Enthalpy (Chemistry)

Question

Calculate the enthalpy of the following chemical reaction:

CS₂(ℓ) + 3O₂(g) ---> CO₂(g) + 2SO₂(g)

Given:

C(s) + O2(g) ---> CO2(g)	ΔH = −393.5 kJ/mol
S(s) + O2(g) ---> SO2(g)	ΔH = −296.8 kJ/mol
C(s) + 2S(s) ---> CS2(ℓ)	ΔH = +87.9 kJ/mol

Answer 1

You should be able to use Hess' Law by now. Here, without explanation is the solution.

CS2(l) ==> C(s) + 2S(s) ... ∆H = -87.9

2S(s) + 2O2(g) ==> 2SO2(g) ... ∆H = -593.6

C(s) + O2(g) ==> CO2(g) ... ∆H = -393.5 kJ

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CS2(l) + 2S(s) + 2O2(g) + C(s) + O2(g) ==> C(s) + 2S(s) + 2SO2(g) + CO2(g)

CS₂(l) + 3O₂(g) ==> 2SO₂(g) + CO₂(g) TARGET EQUATION

∆H = -1075 kJ