Question

Mothballs are composed primarily of naphthalene (C10H8). When 1.025 g of naphthalene burns in a bomb calorimeter, the temperature rises from 24.25 C to 32.33 C. (The Heat Capacity of the Calorimeter is 5.11 kJ/C.) Find DeltaE (in x 103 kJ) for the combustion of naphthalene.

Answer 1

-5.16 × 10³ kJ/mol

Step-by-step explanation:

According to the law of conservation of energy, the sum of the heat released by the combustion of naphthalene (Qcomb) and the heat absorbed by the bomb calorimeter (Qcal) is zero.

Qcomb + Qcal = 0

Qcomb = -Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression:

Qcal = Ccal × ΔT

where,

• Ccal: heat capacity of the calorimeter

• ΔT: change in the temperature

Qcal = Ccal × ΔT = 5.11 kJ/°C × (32.33°C-24.25°C) = 41.3 kJ

From [1],

Qcomb = -41.3 kJ

41.3 kJ are released upon the combustion of 1.025 g of naphthalene (MW 128.17). The change in the internal energy (ΔE) is:

`\Delta E = (-41.3kJ)/(1.025g) * (128.17g)/(mol) =-5.16 * 10^(3) kJ/mol`