find the enthalpy change

Question

Use equations (1), (2), and (3) to find the enthalpy change for the formation of methane, CH 4 , from

chloroform, CHCl 3 .

CHCl 3 (l) + 3HCl(g) → CH 4 (g) + 3Cl 2 (g)

(1) 1/2 H 2 (g) + 1/2 Cl 2 (g) → HCl(g) + 92.3 KJ

(2) C(s) + 2H 2 (g) → CH 4 (g) + 74.8 KJ

(3) C(s) + 1/2 H 2 (g) + 3/2 Cl 2 (g) → CHCl 3 (l) + 134.5 KJ

Answer 1

This is an example of using Hess's Law:

Target equation: CHCl₃(l) + 3HCl(g) ==> CH₄(g) + 3Cl₂(g)

Given:

Eq.1: 1/2 H₂(g) + 1/2 Cl₂(g) ==> HCl(g) ... ∆H = 92.3 kJ

Eq.2: C(s) + 2H₂(g) ==> CH₄(g) ... ∆H = 74.8 kJ

Eq.3: C(s) + 1/2 H₂(g) + 3/2 Cl₂(g) ==> CHCl₃(l) ... ∆H = 134.5 kJ

Procedure:

reverse Eq.3: CHCl₃(l) ==> C(s) + 1/2 H₂(g) + 3/2 Cl₂(g) ... ∆H = -134.5 kJ

reverse Eq.1 x 3: 3HCl(g) ==> 3/2 H2(g) + 3/2 Cl2(g) ... ∆H = -276.9 kJ

copy Eq.2: C(s) + 2H₂(g) ==> CH₄(g) ... ∆H = 74.8 kJ

Add the tree equations and combine/cancel like terms:

CHCl₃(l) + 3HCl(g) + C(s) + 2H₂(g) ==> C(s) + 1/2 H₂(g) + 3/2 Cl₂(g) + 3/2 H2(g) + 3/2 Cl2(g) + CH₄(g)

CHCl₃(l) + 3HCl(g) ==> CH₄(g) + 3Cl₂(g) .. Target equation

∆H = -134.5 kJ + -276.9 kJ + 74.8 kJ

∆H = -337 kJ