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Question

The combustion of liquid quinone (C₆H₄O₂) produces gaseous carbon dioxide and liquid water according to the

following equation:  C₆H₄O₂(l) + 6O₂(g) → 6CO₂(g) + 2H₂O(l)

 When a 2.200 g sample of liquid quinone is combusted with excess oxygen gas in a bomb calorimeter, the

temperature of the calorimeter increases from 23.44^oC to 30.57^oC. In a separate experiment the heat capacity of

the calorimeter was measured to be   7.854 kJ.^oC^–1. Calculate the heat of reaction for the combustion of a

mole of quinone in this calorimeter.

Answer 1

The heat absorbed by the calorimeter is the heat released by the combustion.

The heat absorbed (q) is equal to the change in temperature (∆T) times the heat capacity of the calorimeter (C). In equation form this is

q = C∆T

q = 7.854 kJ/º x 7.13º = 55.999 kJ

This is the heat from 2.200 g of C₆H₄O₂ (molar mass 108 g/mol)

To find q from 1 mole, we convert as follows:

2.200 g x 1 mol/108 g = 0.0204 moles C₆H₄O₂

55.999 kJ / 0.0204 mol = 2745 kJ/mol