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, we need to consider the balanced chemical equation for the formation of benzene:
C6H6 (liquid) + 15/2 O2 (g) -> 6 CO2 (g) + 3 H2O (g)
From the balanced equation, we can see that the formation of benzene involves the combustion of 15/2 moles of O2.
Given:
- The standard heat of combustion of liquid benzene is 40.145 kJ/g.
- The heats of formation of carbon dioxide and water vapor are 393.509 and 241.818 kJ/mol, respectively.
To calculate the enthalpy change for the formation of benzene, we need to calculate the heat of formation of benzene (ΔHf). We can use the following equation:
ΔHf = (ΔHc * MWbenzene) - [(nCO2 * ΔHfCO2) + (nH2O * ΔHfH2O)]
Where:
- ΔHf = heat of formation of benzene
- ΔHc = standard heat of combustion of liquid benzene (40.145 kJ/g)
- MWbenzene = molecular weight of benzene (C6H6) = (6 * MC) + (6 * MH)
MC = atomic weight of carbon = 12.01 g/mol
MH = atomic weight of hydrogen = 1.008 g/mol
- nCO2 = moles of CO2 produced (6 moles)
- ΔHfCO2 = heat of formation of carbon dioxide (393.509 kJ/mol)
- nH2O = moles of H2O produced (3 moles)
- ΔHfH2O = heat of formation of water vapor (241.818 kJ/mol)
Substituting the values into the equation:
MWbenzene = (6 * 12.01) + (6 * 1.008) = 78.114 g/mol
ΔHf = (40.145 kJ/g * 78.114 g/mol) - [(6 moles * 393.509 kJ/mol) + (3 moles * 241.818 kJ/mol)]
Calculating:
ΔHf = 3126.96807 kJ - [(6 * 393.509 kJ) + (3 * 241.818 kJ)]
= 3126.96807 kJ - [2361.054 kJ + 725.454 kJ]
= 3126.96807 kJ - 3086.508 kJ
= 40.46 kJ
Therefore, the enthalpy change for the formation of benzene from the elements at 25°C and one bar is approximately 40.46 kJ.