Answer to Question #151873 in Physical Chemistry for Nancy M

Question

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.229 g
g of naphthalene burns in a bomb calorimeter, the temperature rises from 24.604 ∘C to 30.460 ∘C.

Find ΔrU for the combustion of naphthalene. The heat capacity of the calorimeter, determined in separate experiment, is 5.112 kJ∘C−1.

Answer 1

The heat produced when the naphthalene burns can be calculated from the heat capacity of the calorimeter, 5.112 kJ/°C and the temperature change (30.460-24.604)°C:

$Q = c∆T = 5.112·(30.460-24.604) = 29.94$ kJ.

The bomb calorimetry condition is constant volume condition. Therefore, the heat released corresponds to the internal energy change $∆U$ . To calculate $∆_rU$ , the heat must be divided by the number of the moles of the naphthalene heated.

The number of the moles $n$ of naphthalene is its mass divided by its molar mass $M = 128.17$ g/mol:

$n = \frac{m}{M} = \frac{1.229}{128.17}=0.009589$ mol.

Finally, the internal energy change of the reaction is (the negative sign indicates that the reaction is exothermic):

$\Delta_rU =- \frac{Q}{n} =- \frac{29.94}{0.009589} =- 3122$ kJ/mol.

Answer: $\Delta_rU$ for the combustion of naphthalene is -3122 kJ/mol.