A bomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods. In an experiment, a 0.5967 g sample of 9,10-anthracenedione (C14H8O2) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.367×103 g of water. During the combustion the temperature increases from 24.91 to 27.68 °C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined in a previous experiment to be 765.5 J/°C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of 9,10-anthracenedione based on these data. C14H8O2(s) + 15O2(g) 4 H2O(l) + 14 CO2(g) + Energy

Molar heat of combustion of 9,10-anthracenedione = 6268 kJ

Explanation:

Heat lost by the 9,10-anthracenedione = heat gained by water and the calorimeter

Heat gained by water = mCΔT

m = 1.367×10³ g

C = 4.184 J/g°C

ΔT = 27.68 - 24.91 = 2.77°C

Heat gained by water = (1367×4.184×2.77)

= 15,843.1 J

Heat gained by the calorimeter = c ΔT

c = 765.5 J/°C.

ΔT = 27.68 - 24.91 = 2.77°C

Heat gained by the calorimeter = 765.5 × 2.77 = 2,120.435 J

Heat lost by the 9,10-anthracenedione = heat gained by water and the calorimeter

Heat lost by the 9,10-anthracenedione = 15,843.1 + 2,120.435 = 17,963.535 J

Note that 0.5967 g of the compound was burnt to give 17,963.535 J of heat energy.

To calculate how much heat will be given off by combusting 1 mole of the compound, we convert the sample amount to moles.

Number of moles = (Mass)/(Molar mass)

Mass of 9,10-anthracenedione = 0.5967 g

Molar mass of 9,10-anthracenedione = 208.216 g/mol

Number of moles of 9,10-anthracenedione burnt = (0.5967/208.216) = 0.002866 moles.

Burning 0.002866 moles gives off 17,963.535 J of heat energy.

Burning 1 mole will give off (1×17,963.535)÷(0.002866) J of heat energy.

That is, 6,267,807 J

Molar heat of combustion of 9,10-anthracenedione is the heat given off by burning 1 mole of 9,10-anthracenedione = 6,267,807 J = 6268 kJ

Hope this Helps!!!

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.

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