The steam reforming of methane occurs according to the following chemical equation:

Question

CH₄(g) + 2H₂O(g) ⇌ CO₂(g) + 4H₂(g) Use the following equations to solve for the enthalpy change of steam reforming of methane: H₂(g) + ½O₂(g) → H₂O(g) ∆H° = -242 kJ/mol 2H₂(g) + C(s) → CH₄(g) ∆H° = -75 kJ/mol C(s) + O₂(g) → CO₂(g) ∆H° = -394 kJ/mol

Answer 1

Hess's Law

Target equation: CH₄(g) + 2H₂O(g) <==> CO₂(g) + 4H₂(g)

Given:

(1) H2(g) + 1/2 O2(g) ==> H2O(g) .. ∆H =-242 kJ/mol

(2) 2H2(g) + C(s) ==> CH4(g) .. ∆H = -75 kJ/mol

(3) C(s) + O2(g) ==> CO2(g) .. ∆H = -394 kJ/mol

Procedure:

reverse (1) and x2: 2H2O(g) ==> 2H2(g) + O2(g) .. ∆H = +484 kJ/mol

reverse (2): CH4(g) ==> 2H2(g) + C(s) .. ∆H = +75 kJ/mol

copy (3): C(s) + O2(g) ==> CO2(g) .. ∆H = -394 kJ/mol

Add all 3 together and combine/cancel like terms:

2H2O(g) + CH4(g) + C(s) + O2(g) ==> 2H2(g) + O2(g) + 2H2(g) + C(s) + CO2(g)

2H2O(g) + CH4(g) ==> 4H2(g) + CO2(g) TARGET EQUATION

∆Hrxn = 484 kJ/mol + 75 kJ/mol + (-394 kJ/mol)

∆Hrxn = 165 kJ/mol