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

To solve this type of question, you will need to use the half equations given and manipulate them to re-create the given chemical equation. You can flip them backwards/change the arrow sign and then change the enthalpy from a negative value to a positive. You can also "double up" a half-equation to get the correct coefficients and then multiply the enthalpy by 2. I think of it as puzzle pieces that you have to flip around or multiply in order to get everything to add up or cancel. I always start with the bigger element (C) before I try to balance the O and H.

CH₄ + 2 H₂O ---> CO₂ + 4 H₂

-------------------------------------------------------------------------------------

C + O₂ ------> CO₂ -394 kj/mol (to get the CO₂ on the right)

CH₄ ------> 2 H₂ + C +75 kj/mol (I flipped it around to get the CH₄ on the left)

2 H₂O ------> 2 H₂ + O₂ +484 kj/mol (I flipped it and doubled it to get the 2 H₂O on the left)

You will notice that there were extra elements in those half-equations that are not in the final equation (C and O₂) These will end up canceling out because there is a C on the left in the 1st line and one one the right in the 2nd line. Same for O₂, which has a reactant on the 1st line and a product on the 3rd line). Now, you just add them all up to get that same final equation. Add up your enthalpy values.

-394 + 75 +484 = 165 kj/mol