One gram of liquid benzene is burned in a bomb calorimeter. the temperature before ignition was 20.826 ℃, and the temperature after the combustion was 25.000 ℃. 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 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.

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

the answer would be b i believe