3.95 g of a hydrocarbon, C20H16, were burned in excess oxygen in a constant-volume calorimeter surrounded by

Question

3.95 g of a hydrocarbon, C20H16, were burned in excess oxygen in a constant-volume calorimeter surrounded by 1300. g of water. As a result the temperature of the water and the calorimeter rose from 21.0 oC to 25.7 oC. The heat capacity of the calorimeter was 255.642 J/oC.


Calculate DE (kJ/mol) for the combustion of the hydrocarbon based on these readings.


Enter a numeric answer only, do not include units in your answer.

Answer 1

First, calculate the number of moles of hydrocarbon: C20H16 has a molecular weight of 20*12 + 16*1 = 256 g/mol. Divide the mass by that to get the number of moles: 3.95 g / 256 g/mol = 0.01543 mol.

Then, calculate the change in internal energy. The change in temperature is ΔT = 25.7°C - 21.0°C = 4.7°C. The specific heat capacity of water is 4.184 J/(°C*g). With 1300 g of water, that gives a heat capacity of 4.184 * 1300 J/°C = 5439.2 J/°C. You can add that to the heat capacity of the calorimeter = 5439.2 + 255.642 J/°C = 5694.8 J/°C. Multiply that by the ΔT to get the energy released: 5694.8 J/°C * 4.7°C = 26766 J = 26.76 kJ. Since that energy is leaving the compound and going into the system, the value will be negative, -26.76 kJ.

To get ΔE, divide the energy being released by the number of moles: ΔE = -26.76 kJ / 0.01543 mol = -1734.6 kJ/mol