Answer:
131.29 kJ
Explanation:
Let's consider the following balanced equation.
H₂O(g) + C(graphite)(s) ⇄ H₂(g) + CO(g)
Given the standard enthalpies of formation (ΔH°f), we can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(CO(g)) - 1 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C(graphite)(s))
ΔH°r = 1 mol × (0 kJ/mol) + 1 mol × (-110.53 kJ/mol) - 1 mol × (-241.82 kJ/mol) - 1 mol × (0 kJ/mol)
ΔH°r = 131.29 kJ
