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.
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.
1 answer
1 answer