Question

Calculate the enthalpy of combustion of liquid benzene given that enthalpy of formation of benzene, CO2 and H2O are 48.5 KJ, -393.5 KJ and -285.83 KJ respectively.

Please explain with clear steps.​

Answer 1

Answer:

Enthalpy of combustion of liquid benzene is $-3266.99 KJ$

Explanation:

The balanced chemical equation for the combustion of liquid benzene is

$C_{6}H_{6} +\frac{15}{2} O_{2}$   →   $6CO_{2} +3 H_{2} O$

Enthalpy of combustion is the difference between the sum of enthalpy of formation of products and the sum of enthalpy of formation of reactants.

Enthalpy of formation of benzene, carbondioxide and water are 48.5KJ, -393.5KJ and -285.83KJ

That is,  Δ$H^{0}$ = Δ$H_{f} ^{0}(products) -$ Δ $H_{f} ^{0}(reactants)$

                     $= ((6*-393.5)+(3*-285.83))-(48.5+\frac{15}{2} *0)$

                     $=-3266.99 KJ$