Answer:
ΔH°r = -1562 kJ
Explanation:
Let's consider the following combustion.
C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)
We can calculate the standard heat of reaction (ΔH°r) using the following expression:
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(i) are the standard heats of formation of reactants and products
The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]
ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]
ΔH°r = -1562 kJ
