Use the standard enthalpies of formation values shown in the table below to calculate the enthalpy of the reaction shown below:

Question

2SO2(g)+2H2O(ℓ) ---> 3O2(g)+2H2S(g)

Substance

ΔH°f (kJ/mol)

SO2(g)

84 kJ/mol

H2O(ℓ)

83 kJ/mol

O2(g)

0 kJ/mol

H2S(g)

50 kJ/mol

Answer 1

Reactants:
2SO2(g) + 2H2O(ℓ)

ΔH°f = (2 * 84 kJ/mol) + (2 * 83 kJ/mol)
ΔH°f = 334 kJ/mol

Products:
3O2(g) + 2H2S(g)

ΔH°f = (3 * 0 kJ/mol) + (2 * 50 kJ/mol)
ΔH°f = 100 kJ/mol

Enthalpy of reaction:
ΔH°rxn = ΣΔH°f (products) - ΣΔH°f (reactants)
ΔH°rxn = (3 * 0 kJ/mol) + (2 * 50 kJ/mol) - (2 * 84 kJ/mol) - (2 * 83 kJ/mol)
ΔH°rxn = -469 kJ/mol

Therefore, the enthalpy of the reaction 2SO2(g) + 2H2O(ℓ) ---> 3O2(g) + 2H2S(g) is -469 kJ/mol.