Enthalpy in Reversible Reactions
This topic covers the fundamental principle that the energy change of a reversible reaction in one direction is equal in magnitude but opposite in sign to the energy change in the reverse direction. This is a core concept for understanding chemical equilibrium and energy diagrams.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- In a reversible chemical reaction, the process can proceed in both forward and reverse directions.
- If the forward reaction releases energy (exothermic, ΔH is negative), the reverse reaction must absorb the same amount of energy (endothermic, ΔH is positive).
- The magnitude (the numerical value) of the enthalpy change is identical for both the forward and reverse reactions.
- This relationship can be represented as: ΔHforward = -ΔHreverse.
- On an energy level diagram, if products are at a lower energy level than reactants (exothermic), moving from products back to reactants requires an input of energy.
Formulae
ΔHforward = -ΔHreverse To find the enthalpy change of a reverse reaction when the enthalpy change of the forward reaction is known, or vice versa.
Definitions
- Reversible Reaction
- A chemical reaction that can proceed in both the forward and reverse directions, often denoted by the '⇌' symbol.
- Exothermic Reaction
- A process that releases energy, usually as heat, into the surroundings. It has a negative enthalpy change (ΔH < 0).
- Endothermic Reaction
- A process that absorbs energy, usually as heat, from the surroundings. It has a positive enthalpy change (ΔH > 0).
Worked example
The synthesis of methanol from carbon monoxide and hydrogen is a reversible exothermic reaction with an enthalpy change of -91 kJ/mol. The balanced equation is: CO(g) + 2H₂(g) ⇌ CH₃OH(g). What is the enthalpy change for the decomposition of 2 moles of methanol into carbon monoxide and hydrogen gas?
- 1
Identify the forward reaction and its enthalpy change:
CO(g) + 2H₂(g) → CH₃OH(g), ΔH = -91 kJ/mol.
This value is for the formation of 1 mole of methanol.
- 2
The decomposition of methanol is the reverse reaction:
CH₃OH(g) → CO(g) + 2H₂(g).
- 3
The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to the forward reaction.
So, for the decomposition of 1 mole of methanol, ΔH = -(-91 kJ/mol) = +91 kJ/mol.
- 4
The question asks for the enthalpy change for the decomposition of 2 moles of methanol.
- 5
Multiply the per-mole value by the number of moles:
(+91 kJ/mol) × 2 moles = +182 kJ.
Answer: +182 kJ
Common mistakes
- ×Forgetting to change the sign of the ΔH value when considering the reverse of a given reaction.
- ×Misinterpreting stoichiometric coefficients. The given ΔH value often corresponds to the molar quantities in the equation, so you may need to scale it up or down if the question asks about a different quantity.
- ×Confusing the sign convention: remember that negative ΔH means exothermic (heat exits) and positive ΔH means endothermic (heat enters).
No-calculator tips
- ✓The primary calculation is simply flipping the sign of a number (e.g., -91 becomes +91). This is a conceptual check, not a difficult sum.
- ✓If stoichiometry is involved, you will only be asked to multiply or divide by small, simple integers found in the balanced equation. For example, if ΔH is for 2 moles of product, ΔH for 1 mole is just half the value.
- ✓Pay close attention to keywords. 'Formation' or 'synthesis' implies one direction of the reaction, while 'decomposition' implies the reverse.