Question

Calculate the enthalpy change for the following chemical reaction:

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

1) C (s) + O2 (g) —> CO2 (g)

ΔH°= -393.5 kJ

2) S (s) + O2 (g) —> SO2 (g)

ΔH°= -296.8 kJ

3) C (s) + 2S (s) —> CS2 (l)

ΔH°= +87.9 kJ

Answer 1

eq - equation

(1^st eq): C(s) + O₂(g) → CO₂(g), ∆H = −393.5 kJ/mol

(2^nd eq): S(s) + O₂(g) → SO₂(g), ∆H = −296.8 kJ/mol

(3^d eq): C(s) + 2S(s) → CS₂(l), ∆H = +87.9 kJ/mol

 

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g), ∆H_x = ???

 

Solution:

According to Hess's law, the heat of reaction depends upon initial and final conditions of reactants and does not depend of the intermediate path of the reaction.

 

1) Modify the three given equations to get the target equation:

(1^st eq): do nothing. We need one mole of CO₂(g) on the product side and that's what we have.

(2^nd eq): multiply by two. We need two mole of SO₂(g) on the product side.

(3^d eq): flip it so as to put CS₂(l) on the reactant side.

2) Rewrite all three equations with the changes made (including changes in their enthalpy):

(1^st eq): C(s) + O₂(g) → CO₂(g), ∆H₁ = −393.5 kJ/mol

(2*^nd eq): 2S(s) + 2O₂(g) → 2SO₂(g), ∆H₂ = 2 × (−296.8) = −593.6 kJ/mol

(3*^deq): CS₂(l) → C(s) + 2S(s), ∆H₃ = (−1) × (+87.9) = −87.9 kJ/mol

3) Cancel out the common species on both sides:

C(s) ⇒ (1^st equation) & (3*^d equation)

2S(s) ⇒ (2*^nd equation) & (3*^d equation)

Thus, adding modified equations and canceling out the common species on both sides, we get:

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

4) Add the ΔH values of (1^st), (2*^nd eq) and (3*^d) equations to get your answer:

∆H_x = ∆H₁ + ∆H₂ + ∆H₃ = (−393.5 kJ/mol) + (−593.6 kJ/mol) + (−87.9 kJ/mol) = −1075 kJ/mol

∆H_x = −1075 kJ/mol

Therefore, the enthalpy change for the following chemical reaction is −1075 kJ/mol

Answer: The enthalpy change for the reaction is −1075 kJ/mol