Question

Quinine is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A 0.1964-g sample of quinone (C6H4O2) is burned in a bomb calorimeter with a heat capacity of 1.56kJ/C. The temperature of the calorimeter increases by 3.2 degrees C. Calculate the energy of combustion of quinone per gram and per mole.

Answer 1

Answer:

The heat of combustion is -25 kJ/g = -2700 kJ/mol.

Explanation:

According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.

Qcomb + Qcal = 0

Qcomb = - Qcal

The heat absorbed by the calorimeter can be calculated with the following expression.

Qcal = C × ΔT

where,

C is the heat capacity of the calorimeter

ΔT is the change in temperature

Then,

Qcomb = - Qcal

Qcomb = - C × ΔT

Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ

Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:

$\frac{-5.0kJ}{0.1964g} =-25kJ/g$

The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:

$\frac{-25kJ}{g} .\frac{108g}{1mol} =-2700kJ/mol$