- Define and explain Enthalpy Changes in your own words
- Use key terms such as enthalpy change (ΔH) accurately
- Apply what you have learned to new examples and questions
- Avoid the common mistakes learners make with this topic
Why do some reactions release heat while others absorb it, and how fast do they go? Physical chemistry answers with energy: enthalpy changes measured in kilojoules per mole, collision theory explaining rates, and equilibrium constants predicting where reactions settle. In this chapter you will calculate, predict and explain at A-level depth.
This lesson focuses on Enthalpy Changes: measure heat changes with calorimetry and standard enthalpy definitions.
Measure heat changes with calorimetry and standard enthalpy definitions.
Key ideas
Enthalpy changes are measured — or found with Hess's law
In a simple calorimetry experiment, the heat released warms a known mass of water: q = mcΔT, with c = 4.18 J/g/K for water. Dividing q by the moles of fuel burned gives ΔH in kJ/mol — but heat escapes to the apparatus and air, so experimental values are always less exothermic than data-book values. When a change cannot be measured directly, Hess's law comes to the rescue: the total enthalpy change is independent of the route, so you can add and subtract known enthalpy changes around an energy cycle to find the unknown one.
Equilibria respond to change
At dynamic equilibrium the forward and backward rates are equal, so concentrations stay constant even though both reactions continue. Le Chatelier's principle predicts the response to change: raising the temperature shifts equilibrium in the endothermic direction, and raising pressure favours the side with fewer moles of gas. Kc quantifies the position: a large Kc means the equilibrium lies far to the right.
Key term — enthalpy change (ΔH): The heat energy change of a reaction at constant pressure, in kJ/mol; negative for exothermic reactions.
Burning 0.0500 mol of a liquid fuel raises 200 g of water from 20.0 °C to 32.5 °C. Calculate the enthalpy change of combustion per mole. (c = 4.18 J/g/K).
Temperature rise ΔT = 32.5 − 20.0 = 12.5 K. Heat to water: q = mcΔT = 200 × 4.18 × 12.5 = 10 450 J = 10.45 kJ. Moles of fuel = 0.0500 mol, so ΔH = −q ÷ moles = −10.45 ÷ 0.0500 = −209 kJ/mol. The sign is negative because the reaction is exothermic (heat is released).
Answer: ΔHc ≈ −209 kJ/mol (the true value is more exothermic because of heat losses to the apparatus).
- Forgetting the minus sign on exothermic ΔH Exothermic enthalpy changes are always negative — heat leaving the system. Writing +209 kJ/mol for a combustion is simply wrong, however good the arithmetic.
- Saying a catalyst changes the position of equilibrium A catalyst speeds up both forward and backward reactions equally, so equilibrium is reached faster but Kc and the final concentrations do not change.
Practice
ΔHc = −25.0 ÷ 0.0200 = −1250 kJ/mol.
The peak shifts to higher energy and flattens (fewer molecules at the most probable energy); the total area stays the same, and the area beyond the activation energy grows.
ΔHf = (−394) + 2(−286) − (−890) = −394 − 572 + 890 = −76 kJ/mol.
Particles move faster, colliding more often, but the key effect is that many more collisions have energy at or above the activation energy, so far more collisions are successful.
Quick check
Which of these best defines "enthalpy change (ΔH)"?
For N₂ + 3H₂ ⇌ 2NH₃, equilibrium concentrations are [N₂] = 0.50, [H₂] = 1.20, [NH₃] = 0.80 mol/dm³. Calculate Kc.
- Enthalpy Changes: measure heat changes with calorimetry and standard enthalpy definitions.
- Enthalpy changes are measured — or found with Hess's law: In a simple calorimetry experiment, the heat released warms a known mass of water: q = mcΔT, with c = 4.18 J/g/K for water.
- activation energy: The minimum energy particles need for a collision to cause reaction.
- Watch out for: forgetting the minus sign on exothermic ΔH