12.2 - Standard Enthalpy Change
- 1What standard enthalpy change is
- 2Combustion calorimetry and solution calorimetry
- 3Creating an accurate temperature-time graph
- 4Calculating enthalpy changes experimentally
Standard enthalpy change
The standard enthalpy change for a reaction (ΔH⦵) is the amount of heat transferred at constant pressure under standard conditions and states. It is measured in kilojoules per mole (kJ mol⁻¹).
The superscript plimsoll (⦵) is used after ΔH to denote that the enthalpy change refers to standard conditions:
- 298 K temperature
- 101 kPa pressure
The standard state of a substance is its physical state under standard conditions.
To determine ΔH⦵ for a reaction, you need to calculate the amount of heat released or absorbed during the reaction. This can be done by measuring the temperature change of a pure substance that receives or loses this heat.
Measuring enthalpy changes with calorimetry
We can measure the energy transferred in a reaction as heat using an apparatus called a calorimeter. This allows us to determine enthalpy changes for reactions.
There are two main types of calorimetry experiments:
- Combustion calorimetry - used to find the enthalpy change when a fuel is burned i.e. the enthalpy change of combustion.
- Solution calorimetry - used to find enthalpy changes for reactions occurring in aqueous solutions such as the enthalpy change of neutralisation.
Combustion calorimetry

To find the enthalpy change using combustion calorimetry:
- A weighed fuel sample is burnt underneath a metal canister (calorimeter) containing a known mass of water.
- As the fuel combusts, heat is transferred to the water. A thermometer measures the temperature change.
- The temperature change along with the mass and heat capacity of water are used to calculate the heat energy.
Solution calorimetry

To find the enthalpy change using solution calorimetry:
- Add a measured volume of one reactant to a polystyrene cup calorimeter and record initial temperature.
- Add a measured volume of a second reactant and quickly seal the container with a lid.
- Carefully stir the mixture with the thermometer and record temperatures at regular intervals over time as heat is released or absorbed.
Creating an accurate temperature-time graph
During a calorimetry experiment, the temperature fluctuates throughout - increasing rapidly as heat is released, reaching a peak, then decreasing gradually again.
If we just subtracted the final and initial values, we would underestimate the temperature change by missing the peak and not accounting for heat lost while measuring.
To accurately determine the temperature change:

- Plot the data - Begin by plotting the recorded temperature against time.
- Establish baseline - Draw a line of best fit for the initial temperature readings before the reaction commences. This is your baseline.
- Post-reaction line - After the peak reaction temperature is reached, draw another line of best fit for the cooling period.
- Extrapolation - Extend both lines to intersect at the reaction's start time.
- True temperature change - The difference between the two extrapolated lines of best fit at the start of the reaction gives the accurate temperature change.
Calculating enthalpy change
To measure enthalpy changes in chemical reactions, you need to use two key equations sequentially: first to calculate the heat energy change, and then to use that value to calculate the enthalpy change per mole.
First, the heat energy change can be calculated using the equation:
q = mcΔT
Where:
- q = heat energy (J)
- m = mass of water or solution (g)
- c = specific heat capacity of water, 4.18 J g^-1^ K^-1^
- ΔT = temperature change (K)
The official unit for ΔT is Kelvin (K). However, the Celsius scale gives the same numerical temperature change, so °C can also be used.
Next, use the value of q obtained from the first equation to find the enthalpy change with the equation:
$\Delta\text{H }=-\frac{\text{q}}{\text{n}}$
Where:
- ΔH = enthalpy change (kJ mol-1)
- q = heat energy (kJ)
- n = number of moles of the limiting reactant (mol)
Enthalpy changes are typically expressed in kJ mol-1 so you need to convert q from J to kJ by dividing by 1,000 before using it in the second equation.
Worked example 1 - Calculating standard enthalpy change
In a laboratory experiment, 1.45 g of an organic liquid fuel (Mr = 58.0) were completely burned in oxygen. The heat formed during this combustion raised the temperature of 100 g of water from 293 K to 372 K.
Calculate the standard enthalpy change (ΔH⦵) of this reaction.
The specific heat capacity of water, c = 4.18 J g-1 K-1.
Step 1: Calculate heat energy transferred
q = mcΔT = 100 x 4.18 x (372 - 293) = 33,022 J
Step 2: Conversion of J into kJ
To convert from J into kJ, divide by 1,000
33,022 J = 33.022 kJ
Step 3: Calculate number of moles of fuel burnt
$\text{n }=\frac{\text{m}}{\text{M}_\text{r }}=\frac{1.45}{58.0}=0.0250\text{ mol}$
Step 4: Calculate heat released per mole of fuel
$\Delta\text{H}^{\ominus}=-\frac{\text{q}}{\text{n}}=-\frac{33.022}{0.0250}=-1,320\text{ kJ mol}^{-1}$
The minus sign indicates this is an exothermic reaction, releasing heat to the surroundings.
Worked example 2 - Calculating standard enthalpy change
In a laboratory experiment, 50.0 cm3 of 1.00 mol dm-3 hydrochloric acid (HCl) at 25.0°C is mixed with 50.0 cm3 of 1.00 mol dm-3 sodium hydroxide solution (NaOH) also at 25.0°C. The heat formed during this neutralisation raised the temperature of the mixture to 31.5°C.
Calculate the standard enthalpy change (ΔH⦵) of this reaction.
The specific heat capacity of the mixture, c = 4.18 J g-1 K-1.
The density of the mixture, ρ = 1.00 g cm-3
Step 1: Calculate heat energy transferred
q = mcΔT = (50.0 + 50.0) x 4.18 x (31.5 - 25.0) = 2,717 J
Step 2: Conversion of J into kJ
To convert from J into kJ, divide by 1,000
2,717 J = 2.717 kJ
Step 3: Conversion of cm3 into dm3
To convert from cm3 into dm3, divide by 1,000
50.0 cm3 = 0.0500 dm3
Step 4: Calculate number of moles of HCl used
n = c x V = 1.00 x 0.0500 = 0.0500 mol
Step 5: Calculate heat released per mole of HCl
$\Delta\text{H}^{\ominus}=-\frac{\text{q}}{\text{n}}=-\frac{2.717}{0.0500}=-54.3\text{ kJ mol}^{-1}$
The minus sign indicates this is an exothermic reaction, releasing heat to the surroundings.