Answer to Question #145975 in General Chemistry for Ella MacIsaac

Question

Using equations 1), 2) and 3), calculate the change in the enthalpy of the formation of methane, CH4, using chloroform, CHC13. CHC13 () + 3HCI (g) → CH4 (g) + 3C12 (g) 1) H2 (g) 2) Č (s) + 2H2 (g) HCl8) AH ° = -92.3 kJ CH4 (g) AH = -74.8 kJ + Cl2 (g) 3) C (s) + Hz1g) + Cl2) → CHC13 () 3) C (s) + H2 (g) + Cl2 (g) AH = -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

So it becomes,

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

We use all 3 equations, are

(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

∆H° = +336.6 kJ

the change in the enthalpy of the formation of methane, CH4, using chloroform, CHC13, ∆H° = +336.6 kJ.