Question
Suppose that 1.048 g of benzene is combusted in a bomb calorimeter which contains 945 g of
water. The temperature of the water increases from 23.640 OC to 32.681 oC. The heat capacity of the
empty calorimeter is 891 J/ OC .
C6H6 (l) + 15/2 O2 (g) > 3 H2O (l) + 6 CO2 (g)
(a) Calculate DE for the combustion of benzene in kJ/mol .
(b) Calculate DH for the combustion of benzene in kJ/mol. Assume that the volumes of all liquids are
negligible compared to the volumes of the gases and that the temperature is 298 K.
water. The temperature of the water increases from 23.640 OC to 32.681 oC. The heat capacity of the
empty calorimeter is 891 J/ OC .
C6H6 (l) + 15/2 O2 (g) > 3 H2O (l) + 6 CO2 (g)
(a) Calculate DE for the combustion of benzene in kJ/mol .
(b) Calculate DH for the combustion of benzene in kJ/mol. Assume that the volumes of all liquids are
negligible compared to the volumes of the gases and that the temperature is 298 K.
Answers
At constant volume you calculate dE.
dE = [(mass H2O x specific heat H2O x (Tfinal-Tinitial)] + Ccal*(Tfinal-Tinitial)
Then dE x (g C6H6/molar mass C6H6) = J/mol. Convert to kJ/mol
For dH, use dE = dH + w and w can be obtained from pdV = delta n(RT)
You will need to assign the proper sign to work.
dE = [(mass H2O x specific heat H2O x (Tfinal-Tinitial)] + Ccal*(Tfinal-Tinitial)
Then dE x (g C6H6/molar mass C6H6) = J/mol. Convert to kJ/mol
For dH, use dE = dH + w and w can be obtained from pdV = delta n(RT)
You will need to assign the proper sign to work.