Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:

2 S(s) + 3 O2(g) → 2 SO3(g) ΔH°rxn = ?
Given:
SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ
2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -197.8 kJ

a. -494.6 kJ
b. -692.4 kJ
c. -791.4 kJ
d. 1583 kJ

The answer is -791.5 kJ (option c)

Explanation:

You know:

SO₂ (g) → S (s) + O₂ (g) ΔH°rxn = +296.8 kJ

2 SO₂ (g) + O₂ (g) → 2 SO₃ (g) ΔH°rxn = -197.8 kJ

You must add them to obtain the desired equation:

2 S(s) + 3 O₂(g) → 2 SO₃(g) ΔH°rxn = ?

You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. The calculation is made using Hess's law. This law states: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps.

In Hess's law he explains that the enthalpy changes are additive, ΔHneta = ΣΔHr and contains three rules:

The sum of the fitted equations should give the problem equation. In this case:

2*( S(s) +  O₂(g)  → SO₂ )     To obtain the desired reaction, this equation must be inverted, so the enthalpy value is also inverted.  It must also be multiplied by 2, then the whole equation is multiplied, both reactants and products and the value of the enthalpy.  So ΔH°rxn = (-296.8 kJ)*2= -593.6 kJ

2 SO₂(g) + O₂(g) → 2 SO₃ (g)               ΔH°rxn = -197.8 kJ

2 S(s) + 3 O₂(g) → 2 SO₃(g)    ΔH°rxn =  -791.4 kJ ( Enthalpies are added algebraically)

The answer is -791.5 kJ (option c)

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