CCEA A-Level Life and Health Sciences 0008
Experimental Techniques
Human Body Systems
Physical Chemistry in Industrial Processes
Brain ScienceDouble Award
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Scientific Method, Investigation, Analysis and Evaluation
Organic Chemistry
Medical PhysicsA2 option
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Physical Chemistry in Industrial Processes
Hess's law and average bond enthalpies
- 3 things to know
- AS 3
Overview
Conservation of energy and Hess's law, enthalpy cycles from combustion and formation data, average bond enthalpies, and worked calculations for ammonia, hydrogen and hydrogen chloride.
What you need to know
Answer each one before you reveal it.
State Hess's law.
How does Hess's law follow from the conservation of energy?
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the principle of conservation of energy.
Model answer
Energy cannot be created or destroyed, only transferred from one form to another. The total energy of a reaction and its surroundings stays constant.
- 2
State Hess's law.
Model answer
Hess's law states that the enthalpy change for a reaction is independent of the route taken. The change depends only on the initial and final states.
- 3
How does Hess's law follow from the conservation of energy?
Model answer
Two routes between the same reactants and products must involve the same total energy change. A different total would create or destroy energy on a round trip.
- 4
Why are enthalpy cycles needed for reactions such as C(s) + ½O₂(g) → CO(g)?
Model answer
The enthalpy change cannot be measured directly because some carbon dioxide always forms as well. An enthalpy cycle calculates it from combustion data instead.
- 5
Use the enthalpies of combustion of carbon (−393.5 kJ mol⁻¹) and carbon monoxide (−283.0 kJ mol⁻¹) to find ΔH for C(s) + ½O₂(g) → CO(g).
Model answer
The enthalpy change is −110.5 kJ mol⁻¹. The working is −393.5 − (−283.0) = −110.5 kJ mol⁻¹.
- 6
Calculate ΔH for CaCO₃(s) → CaO(s) + CO₂(g), given enthalpies of formation of −1207 kJ mol⁻¹ for CaCO₃, −635 kJ mol⁻¹ for CaO and −394 kJ mol⁻¹ for CO₂.
Model answer
The enthalpy change is +178 kJ mol⁻¹. The working is (−635 + −394) − (−1207) = +178 kJ mol⁻¹.
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