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The equilibrium constant, Kc

Covers AS91166

What $K_c$ tells you

The equilibrium constant $K_c$ measures how far a reaction goes before reaching equilibrium — the position of the equilibrium.
It is calculated from the concentrations at equilibrium.

For the general reaction $aA + bB \rightleftharpoons cC + dD$ :

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

Square brackets mean concentration in $mol\ L^{-1}$ .
Products go on top, reactants underneath.
Each concentration is raised to the power of its coefficient in the balanced equation.

The standard limits this to homogeneous systems — everything in the same phase (all gases, or all in solution).

Interpreting the value

$K_c$	Meaning	Position of equilibrium
large (≫1)	mostly products at equilibrium	lies to the right
about 1	comparable amounts of both	roughly central
small (≪1)	mostly reactants at equilibrium	lies to the left

A large $K_c$ does not mean the reaction is fast. $K_c$ describes how far, not how quickly — rate is a separate question entirely.

What changes $K_c$

Only temperature changes the value of $K_c$ .
Changing concentration or pressure shifts the position of the equilibrium, but the system adjusts so that $K_c$ ends up at the same value.
A catalyst changes neither.
So if a question says $K_c$ has changed, the temperature must have changed.

Writing the expression

Balance the equation first — the coefficients become the powers.
Put the products on top and the reactants underneath.
Raise each concentration to its coefficient.

Worked through:

$N_2 + 3H_2 \rightleftharpoons 2NH_3$ gives $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$
$H_2 + I_2 \rightleftharpoons 2HI$ gives $K_c = \frac{[HI]^2}{[H_2][I_2]}$

Units

The units of $K_c$ depend on the equation, so work them out from the expression each time.
For $H_2 + I_2 \rightleftharpoons 2HI$ the units cancel — $K_c$ has no units.
For $N_2 + 3H_2 \rightleftharpoons 2NH_3$ they do not cancel:

$\frac{(mol\ L^{-1})^2}{(mol\ L^{-1})(mol\ L^{-1})^3} = \frac{1}{(mol\ L^{-1})^2} = mol^{-2}\ L^{2}$

Worked ExampleWriting and interpreting Kc

For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ :

(a) Write the expression for $K_c$ and determine its units. (b) At 500 °C, $K_c = 2.8 \times 10^{2}$ . State what this tells you about the position of equilibrium. (c) State and explain what happens to $K_c$ if more $O_2$ is added.

(a) The expression and its units

Step 1 — Put the products on top, reactants underneath. The product is $SO_3$ ; the reactants are $SO_2$ and $O_2$ .

Step 2 — Raise each to the power of its coefficient. $SO_3$ has coefficient 2, $SO_2$ has coefficient 2, $O_2$ has coefficient 1.

$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$

Step 3 — Work out the units. Substituting $mol\ L^{-1}$ for each concentration:

$\frac{(mol\ L^{-1})^2}{(mol\ L^{-1})^2 (mol\ L^{-1})}$

The squared terms cancel, leaving:

$\frac{1}{mol\ L^{-1}} = mol^{-1}\ L$

(b) Interpreting the value

$K_c = 2.8 \times 10^{2} = 280$ , which is much greater than 1.

Since $K_c$ is products over reactants, a large value means the numerator is much larger than the denominator — there is a much higher concentration of products than reactants at equilibrium.

So the position of equilibrium lies well to the right, and the reaction produces a high yield of $SO_3$ .

Note this says nothing about the rate — the reaction may still be slow, which is why a catalyst is used industrially.

(c) The effect of adding more $O_2$

The value of $K_c$ does not change.

Adding $O_2$ increases the denominator, so momentarily the ratio falls below $K_c$ . By Le Chatelier's principle the system responds by shifting right, converting some of the added $O_2$ and $SO_2$ into $SO_3$ .

This increases the numerator and decreases the denominator until the ratio returns to exactly the same value as before.

So the position of equilibrium shifts right (more $SO_3$ is formed), but $K_c$ is unchanged. Only a change in temperature would alter the value of $K_c$ .

The equilibrium constant, Kc

Covers AS91166

What $K_c$ tells you

The equilibrium constant $K_c$ measures how far a reaction goes before reaching equilibrium — the position of the equilibrium.
It is calculated from the concentrations at equilibrium.

For the general reaction $aA + bB \rightleftharpoons cC + dD$ :

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

Square brackets mean concentration in $mol\ L^{-1}$ .
Products go on top, reactants underneath.
Each concentration is raised to the power of its coefficient in the balanced equation.

The standard limits this to homogeneous systems — everything in the same phase (all gases, or all in solution).

Interpreting the value

$K_c$	Meaning	Position of equilibrium
large (≫1)	mostly products at equilibrium	lies to the right
about 1	comparable amounts of both	roughly central
small (≪1)	mostly reactants at equilibrium	lies to the left

A large $K_c$ does not mean the reaction is fast. $K_c$ describes how far, not how quickly — rate is a separate question entirely.

What changes $K_c$

Only temperature changes the value of $K_c$ .
Changing concentration or pressure shifts the position of the equilibrium, but the system adjusts so that $K_c$ ends up at the same value.
A catalyst changes neither.
So if a question says $K_c$ has changed, the temperature must have changed.

Writing the expression

Balance the equation first — the coefficients become the powers.
Put the products on top and the reactants underneath.
Raise each concentration to its coefficient.

Worked through:

$N_2 + 3H_2 \rightleftharpoons 2NH_3$ gives $K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$
$H_2 + I_2 \rightleftharpoons 2HI$ gives $K_c = \frac{[HI]^2}{[H_2][I_2]}$

Units

The units of $K_c$ depend on the equation, so work them out from the expression each time.
For $H_2 + I_2 \rightleftharpoons 2HI$ the units cancel — $K_c$ has no units.
For $N_2 + 3H_2 \rightleftharpoons 2NH_3$ they do not cancel:

$\frac{(mol\ L^{-1})^2}{(mol\ L^{-1})(mol\ L^{-1})^3} = \frac{1}{(mol\ L^{-1})^2} = mol^{-2}\ L^{2}$

Worked ExampleWriting and interpreting Kc

For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ :

(a) The expression and its units

Step 1 — Put the products on top, reactants underneath. The product is $SO_3$ ; the reactants are $SO_2$ and $O_2$ .

Step 2 — Raise each to the power of its coefficient. $SO_3$ has coefficient 2, $SO_2$ has coefficient 2, $O_2$ has coefficient 1.

$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$

Step 3 — Work out the units. Substituting $mol\ L^{-1}$ for each concentration:

$\frac{(mol\ L^{-1})^2}{(mol\ L^{-1})^2 (mol\ L^{-1})}$

The squared terms cancel, leaving:

$\frac{1}{mol\ L^{-1}} = mol^{-1}\ L$

(b) Interpreting the value

$K_c = 2.8 \times 10^{2} = 280$ , which is much greater than 1.

So the position of equilibrium lies well to the right, and the reaction produces a high yield of $SO_3$ .

Note this says nothing about the rate — the reaction may still be slow, which is why a catalyst is used industrially.

(c) The effect of adding more $O_2$

The value of $K_c$ does not change.

This increases the numerator and decreases the denominator until the ratio returns to exactly the same value as before.

So the position of equilibrium shifts right (more $SO_3$ is formed), but $K_c$ is unchanged. Only a change in temperature would alter the value of $K_c$ .

The equilibrium constant, Kc

What KcK_cKc​ tells you

Interpreting the value

What changes KcK_cKc​

Writing the expression

Units

(a) The expression and its units

(b) Interpreting the value

(c) The effect of adding more O2O_2O2​

The equilibrium constant, Kc

What KcK_cKc​ tells you

Interpreting the value

What changes KcK_cKc​

Writing the expression

Units

(a) The expression and its units

(b) Interpreting the value

(c) The effect of adding more O2O_2O2​

What $K_c$ tells you

What changes $K_c$

(c) The effect of adding more $O_2$

What $K_c$ tells you

What changes $K_c$

(c) The effect of adding more $O_2$