Question

(8 marks) A chemist studies the four chemical reactions
1. CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
2. 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l)
3. CO2(g) + H2(g)  CO(g) + H2O(g)
4. C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
With corresponding enthalpy of reactants and products:

Reaction Enthalpy of reactants Enthalpy of products
1. –74.8 kJ/mol –965.1 kJ/mol
2. –184 kJ/mol –1716 kJ/mol
3. –393.5 kJ/mol –353.1 kJ/mol
4. +38 kJ/mol –1285 kJ/mol

Arrange the reactions in order of increasing order of H. Explain the theory that allows you to complete this task.

Answer 1

First compute ΔH = H(products) − H(reactants) for each reaction:

- Reaction 1: ΔH1 = −965.1 − (−74.8) = −890.3 kJ mol−1
- Reaction 2: ΔH2 = −1716 − (−184) = −1532 kJ mol−1
- Reaction 3: ΔH3 = −353.1 − (−393.5) = +40.4 kJ mol−1
- Reaction 4: ΔH4 = −1285 − (+38) = −1323 kJ mol−1

Arrange in increasing order (most negative → most positive):

Reaction 2 (−1532) < Reaction 4 (−1323) < Reaction 1 (−890.3) < Reaction 3 (+40.4).

Theory used: ΔH is the enthalpy change of the process and equals the difference between product and reactant enthalpies. Enthalpy is a state function (H depends only on initial and final states), so ΔH can be obtained directly by subtraction; this is a consequence of the first law of thermodynamics and Hess’s law (path independence of enthalpy). Negative ΔH = exothermic, positive ΔH = endothermic.