Answer:
ΔH°f(C₈H₈(l)) = 98.80 kJ/mol
Explanation:
Let's consider the combustion of styrene.
C₈H₈(l) + 10 O₂(g) → 8 CO₂(g) + 4 H₂O(l)
42.15 kJ are released per gram of styrene, that is, -42.15 kJ/g. In the balanced equation there is 1 mole of C₈H₈, so the heat released per mole is:
[tex]\Delta H\°rxn = \frac{-42.15kJ}{g} .\frac{104.15g}{mol} =-4390kJ/mol[/tex]
In order to find the standard enthalpy of formation of styrene (ΔH°f(C₈H₈(l))) we will use the following expression.
ΔH°rxn = Σ np . ΔH°f(p) - Σ nr . ΔH°f(r)
where,
ni: moles of reactants and products
ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₈H₈(l)) - 10 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₈H₈(l)) = 8 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l)) - 10 mol × ΔH°f(O₂(g)) - ΔH°rxn
1 mol × ΔH°f(C₈H₈(l)) = 8 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol) - 10 mol × 0 kJ/mol - (-4390 KJ)
ΔH°f(C₈H₈(l)) = 98.80 kJ/mol
