Use Hess’ Law and equations 1, 2, and 3 to find ΔH for the formation of methane from chloroform CHCl3(l) + 3HCl(g) → CH4(g) + 3Cl2(g)

Question

(1) ½H2(g) + ½Cl2(g) → HCl(g) ΔH° = -92.3 kJ

(2) C(s) + 2H2(g) → CH4(g) ΔH° = -74.8 kJ

(3) C(s) + ½H2(g) + 3/2 Cl2(g) → CHCl3(l) ΔH° = -134.5 kJ

Answer 1

(1) ½ H₂ (g) + ½ Cl₂ (g) → HCl (g) ΔH° = -92.3 kJ

(2) C (s) + 2 H₂ (g) → CH₄ (g) ΔH° = -74.8 kJ

(3) C (s) + ½ H₂ (g) + 3/2 Cl₂ (g) → CHCl₃(l) ΔH° = -134.5 kJ

In the formation of methane from chloroform, we need to reverse equation (3) since it has chloroform as a product, and we want it as a reactant. When you reverse this reaction, the sign of ΔH° will change from - to +.

(3) CHCl₃ (l) → C (s) + ½ H₂ (g) + 3/2 Cl₂ (g) ΔH° = +134.5 kJ

Equation (2) will remain as is since methane is on the product side.

Reverse equation (1) since HCl has to be on the reactant side.

(1) HCl (g) → ½ H₂ (g) + ½ Cl₂ (g) ΔH° = +92.3 kJ

Multiply equation (1) by a factor of 3 since we need 3 moles of HCl in the formation of methane.

(1) [ HCl (g) → ½ H₂ (g) + ½ Cl₂ (g) ΔH° = +92.3 kJ ] x 3

(1) 3 HCl (g) → 3/2 H₂ (g) + 3/2 Cl₂ (g) ΔH° = +276.9 kJ

Add all 3 equations, you get

(1) 3 HCl (g) → 3/2 H₂ (g) + 3/2 Cl₂ (g) ΔH° = +276.9 kJ

(2) C (s) + 2 H₂ (g) → CH₄ (g) ΔH° = -74.8 kJ

(3) CHCl₃ (l) → C (s) + ½ H₂ (g) + 3/2 Cl₂ (g) ΔH° = +134.5 kJ

3 HCl (g) + C (s) + 2 H₂ (g) + CHCl₃ (l) → 3/2 H₂ (g) + 3/2 Cl₂ (g) + CH₄ (g) + C (s) + ½ H₂ (g) + 3/2 Cl₂ (g)

The net reaction after canceling and adding the coefficients of the common reactants / products will give:

CHCl₃ (l) + 3 HCl (g) → CH₄ (g) + 3 Cl₂ (g) ΔH° = +276.9 kJ - 74.8 kJ + 134.5 kJ = +336.6 kJ