The standard heat of combustion of liquid benzene (giving water
vapor) is 40.145 kJ/g at 25°C. the heats of formation of carbon
dioxide and water vapor are 393.509 and 241.818 kJ/mol,
respectively, at 25°C and one bar pressure. Calculate the enthalpy
change for the formation of benzene (C6H6) from the elements at
25°C and one bar
To calculate the enthalpy change for the formation of benzene (C6H6) from the elements at 25°C and one bar pressure, we need to determine the balanced chemical equation for the formation of benzene and use the heat of combustion and heats of formation values.
The balanced chemical equation for the combustion of benzene is:
C6H6(l) + O2(g) -> 6CO2(g) + 3H2O(g)
From the equation, we can see that the formation of benzene from the elements can be represented as:
6CO2(g) + 3H2O(g) -> C6H6(l) + O2(g)
The enthalpy change for this reaction can be calculated using Hess's law. Hess's law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for a series of reactions that lead to the overall reaction.
The enthalpy change for the formation of benzene can be calculated as follows:
Enthalpy change = (6 x enthalpy of formation of CO2) + (3 x enthalpy of formation of H2O) - enthalpy of combustion of benzene
Given:
Enthalpy of formation of CO2 = 393.509 kJ/mol
Enthalpy of formation of H2O = 241.818 kJ/mol
Enthalpy of combustion of benzene = 40.145 kJ/g
First, we need to convert the enthalpy of combustion of benzene from kJ/g to kJ/mol.
The molar mass of benzene (C6H6) = (12.01 g/mol x 6) + (1.008 g/mol x 6) = 78.11 g/mol
Enthalpy of combustion of benzene = 40.145 kJ/g x (78.11 g/mol) = 3133.225 kJ/mol
Plugging these values into the equation, we get:
Enthalpy change = (6 x 393.509 kJ/mol) + (3 x 241.818 kJ/mol) - 3133.225 kJ/mol
Enthalpy change = 2361.054 kJ + 725.454 kJ - 3133.225 kJ
Enthalpy change = - 46.717 kJ
Therefore, the enthalpy change for the formation of benzene from the elements at 25°C and one bar pressure is -46.717 kJ.