2. Calculate the enthalpy of the following chemical reaction: Dalta H rxn =?
CS (l) + 3O2(8) ---> CO2(g) + 2SO (g)
Given:
C(s) + O2(g) ---> CO2(g)
S(s) + O2(g) --> SO2(g)
C(s) + 2S(s)---> CS (1)
DaltaH = -393.5 kJ/mol
DaltaH = -296.8 kJ/mol
DaltaH = +87.9 kJ/mol

The enthalpy of the chemical reaction: ΔH rxn = -1075 kJ/mol

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways

Reaction

CS₂ (l) + 3O₂(g) ---> CO₂(g) + 2SO₂ (g)

1. C(s)+O₂(g)⇒CO₂(g) ΔH=-393.5 kJ/mol

2. S(s)+O₂(g)⇒SO₂(g)  ΔH=-296.8 kJ/mol x 2

2S(s)+2O₂(g)⇒2SO₂(g)  ΔH=-593.6 kJ/mol

3. C(s)+2S(s)⇒CS₂(l)  ΔH=+87.9 kJ/mol

Reverse

CS₂(l)⇒C(s)+2S(s)  ΔH=-87.9 kJ/mol

Add all and remove the same compound on different sides

C(s)+O₂(g)⇒CO₂(g) ΔH=-393.5 kJ/mol

2S(s)+2O₂(g)⇒2SO₂(g)  ΔH=-593.6 kJ/mol

CS₂(l)⇒C(s)+2S(s)  ΔH=-87.9 kJ/mol

+

CS₂ (l) + 3O₂(g) ---> CO₂(g) + 2SO₂ (g) ΔH=-1075 kJ/mol

14 days is the half-life of phosphorus-32