Question

When 1.365 g of anthracene, C14H10, is combusted in a bomb calorimeter that has a water jacket containing 500.0 g of water, the temperature of the water increases by 25.89°C. Assuming that the specific heat of water is 4.18 J/(g ∙°C), and that the heat absorption by the calorimeter is negligible, estimate the enthalpy of combustion per mole of anthracene.

Answer 1

The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)

Further explanation

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

Heat released by anthracene= Heat absorbed by water

Heat absorbed by water =

`\tt Q=500* 4.18* 25.89=54110.1~J`

mol of anthracene (MW=178,23 g/mol)

`\tt (1.365)/(178.23)=0.00766`

The enthalpy of combustion per mole of anthracene :

`\tt \Delta H=-(Q)/(n)=(54110.1)/(0.00766)=-7063981.7~J/mol\approx -7064~kJ/mol`