Q.16. Calculate the enthalpy of formation of Cols (1) given that the e

1.	Q.16. Calculate the enthalpy of formation of Cols (1) given that the enthalpy of combustion of benzene is-3267.7 kJ and the enthalpies of format ion of CO-(e) and H20(0) are 393.3Kj and -286.6 kJrespectively
Answer» Calculate enthalpy of formation for benzene, C6H6, from the following data.2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l) Enthalpy of combustion of C6H6 = -6534 kJEnthalpy of formation (H2O) = -285.8 kJEnthalpy of formation (CO2) = -393.5 kJ delta H for the reaction = [sum of heat of formation for products] - [sum of heat of formation for reactants] (-6534) = [(12)(-393.5) + (6)(-285.8)] - (2X) (-6534) = -6436.8 - 2X 2X = 97.2 X = +48.6 kJ, which is the heat of formation for benzene.

Q.16. Calculate the enthalpy of formation of Cols (1) given that the enthalpy of combustion of benzene is-3267.7 kJ and the enthalpies of format ion of CO-(e) and H20(0) are 393.3Kj and -286.6 kJrespectively

Answer»

Calculate enthalpy of formation for benzene, C6H6, from the following data.2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l)

Enthalpy of combustion of C6H6 = -6534 kJEnthalpy of formation (H2O) = -285.8 kJEnthalpy of formation (CO2) = -393.5 kJ

delta H for the reaction = [sum of heat of formation for products] - [sum of heat of formation for reactants]

(-6534) = [(12)(-393.5) + (6)(-285.8)] - (2X)

(-6534) = -6436.8 - 2X

2X = 97.2

X = +48.6 kJ, which is the heat of formation for benzene.