Each concentration in $K$ is raised to the power of:

Its coefficient in the balanced equation

Calculating the Equilibrium Constant | HSC Chemistry Module 5

In short: The equilibrium constant $K$ is the concentrations of the products divided by those of the reactants, each raised to the power of its coefficient. A large $K$ means mostly products at equilibrium, a small $K$ means mostly reactants, and only temperature changes $K$ .

The equilibrium expression

For a general reaction

$a A + b B ⇌ c C + d D$

the equilibrium constant is

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

Square brackets mean concentration in mol/L.
Products on top, reactants on the bottom.
Each concentration is raised to the power of its coefficient in the balanced equation.
Solids and pure liquids are left out (their "concentration" never changes).

What the size of K tells you

Value of $K$	Meaning
Very large (like $1 0^{10}$ )	Mostly products at equilibrium. The reaction nearly goes to completion.
About 1	Similar amounts of reactants and products.
Very small (like $1 0^{- 10}$ )	Mostly reactants. Very little product forms.

$K$ tells you how far a reaction goes, not how fast.

Only temperature changes K

Adding chemicals, changing pressure or adding a catalyst can shift the position, but $K$ stays the same. Only temperature changes $K$ :

Endothermic forward reaction: heating increases $K$ .
Exothermic forward reaction: heating decreases $K$ .

Calculating K from equilibrium concentrations

If you know every concentration at equilibrium, just substitute.

Remember to convert moles to concentration first: $c = \frac{n}{V}$ , with $V$ in litres.

ICE tables

Often you only know the starting amounts and one equilibrium amount. An ICE table (Initial, Change, Equilibrium) fills in the rest, using the mole ratio.

Worked exampleDinitrogen tetroxide

1.00 mol of $N_{2} O_{4}$ is put in a 2.0 L flask. At equilibrium, 20% has broken down: $N_{2} O_{4} ⇌ 2N O_{2}$ .

Starting concentration $= 1.00 \div 2.0 = 0.50$ mol/L. 20% of 0.50 is 0.10.

	$N_{2} O_{4}$	$N O_{2}$
Initial	0.50	0
Change	−0.10	+0.20
Equilibrium	0.40	0.20

$K = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{0.2 0 ^{2}}{0.40} = 0.10$

The Change row always follows the coefficients: here 1 : 2.

The reaction quotient Q

$Q$ uses the same formula as $K$ , but with the concentrations right now, which may not be at equilibrium.

Compare	What happens
$Q < K$	Too few products. The reaction shifts right.
$Q = K$	At equilibrium. No change.
$Q > K$	Too many products. The reaction shifts left.

Equilibrium constant

Expression

Products over reactants
Powers = coefficients
Leave out solids and liquids

Size of K

Large: mostly products
Small: mostly reactants

Changes K

Only temperature

ICE table

Initial
Change by mole ratio
Equilibrium

Q < K: shifts right
Q > K: shifts left

Year 12Harder K calculations

Changing the equation changes K.

Reverse the equation: $K_{n e w} = \frac{1}{K}$ .
Double every coefficient: $K_{n e w} = K^{2}$ .
Halve every coefficient: $K_{n e w} = K$ .

Finding an unknown with a square root. For $CO + H_{2} O ⇌ C O_{2} + H_{2}$ , $K = 4.0$ , starting with 0.100 mol/L of $CO$ and $H_{2} O$ only. Let $x$ react:

$\frac{x ^{2}}{( 0.100 - x ) ^{2}} = 4.0 \Rightarrow \frac{x}{0.100 - x} = 2.0 \Rightarrow x = 0.0667$

So $[C O_{2}] = [H_{2}] = 0.067$ mol/L.

Finding one missing concentration. For $N_{2} + 3 H_{2} ⇌ 2N H_{3}$ , $K = 0.50$ , $[N_{2}] = 0.20$ , $[H_{2}] = 0.40$ :

$[N H_{3}]^{2} = 0.50 \times 0.20 \times 0.4 0^{3} = 0.0064$ , so $[N H_{3}] = 0.080$ mol/L.

Mixing two solutions. Equal volumes halve both concentrations. Mixing $2.00 \times 1 0^{- 3}$ mol/L $F e^{3 +}$ and $SC N^{-}$ gives $1.00 \times 1 0^{- 3}$ of each. If $[FeSC N^{2 +}] = 1.45 \times 1 0^{- 4}$ , each reactant falls to $8.55 \times 1 0^{- 4}$ , so $K = \frac{1.45 \times 1 0 ^{- 4}}{( 8.55 \times 1 0 ^{- 4} ) ^{2}} = 198$ .

A very small K. If $K$ is tiny, hardly anything reacts, so the starting amount barely changes. For $COC l_{2} ⇌ CO + C l_{2}$ with $K = 2.1 \times 1 0^{- 8}$ and $[COC l_{2}] = 0.50$ : $x^{2} = 2.1 \times 1 0^{- 8} \times 0.50$ , so $x = 1.0 \times 1 0^{- 4}$ mol/L.

Combining two equilibria. Add two equations and multiply their $K$ values. Reverse one and use $\frac{1}{K}$ . For $C H_{3} COOH + HCO O^{-} ⇌ C H_{3} CO O^{-} + HCOOH$ : $K = \frac{K _{a} ( acetic )}{K _{a} ( methanoic )} = \frac{1 0 ^{- 4.76}}{1 0 ^{- 3.75}} = 0.098$ .

Changing one amount to reach a target. Fix the other concentrations at their new values, then solve $K$ for the one you are adding. Remember the amount added also has to cover what reacts.

Quick summary

$K = \frac{[ products ] ^{coefficients}}{[ reactants ] ^{coefficients}}$ . Leave out solids and pure liquids.
Large $K$ : mostly products. Small $K$ : mostly reactants.
Only temperature changes $K$ .
Convert moles to mol/L. Use an ICE table with the mole ratio.
$Q < K$ : shifts right. $Q > K$ : shifts left. $Q = K$ : at equilibrium.

Quiz

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Try 3 questions now

In an equilibrium expression, the products go:
1. A On the top
2. B On the bottom
3. C Both top and bottom
4. D Nowhere
Show answer
A: On the top
$K$ is products over reactants.
Each concentration in $K$ is raised to the power of:
1. A Its molar mass
2. B Its coefficient in the balanced equation
3. C Its charge
4. D 2
Show answer
B: Its coefficient in the balanced equation
For $2N H_{3}$ you write $[N H_{3}]^{2}$ .
Which is left out of an equilibrium expression?
1. A Gases
2. B Dissolved ions
3. C Solids and pure liquids
4. D Products
Show answer
C: Solids and pure liquids
Their concentration does not change, so they are left out.

Calculating the Equilibrium Constant