Question

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature after the combustion was 25.000 C. This was an adiabatic calorimeter. The heat capacity of the bomb, the water around it, and the contents of the bomb before the combustion was 10 000 J K 1. Calculate fH for C6H6(l) at 298.15 K from these data. Assume that the water produced in the combustion is in the liquid state and the carbon dioxide produced in the combustion is in the gas state.

Answer 1

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol