determine the enthalpy of the reaction

Question

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

This is a problem dealing with Hess's Law

TARGET EQUATION

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

GIVEN

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

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

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

PROCEDURE

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

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

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

add all 3 together and combine/cancel like terms to get CH4(g) + 2H2O(g) <==> CO2(g) + 4H2(g)

This is the target equation, so to get ∆H, simply add the appropriate values:

75 + 484 - 394 = 165 kJ/mol = ∆Hº