Equilibrium Law Problem Solving (HL) | HL IB Chemistry Revision Notes 2025

Equilibrium Law Problem Solving

Calculations involving K

In the equilibrium expression, each term inside a square bracket represents the concentration of that chemical in mol dm^-3
Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture
The concentrations of the reactants and products can then be calculated from the number of moles and total volume, using the equation:

$concentration (mol {dm}^{- 3}) = \frac{amount of substance (mol)}{volume ({dm}^{3})}$

Worked example

Calculating K of ethanoic acid

Ethanoic acid and ethanol react to form the ester ethyl ethanoate and water as follows:

CH₃COOH (I) + C₂H₅OH (I) ⇌ CH₃COOC₂H₅ (I) + H₂O (I)

At equilibrium, 500 cm³ of the reaction mixture contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.

Use this data to calculate a value of Kfor this reaction.

Answer:

Step 1: Calculate the concentrations of the reactants and products:

[CH₃COOH (l)]	=	$\frac{0.235}{0.500}$	=	0.470 mol dm^-3

[C₂H₅OH (l)]	=	$\frac{0.035}{0.500}$	=	0.070 mol dm^-3

[CH₃COOC₂H₅ (l)]	=	$\frac{0.182}{0.500}$	=	0.364 mol dm^-3

[H₂O (l)]	=	$\frac{0.182}{0.500}$	=	0.364 mol dm^-3

Step 2: Write out the balanced symbol equation with the concentrations of each chemical underneath:

CH₃COOH (l)	+	C₂H₅OH (l)	⇌	CH₃COOC₂H₅ (l)	+	H₂O (l)
0.470 mol dm^-3		0.070 mol dm^-3		0.364 mol dm^-3		0.364 mol dm^-3

Step 3: Write out the equilibrium constant for the reaction:

$K = \frac{[H_{2} O] [{CH}_{3} {COOC}_{2} H_{5}]}{[C_{2} H_{5} OH] [{CH}_{3} COOH]}$

Step 4: Substitute the equilibrium concentrations into the expression and calculate the answer:

$K = \frac{0.364 \times 0.364}{0.070 \times 0.470} = 4.03$

Note: The smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures

ICE tables

Some questions do not give the initial and equilibrium concentrations of all of the reactants and products
An initial, change and equilibrium (ICE) table should be used to determine the equilibrium concentration of the remaining reactants and products using the molar ratio in the stoichiometric equation
Step 1 in the worked example below shows how to use an ICE table to find the equilibrium concentrations

Worked example

Calculating K of ethyl ethanoate

Ethyl ethanoate is hydrolysed by water:

CH₃COOC₂H₅ (I) + H₂O (I) ⇌ CH₃COOH (I) + C₂H₅OH (I)

0.1000 mol of ethyl ethanoate is added to 0.1000 mol of water. A little acid catalyst is added and the mixture is made up to 1 dm³. At equilibrium 0.0654 mol of water are present. Use this data to calculate a value of K for this reaction.

Answer:

Step 1: Complete the ICE table for the reaction:
- Write out the balanced chemical equation with the number of moles of each substance given in the question beneath using an initial, change and equilibrium table:

- Calculate the change in moles of water and add to the table (an increase is shown by + and a decrease is shown by -)
  - Equilibrium amount = Initial amount + Change in amount
  - 0.0654 = 0.100 + Change in amount
  - Change in amount = 0.0654 - 0.100 = –0.0346

- Use the stoichiometry of the equation to calculate the change in amounts of the remaining reactants/products and add to the table
  - There is a 1 : 1 reacting ratio between H₂O and all other reactants/products
  - As H₂O has decreased by 0.0346 mol, the other reactant CH₃COOC₂H₅will decrease by 0.0346 mol
  - Since CH₃COOH and C₂H₅OH are products, they will both increase by 0.0346 mol

- Calculate the number of moles at equilibrium of the remaining reactants / products to complete the table
  - Equilibrium amount = Initial amount + Change in amount
  - Equilibrium amount of CH₃COOC₂H₅ = 0.100 + (-0.0346) = 0.0654 mol
  - Equilibrium amount of CH₃COOH = 0.000 + 0.0346 = 0.0346 mol
  - Equilibrium amount of C₂H₅OH = 0.000 + 0.0346 = 0.0346 mol

	CH₃COOC₂H₅ (I) +	H₂O (I) ⇌	CH₃COOH (I) +	C₂H₅OH (I)
Initial moles	0.100	0.100	0.000	0.000
Change	-0.0346	-0.0346	+0.0346	+0.0346
Equilibrium moles	0.0654	0.0654	0.0346	0.0346

Step 2: Calculate the concentrations of the reactants and products:

[CH₃COOH (l)]	=	$\frac{0.0654}{1.000}$	=	0.0654 mol dm^-3

[C₂H₅OH (l)]	=	$\frac{0.0654}{1.000}$	=	0.0654 mol dm^-3

[CH₃COOC₂H₅ (l)]	=	$\frac{0.0346}{1.000}$	=	0.0346 mol dm^-3

[H₂O (l)]	=	$\frac{0.0346}{1.000}$	=	0.0346 mol dm^-3

Step 3: Write the equilibrium constant for this reaction in terms of concentration:

$K = \frac{[C_{2} H_{5} OH] [{CH}_{3} COOH]}{[H_{2} O] [{CH}_{3} {COOC}_{2} H_{5}]}$

Step 4: Substitute the equilibrium concentrations into the expression:

$K = \frac{0.0346 \times 0.0346}{0.0654 \times 0.0654} = 0.280$

Examiner Tip

For reactions which have the same number of concentration terms in both the numerator and denominator in their equilibrium constant expression, such as the reaction in the worked example above, you do not need to know the volume to be able to calculate K.
As concentration = moles ÷ volume, the volume terms will cancel in the K expression, and K can be calculated directly from the number of moles of reactants and products at equilibrium:

$\begin{array}{rcl} K & = & \frac{(\frac{{mol}_{C}}{V}) (\frac{{mol}_{D}}{V})}{(\frac{{mol}_{A}}{V}) (\frac{{mol}_{B}}{V})} \end{array}$ = $\begin{array}{rcl} \frac{{mol}_{C} \times {mol}_{D}}{{mol}_{A} \times {mol}_{B}} \end{array}$

Calculating K when K is very small

When K < 10^-3, the reaction lies far to the left and the equilibrium mixture contains mainly reactants
The change from the initial amount of reactant to the equilibrium amount is close to zero
The initial amount of reactant and the equilibrium amount of reactants are approximately the same
Therefore the following approximation can also be made:

[reactant]_initial≈ [reactant]_equilibrium

This approximation can be used in calculations involving K

Worked example

Ammonia decomposes to produce hydrogen and nitrogen. At 450 K, the equilibrium constant, K, for the reaction is 3.85 x 10^-4.

2NH₃ (g) ⇌ 3H₂ (g) + N₂ (g)

A reaction is set up at 450 K, where the initial concentration of ammonia is 0.20 mol dm^-3. Calculate the concentration of hydrogen at equilibrium.

Answer:

Complete an ICE table. As the change in concentration is unknown, we can use algebra to find the equilibrium concentrations:
- Assign the changes in concentration according to the stoichiometry of the reaction:
  - Change in concentration of NH₃ = -2 $x$
  - Change in concentration of 3H₂ = +3 $x$
  - Change in concentration of N₂ = + $x$

	2NH₃ (g) ⇌	3H₂ (g) +	N₂ (g)
Initial (mol dm^-3)	0.20	0.00	0.00
Change (mol dm^-3)	-2 $x$	+3 $x$	+ $x$
Equilibrium (mol dm^-3)	0.20 - 2 $x$ ≈ 0.20	3 $x$	$x$

As K is small, the change in concentration is small and 0.20 $≫$ 2 $x$ , so the assumption 0.20 - 2 $x$ ≈ 0.20 is justified
Write the expression for K and substitute in the equilibrium concentration:

$K = \frac{{[H_{2}]}^{3} [N_{2}]}{{[{NH}_{3}]}^{2}}$ $= \frac{{(3 x)}^{3} x}{{(0.20)}^{2}} =$ 3.85 x 10^-4

Simplify and rearrange for $x$ :

3 $x^{4}$ = 3.85 x 10^-4x (0.20)²

3 $x^{4}$ = 1.54 x 10^-5

$x^{4}$ = 5.1333 x 10^-6

$x$ = 0.0476

Calculate the equilibrium concentration of H₂:

[H₂]_equilibrium= 3 $x$ = 0.143 mol dm^-3

Examiner Tip

When using this method, you must state in your answer the approximation that you have made and why you have been able to use it, i.e. because K < 10^-3.

Equilibrium Law Problem Solving (HL) (HL IB Chemistry)

Revision Note

Calculations involving K

Calculating K of ethanoic acid

ICE tables

Calculating K of ethyl ethanoate

Calculating K when K is very small

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