Question

The standard enthalpy of formation of benzoic acid, C₆H₅COOH, is −385kJ mol⁻¹ at 298 K. Calculate the standard enthalpy of combustion of benzoic acid at this temperature, given that the standard enthalpy of formation of liquid water, H₂O is −285.8 kJ mol⁻¹ and gaseous carbon dioxide, CO₂, is −393.51 kJ mol⁻¹.

A		−294.3 kJ mol−1
B		−3997kJ mol−1
C		−3227.0 kJ mol−1
D		2282.2 kJ mol−1

Answer 1

The balance reaction isL:

2 C6H5COOH + 15 O2 —> 14 CO2 + 6 H2O

we have:

Hof(C6H5COOH) = -385.0 KJ/mol

Hof(O2) = 0.0 KJ/mol

Hof(CO2) = -393.51 KJ/mol

Hof(H2O) = -285.8 KJ/mol

we have the Balanced chemical equation as:

2 C6H5COOH + 15 O2 ---> 14 CO2 + 6 H2O

deltaHo rxn = 14*Hof(CO2) + 6*Hof(H2O) - 2*Hof( C6H5COOH) - 15*Hof(O2)

deltaHo rxn = 14*(-393.51) + 6*(-285.8) - 2*(-385.0) - 15*(0.0)

deltaHo rxn = -6453.94 KJ

This is when 2 moles of C6H5COOH reacts

So,

enthalpy of combustion = -6453.94 KJ / 2 mol = -3227.0 KJ/mol

Answer: C