{3.20} \times 2.00 \times 10^{4} = 3750\ \text{kPa}

p_{\ce{H2}} = \frac{1.80}{3.20} \times 2.00 \times 10^{4} = 11\,250\ \text{kPa}

p_{\ce{NH3}} = \frac{0.80}{3.20} \times 2.00 \times 10^{4} = 5000\ \text{kPa}

K_p = \frac{5000^2}{3750 \times 11\,250^3} = 4.68 \times 10^{-9}\ \text{kPa}^{-2}

(b) Forming $0.80\ \text{mol}$ $\ce{NH3}$ used $0.40\ \text{mol}$ $\ce{N2}$ and $1.20\ \text{mol}$ $\ce{H2}$ . Initial: $\ce{N2}$ $= 0.60 + 0.40 = 1.00\ \text{mol}$ ; $\ce{H2}$ $= 1.80 + 1.20 = 3.00\ \text{mol}$ .

(c) $K_p$ decreases. The forward reaction is exothermic, so increasing the temperature moves the position of equilibrium to the left (the endothermic direction); the equilibrium partial pressure of ammonia falls and those of nitrogen and hydrogen rise, so the ratio $K_p$ becomes smaller.

Watch out

Using initial amounts in $K$ . Only equilibrium amounts go into the expression.
Using moles instead of concentrations when the moles of reactants and products differ. Divide by the volume.
Wrong powers. $[\ce{H2}]^3$ in the Haber expression, not $3[\ce{H2}]$ .
Using total pressure in $K_p$ . Each term is a partial pressure, found from the mole fraction.
Saying adding a reactant increases $K$ . It shifts the position; $K$ is constant at constant temperature.
Forgetting units, or stating units when there are none.

Exam tip

Write the expression first, then the ICE table, then substitute. Each is usually a mark.
Units are often a separate mark: show the cancelling.
For $K_p$ , show mole fractions and partial pressures as separate steps.
When asked about the effect of a change on $K$ , state clearly whether $K$ changes, and if it is temperature, link the direction to the sign of $\Delta H$ .
Give answers to 3 significant figures unless the data justify fewer.

Summary

$K_c = \dfrac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}$ using equilibrium concentrations.
Mole fraction $x_{\ce{A}} = n_{\ce{A}}/n_{\text{total}}$ ; partial pressure $p_{\ce{A}} = x_{\ce{A}} \times p_{\text{total}}$ .
$K_p$ uses equilibrium partial pressures of gases only.
Work out units by substitution; they may be none.
ICE table: initial, change (from the mole ratio), equilibrium; then convert to concentrations or partial pressures.
Only temperature changes $K$ . Exothermic forward reaction: $K$ falls as temperature rises. Endothermic: $K$ rises.
Large $K$ : products favoured; small $K$ : reactants favoured.

Practice

Question

Write $K_c$ expressions, with units, for: (a) $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ ; (b) $\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}$ ; (c) $\ce{N2O4(g) <=> 2NO2(g)}$ .
Write $K_p$ expressions, with units in kPa, for: (a) $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ ; (b) $\ce{2NO2(g) <=> N2O4(g)}$ .
A gas mixture contains $2.0\ \text{mol}$ $\ce{N2}$ , $6.0\ \text{mol}$ $\ce{H2}$ and $2.0\ \text{mol}$ $\ce{NH3}$ at a total pressure of $200\ \text{kPa}$ . Calculate the mole fraction and partial pressure of each gas.
$1.00\ \text{mol}$ of hydrogen iodide was heated in a sealed $1.00\ \text{dm}^3$ flask. At equilibrium, $0.11\ \text{mol}$ of hydrogen was present. Calculate $K_c$ for $\ce{2HI(g) <=> H2(g) + I2(g)}$ .
For $\ce{N2O4(g) <=> 2NO2(g)}$ , $K_c = 4.0 \times 10^{-3}\ \text{mol dm}^{-3}$ at a certain temperature. Calculate $[\ce{NO2}]$ in an equilibrium mixture in which $[\ce{N2O4}] = 0.0400\ \text{mol dm}^{-3}$ .
For an exothermic reaction, state the effect of increasing temperature on the position of equilibrium and on $K_c$ .
$1.00\ \text{mol}$ of $\ce{PCl5}$ was heated to a temperature at which it is $40.0\%$ dissociated at a total pressure of $200\ \text{kPa}$ : $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ . Calculate $K_p$ .
For the esterification of ethanoic acid with ethanol, $K_c = 4.0$ . An equilibrium mixture contains $0.80\ \text{mol}$ ester, $0.80\ \text{mol}$ water and $0.40\ \text{mol}$ ethanol. Calculate the amount of ethanoic acid present.
$1.00\ \text{mol}$ of hydrogen and $1.00\ \text{mol}$ of carbon dioxide were heated in a sealed vessel: $\ce{H2(g) + CO2(g) <=> H2O(g) + CO(g)}$ . At equilibrium, $0.44\ \text{mol}$ of CO was present. Calculate $K_c$ , and explain why it was not necessary to know the volume of the vessel. State, with a reason, the effect of halving the volume of the vessel on the amount of CO at equilibrium.
$1.00\ \text{mol}$ of nitrogen and $3.00\ \text{mol}$ of hydrogen reached equilibrium at a total pressure of $1.50 \times 10^{4}\ \text{kPa}$ . The equilibrium mixture contained $0.50\ \text{mol}$ of ammonia. Calculate $K_p$ , with units, and explain why the percentage of ammonia in the equilibrium mixture would be greater at a higher total pressure even though $K_p$ is unchanged.

Answers

(a) $K_c = \dfrac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$ ; $\text{mol}^{-1}\ \text{dm}^3$ . (b) $K_c = \dfrac{[\ce{CH3COOC2H5}][\ce{H2O}]}{[\ce{CH3COOH}][\ce{C2H5OH}]}$ ; no units. (c) $K_c = \dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$ ; $\text{mol dm}^{-3}$ .
(a) $K_p = \dfrac{p_{\ce{NH3}}^{\,2}}{p_{\ce{N2}}\,p_{\ce{H2}}^{\,3}}$ ; $\text{kPa}^{-2}$ . (b) $K_p = \dfrac{p_{\ce{N2O4}}}{p_{\ce{NO2}}^{\,2}}$ ; $\text{kPa}^{-1}$ .
Total $= 10.0\ \text{mol}$ . $\ce{N2}$ : $x = 0.20$ , $p = 40\ \text{kPa}$ . $\ce{H2}$ : $x = 0.60$ , $p = 120\ \text{kPa}$ . $\ce{NH3}$ : $x = 0.20$ , $p = 40\ \text{kPa}$ .
$0.11\ \text{mol}$ $\ce{H2}$ and $0.11\ \text{mol}$ $\ce{I2}$ formed from $0.22\ \text{mol}$ HI; HI remaining $= 0.78\ \text{mol}$ . $K_c = \dfrac{0.11 \times 0.11}{0.78^2} = 0.0199$ (no units).
$[\ce{NO2}]^2 = K_c \times [\ce{N2O4}] = 4.0 \times 10^{-3} \times 0.0400 = 1.6 \times 10^{-4}$ ; $[\ce{NO2}] = 0.0126\ \text{mol dm}^{-3}$ .
The position moves to the left (the endothermic, reverse direction); $K_c$ decreases.
$\ce{PCl5}$ $0.600$ , $\ce{PCl3}$ $0.400$ , $\ce{Cl2}$ $0.400\ \text{mol}$ ; total $1.400$ . Partial pressures: $\ce{PCl5}$ $\tfrac{0.600}{1.400} \times 200 = 85.7$ ; $\ce{PCl3}$ and $\ce{Cl2}$ each $57.1\ \text{kPa}$ . $K_p = \dfrac{57.1 \times 57.1}{85.7} = 38.1\ \text{kPa}$ .
$4.0 = \dfrac{0.80 \times 0.80}{x \times 0.40}$ , so $x = \dfrac{0.64}{1.6} = 0.40\ \text{mol}$ .
Equilibrium: $\ce{H2O}$ $0.44$ , CO $0.44$ , $\ce{H2}$ $0.56$ , $\ce{CO2}$ $0.56\ \text{mol}$ . $K_c = \dfrac{0.44 \times 0.44}{0.56 \times 0.56} = 0.617$ (no units). The volume cancels because there are equal numbers of moles on each side (each concentration is $n/V$ , and $V^2$ cancels top and bottom). Halving the volume has no effect on the amount of CO: there are 2 moles of gas on each side, so the pressure change does not shift the position.
Formed $0.50\ \text{mol}$ $\ce{NH3}$ from $0.25\ \text{mol}$ $\ce{N2}$ and $0.75\ \text{mol}$ $\ce{H2}$ . Equilibrium: $\ce{N2}$ $0.75$ , $\ce{H2}$ $2.25$ , $\ce{NH3}$ $0.50$ ; total $3.50\ \text{mol}$ . Partial pressures: $\ce{N2}$ $3214$ , $\ce{H2}$ $9643$ , $\ce{NH3}$ $2143\ \text{kPa}$ . $K_p = \dfrac{2143^2}{3214 \times 9643^3} = 1.59 \times 10^{-9}\ \text{kPa}^{-2}$ . At higher total pressure the position moves to the right (fewer gas moles), increasing the proportion of ammonia; $K_p$ is unchanged because the partial pressures adjust so that their ratio in the expression still equals $K_p$ (the denominator has a higher power of pressure than the numerator, so the mole fraction of ammonia must rise to keep the ratio constant).

The equilibrium constant Kc

Units of Kc

What the size of K tells you

Partial pressures and Kp

Calculating equilibrium amounts

What changes the value of K

Worked examples

Practice

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