# Calculating the Equilibrium Constant

> Writing equilibrium expressions, calculating K with ICE tables, using the reaction quotient Q to predict direction, and how temperature changes K.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/equilibrium-constant/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The recipe that never changes
Every winter, Leo's family makes a big pot of soup. No matter how big the pot, the soup always tastes the same, because Leo's mum keeps the ratio of stock to vegetables fixed.

In 1864, two Norwegian scientists, Cato Guldberg and Peter Waage, found something similar inside reactions. However much of each chemical you start with, once equilibrium is reached, one special ratio of the concentrations always comes out to the same number at a given temperature. They called it the **equilibrium constant**, $K$. With it, chemists can predict exactly how much product they will get.

**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

$$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$$

the equilibrium constant is

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

- 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).

### Worked example: Writing expressions
$\mathrm{N_2(g)} + \mathrm{3H_2(g)} \rightleftharpoons \mathrm{2NH_3(g)}$:
$$K_{eq} = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$$

$\mathrm{Fe_2O_3(s)} + \mathrm{3CO(g)} \rightleftharpoons \mathrm{2Fe(s)} + \mathrm{3CO_2(g)}$: the solids drop out, so
$$K_{eq} = \frac{[\mathrm{CO_2}]^3}{[\mathrm{CO}]^3}$$

## What the size of K tells you

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

### Worked example: Hydrogen iodide
At equilibrium, $[\mathrm{H_2}] = 0.10$, $[\mathrm{I_2}] = 0.10$ and $[\mathrm{HI}] = 0.70$ mol/L.

$$K = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \frac{0.70^2}{0.10 \times 0.10} = \frac{0.49}{0.010} = 49$$

**Remember to convert moles to concentration first**: $c = \dfrac{n}{V}$, 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 example: Dinitrogen tetroxide
1.00 mol of $\mathrm{N_2O_4}$ is put in a 2.0 L flask. At equilibrium, 20% has broken down: $\mathrm{N_2O_4} \rightleftharpoons \mathrm{2NO_2}$.

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

| | $\mathrm{N_2O_4}$ | $\mathrm{NO_2}$ |
|---|---|---|
| Initial | 0.50 | 0 |
| Change | −0.10 | +0.20 |
| Equilibrium | 0.40 | 0.20 |

$$K = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = \frac{0.20^2}{0.40} = 0.10$$

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**. |

### Worked example: Is it at equilibrium?
For $\mathrm{PCl_5} \rightleftharpoons \mathrm{PCl_3} + \mathrm{Cl_2}$, $K = 0.040$. A 2.0 L flask has 0.50 mol $\mathrm{PCl_5}$, 0.10 mol $\mathrm{PCl_3}$ and 0.10 mol $\mathrm{Cl_2}$.

Concentrations: 0.25, 0.050 and 0.050 mol/L.

$Q = \dfrac{0.050 \times 0.050}{0.25} = 0.010$

$Q < K$, so the reaction is not at equilibrium. It will shift right, making more $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$.

**Mind map**

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

### Year 11: Moles, volume and concentration
Equilibrium questions mix up moles and concentrations on purpose. Always change everything to mol/L before you substitute:

$$c = \frac{n}{V} \qquad n = \frac{m}{MM}$$

For example, 3.20 g of $\mathrm{SO_2}$ (molar mass 64.07 g/mol) is 0.0500 mol. In a 500 mL flask, that is $0.0500 \div 0.500 = 0.100$ mol/L.

Doing the ICE table in **moles** is fine as long as you divide by the volume at the end, before using $K$.

### Year 12: Harder K calculations
**Changing the equation changes K.**

- Reverse the equation: $K_{new} = \dfrac{1}{K}$.
- Double every coefficient: $K_{new} = K^2$.
- Halve every coefficient: $K_{new} = \sqrt{K}$.

**Finding an unknown with a square root.** For $\mathrm{CO} + \mathrm{H_2O} \rightleftharpoons \mathrm{CO_2} + \mathrm{H_2}$, $K = 4.0$, starting with 0.100 mol/L of $\mathrm{CO}$ and $\mathrm{H_2O}$ 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 $[\mathrm{CO_2}] = [\mathrm{H_2}] = 0.067$ mol/L.

**Finding one missing concentration.** For $\mathrm{N_2} + \mathrm{3H_2} \rightleftharpoons \mathrm{2NH_3}$, $K = 0.50$, $[\mathrm{N_2}] = 0.20$, $[\mathrm{H_2}] = 0.40$:

$[\mathrm{NH_3}]^2 = 0.50 \times 0.20 \times 0.40^3 = 0.0064$, so $[\mathrm{NH_3}] = 0.080$ mol/L.

**Mixing two solutions.** Equal volumes halve both concentrations. Mixing $2.00 \times 10^{-3}$ mol/L $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^-}$ gives $1.00 \times 10^{-3}$ of each. If $[\mathrm{FeSCN^{2+}}] = 1.45 \times 10^{-4}$, each reactant falls to $8.55 \times 10^{-4}$, so $K = \dfrac{1.45 \times 10^{-4}}{(8.55 \times 10^{-4})^2} = 198$.

**A very small K.** If $K$ is tiny, hardly anything reacts, so the starting amount barely changes. For $\mathrm{COCl_2} \rightleftharpoons \mathrm{CO} + \mathrm{Cl_2}$ with $K = 2.1 \times 10^{-8}$ and $[\mathrm{COCl_2}] = 0.50$: $x^2 = 2.1 \times 10^{-8} \times 0.50$, so $x = 1.0 \times 10^{-4}$ mol/L.

**Combining two equilibria.** Add two equations and multiply their $K$ values. Reverse one and use $\dfrac{1}{K}$. For $\mathrm{CH_3COOH} + \mathrm{HCOO^-} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{HCOOH}$: $K = \dfrac{K_a(\text{acetic})}{K_a(\text{methanoic})} = \dfrac{10^{-4.76}}{10^{-3.75}} = 0.098$.

**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

- $K = \dfrac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}$. 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

Try these questions from the chapter quiz.

1. In an equilibrium expression, the products go:
   - A. On the top
   - B. On the bottom
   - C. Both top and bottom
   - D. Nowhere
   - Answer: A. On the top. $K$ is products over reactants.

2. Each concentration in $K$ is raised to the power of:
   - A. Its molar mass
   - B. Its coefficient in the balanced equation
   - C. Its charge
   - D. 2
   - Answer: B. Its coefficient in the balanced equation. For $\mathrm{2NH_3}$ you write $[\mathrm{NH_3}]^2$.

3. Which is left out of an equilibrium expression?
   - A. Gases
   - B. Dissolved ions
   - C. Solids and pure liquids
   - D. Products
   - Answer: C. Solids and pure liquids. Their concentration does not change, so they are left out.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/equilibrium-constant/#quiz