Question

A 6.55 g sample of aniline (C₆H₅NH₂, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter. If the temperature rose by 32.9°C, use the information below to determine the heat capacity of the calorimeter.

4 C₆H₅NH₂(l) + 35 O₂(g) → 24 CO₂(g) + 14 H₂O(g) + 4 NO₂(g)

ΔH°_rxn = -1.28 x 10⁴ kJ

Answer 1

4 C₆H₅NH₂(l) + 35 O₂(g) → 24 CO₂(g) + 14 H₂O(g) + 4 NO₂(g)

ΔH°_rxn = -1.28 x 10⁴ kJ

First calculate the number of moles of aniline as follows:

(6.55 g C6H5NH2) / (93.13 g C6H5NH2/mol)= 0.070 mol

Now calculate the total ΔH°_rxn as follows:

1.28 x 10⁴ kJ/ 4 mol C6H5NH2 * 0.07033 mol = 225.056 kJ

the heat capacity of the calorimeter = total amount of heat / temperature change

(225.056 kJ) / (32.9 °C) = 6.84 kJ/°C