Answer:
ΔH°r = -483.64 kJ
Explanation:
Let's consider the following balanced equation.
2 H₂(g) + O₂(g) ⇒ 2 H₂O(g)
We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.
ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr
where
ΔH°f: standard heat of formation
n: moles
p: products
r: reactants
ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol
ΔH°r = (-241.82 kJ/mol) × 2 mol - 0 kJ/mol × 2 mol - 0 kJ/mol × 1 mol
ΔH°r = -483.64 kJ
