Styrene, c8h8, is one of the substances used in the production of synthetic rubber. when styrene burns in oxygen to form carbon dioxide and liquid water under standard-state conditions at 25°c, 42.15 kj are released per gram of styrene. find the standard enthalpy of formation of styrene at 25°c.
(given: ? h°f[co2(g)] = –393.5 kj/mol, ? h°f[h2o(l)] = –285.8 kj/mol, ? h°f[h2o(g)] = –241.8 kj/mol)

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:

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

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explanation: