Calorimetry (AQA A Level Chemistry): Revision Note

You must report calculations to an appropriate number of significant figures. Calculated results can only be reported to the limits of the least accurate measurement.

In the example below, the least accurate measurement is the temperature, which is two significant figures, so the final answer is adjusted to the same.

Calculating the energy released in combustion

In a calorimetry experiment, 2.50 g of methane is burnt in excess oxygen.

30% of the energy released during the combustion is absorbed by 500 g of water, the temperature of which rises from 25 °C to 68 °C

The specific heat capacity of water is 4.18 J g^-1 °C^-1  

What is the total energy released per gram of methane burnt?

Answer

Step 1: Calculate the temperature change

             ΔT (of water) = 68 °C - 25 °C

                                  = 43 °C

Step 2: Calculate the heat energy transferred (q)

q = m x c x ΔT

q = 500 x 4.18 x 43

       = 89 870 J

Step 3: Calculate the total heat energy transferred

This is only 30% of the total energy released by methane

Total energy x 0.3 = 89 870 J

Total energy = 299 567 J

Step 4: Calculate the energy transferred per gram

This is released by 2.50 g of methane

Energy released by 1.00 g of methane = 299 567 ÷ 2.50

= 119 827 J = 120 000 J

= 120 kJ g^-1 (2 significant figures)

A student mixes 50.0 cm³ of 1.00 mol dm^-3 hydrochloric acid with 50.0 cm^-3 of 1.00 mol dm^-3 sodium hydroxide in a polystyrene cup.

The temperature rises from 20.5°C to 27.3°C.

Assume:

• Density of solution = 1.00 g cm^-3

• Specific heat capacity, c = 4.18 J g⁻¹ K^-¹

• No heat is lost to the surroundings.

Calculate the enthalpy change of neutralisation (ΔH) in kJ mol⁻¹.

Answer

Step 1: Calculate the temperature change

$\mathrm{\Delta }$T= 27.3 − 20.5 = 6.8 °C

Since a temperature difference in °C is the same as in K,

$\mathrm{\Delta }$T= 6.8 K

Step 2: Calculate the mass of solution

Total volume

50.0+50.0 = 100.0 cm³

Using the density:

mass = 100.0 g

Step 3: Calculate the heat energy transferred (q)

Use

q=mc$\mathrm{\Delta }$T

Substitute the values:

q = 100.0 × 4.18 × 6.8

q = 2842.4 J

or

q = 2.84 kJ

Step 4: Calculate the number of moles reacting

Reaction:

HCl + NaOH $\to$ NaCl + H2​O

Moles of HCl:$\frac{50.0}{1000}$​×1.00 = 0.0500 mol

The reaction is 1 : 1, so

Moles of NaOH: 0.0500 mol

Step 5: Calculate ΔH

Heat released per mole:

$\mathrm{\Delta H}=\frac{-2.84}{0.0500}$

= −56.8 kJ mol⁻¹

Aqueous solutions of acids, alkalis, and salts are assumed to be largely water, so you can just use the m and c values of water when calculating the energy transferred.

When there is a rise in temperature, the value for ΔH becomes negative, suggesting that the reaction is exothermic, and when the temperature falls, the value for ΔH becomes positive, suggesting that the reaction is endothermic.

Students will not be expected to recall the value of the specific heat capacity, c (4.18 J g⁻¹ °C⁻¹ for water).