Understanding Standard Enthalpy Change and Heat Calculations
What is Standard Enthalpy Change ($ΔH^\ominus$)?
Enthalpy change
The enthalpy change (ΔH) of a reaction quantifies the heat exchanged at constant pressure during a chemical or physical change.
- Chemical reactions are accompanied by energy transfers between the system (reactants and products) and the surroundings.
- When this measurement is conducted under standard conditions, defined as 298 K (25°C), 100 kPa, and reactants/products in their standard states, it is referred to as the standard enthalpy change($ΔH^\ominus$).
- Exothermic reactions: Release heat into the surroundings, resulting in a negative $ΔH$ (e.g., combustion of fuels).
- Endothermic reactions: Absorb heat from the surroundings, resulting in a positive $ΔH$ (e.g., melting of ice).
Units of Enthalpy Change
- The standard unit for $\Delta H$ is kilojoules per mole (kJ mol⁻¹).
- This unit reflects the energy change associated with one mole of reactant or product under standard conditions.
- Always ensure your final answer for $\Delta H$ is in kJ mol⁻¹.
- If $Q$ is calculated in joules, convert it to kilojoules by dividing by 1000.
Calculating Heat Transfer $Q$
The heat transferred during a reaction can be calculated using the formula: $$Q = mc\Delta T$$ where:
- $Q$: Heat transferred (in joules, J)
- $m$: Mass of the substance being heated (in grams, g)
- $c$: Specific heat capacity of the substance (in J g⁻¹ K⁻¹)
- $\Delta T$: Temperature change (in kelvin, K, or degrees Celsius, °C)
Breaking Down the Formula
- Mass $m$: This is the mass of the substance (often water in calorimetry experiments) that absorbs or loses heat.
- Specific Heat Capacity $c$: A property of the substance that indicates how much heat is required to raise the temperature of 1 g of the substance by 1 K. For water, $c = 4.18 \, \mathrm{J \ g^{-1} \ K^{-1}}$.
- Temperature Change $\Delta T $: Calculated as $\Delta T = T_{\text{final}} - T_{\text{initial}}$.
Suppose 50 g of water is heated from 20°C to 30°C. How much heat is absorbed by the water?
Solution
$$Q = mc\Delta T$$
$$Q = (50 \, \mathrm{g})(4.18 \, \mathrm{J g^{-1} K^{-1}})(30 - 20 \, \mathrm{K})$$
$$Q = 2090 \, \mathrm{J} \, \text{(or 2.09 kJ)}$$



