Solve for the enthalpy change of steam reforming methane

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

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    (i)

2H₂(g) + C(s) → CH₄(g) ∆H° = -75 kJ/mol        (ii)

C(s) + O₂(g) → CO₂(g) ∆H° = -394 kJ/mol       (iii)

Using the concepts that we can manipulate enthalpy reactions (in summary, reverse rxns, enthalpy value sign changes; change the # of moles, the enthalpy changes appropriately)

We can formulate the reaction using the given reactions

[Reverse rxn (ii) ] +[ Reverse rxn #(i) x2] + rxn (iii) gives you the reaction asked for. Adjusting the enthalpy given gives you the result

Enthalpy = +75 kJ/mol + (2x242 kJ/mol) + (-394 kJ/mol) = +165 kJ/mol

Hope this helps!