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:
[tex]\frac{-5.0kJ}{0.1964g} =-25kJ/g[/tex]
The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:
[tex]\frac{-25kJ}{g} .\frac{108g}{1mol} =-2700kJ/mol[/tex]
Based on the dat provided, the heat of combustion is -25 kJ/g = -2,749.68 kJ.
The heat of combustion of a substance is the amount of heat evolved when one mole of that substance reacts with oxygen completely.
Based on the law of conservation of energy, the sum of the heat evolved by the combustion reaction and the heat absorbed by the bomb calorimeter equals zero.
Therefore,
Qcomb = - Qcal
The heat absorbed by the calorimeter is given by;
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
Qcomb = -5.0 kJ
This is the heat released by 0.1964 g
Then, the heat released quinone per gram of quinome is:
-5.0 kJ/0.1964 = -25.46 kJ
Molar mass of quinone (C₆H₄O₂) is 108 g/mol.
Therefore,
Heat of combustion per mole = 25.46 × 108 = -2,749.68 kJ
Heat of combustion per mole = 2,749.68 kJ
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