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Benzene (C6H6) burns according to the following balanced equation: C6H6(l)+152O2(g)→6CO2(g)+3H2O(g) Part A Calculate ΔH∘rxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid benzene is 49.1 kJ/mol.)

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Answer:

ΔHrxn = -3135.5 KJ

Explanation:

Given data:

Standard enthalpy of formation of benzene = 49.1 Kj/mol

Enthalpy of reaction = ?

Solution:

Chemical equation:

C₆H₆ + 15/2 O₂  →    6CO₂ + 3H₂O

The standard enthalpies of CO₂ and water are:

ΔHF of CO₂ = -393.5 kj/mole × 6 moles = -2361 kj

ΔHF of H₂O = -241.8 kj/mole × 3 moles = -725.4 KJ

ΔHrxn = Product - Reactant

ΔHrxn = (ΔHF CO₂ + ΔHF H₂O) - ΔHF C₆H₆

ΔHrxn = -2361 kj -725.4 KJ - 49.1 Kj

ΔHrxn = -3135.5 KJ

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