Question

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

4 S(s) + 6 O2(g) → 4 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

Group of answer choices

-494.6 kJ

-692.4 kJ

-791.4 kJ

1583 kJ

-1583 kJ

Answer 1

Step-by-step explanation:

To find the enthalpy change of the given reaction, we can use Hess's Law which states that the total enthalpy change of a reaction is independent of the pathway between the initial and final states.

The given equation can be obtained by adding the following reactions:

S(s) + O2(g) → SO2(g) ΔH° = -296.8 kJ/mol

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

Note that the first reaction needs to be reversed to get S(s) on the left side, and then the two equations are added to obtain the desired reaction:

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

S(s) + O2(g) → SO2(g) reverse: SO2(g) → S(s) + O2(g) ΔH° = +296.8 kJ/mol

4 S(s) + 6 O2(g) → 4 SO3(g) ΔH° = [4 x (-197.8)] + [296.8] = -791.4 kJ/mol

Therefore, the enthalpy change of the given reaction is -791.4 kJ/mol. Answer: **-791.4 kJ**