Question

Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined with an excess of O2 and ignited in a bomb calorimeter. After the reaction, it was found that the temperature of the calorimeter had increased from 25.000C to 26.061C. The calorimeter contained 1.000 kg of water. The heat capacity of the calorimeter was 4.643 kJ/C. Determine the heat of reaction, in kJ/mol propane. The reaction was:

Answer 1

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

$q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ$

Moles of propane =$\frac{1.860 g}{44 g/mol}=0.0422 mol$

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

$\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol$