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.

a) -341 kJ/mol
b) -511 kJ/mol
c) -671 kJ/mol
d) -839 kJ/mol

Answer 1

Using Hess's Law, the enthalpy change for the steam reforming of methane is calculated by reversing and manipulating given equations and summing their enthalpy changes. The correct enthalpy change is found to be -165 kJ/mol which does not match any of the given options.

Step-by-step explanation:

To calculate the enthalpy change for the steam reforming of methane, we need to apply Hess's Law. This principle states that the total enthalpy change for a reaction is the sum of all changes and is independent of the multiple stages of a reaction. Given the following individual reactions, we can find the enthalpy change for the overall reaction.

• H₂(g) + ½O₂(g) → H₂O(g) ΔH° = -242 kJ/mol
• 2H₂(g) + C(s) → CH₄(g) ΔH° = -75 kJ/mol
• C(s) + O₂(g) → CO₂(g) ΔH° = -394 kJ/mol

To solve, reverse the second equation and adjust the coefficients of the first equation, so they match the steam reforming reaction. Then, sum the enthalpy changes:

• 2H₂O(g) → 2H₂(g) + O₂(g) ΔH° = 2 x 242 kJ/mol = 484 kJ/mol (reversed and multiplied by 2)
• CH₄(g) → 2H₂(g) + C(s) ΔH° = 75 kJ/mol (reversed)
• C(s) + O₂(g) → CO₂(g) ΔH° = -394 kJ/mol

Adding the enthalpy changes together: 484 kJ/mol + 75 kJ/mol - 394 kJ/mol = 165 kJ/mol. However, the overall reaction is endothermic, so we take the negative of the calculated value, which gives us ΔH° = -165 kJ/mol.

Therefore, the correct answer is not present in the options given, so there may be a mistake in the question or its options.