At constant volume, the heat of combustion of a particular compound, compound A,

Question

At constant volume, the heat of combustion of a particular compound, compound A, is -3359.0 kJ/mol. When 1.409 g of compound A (molar mass 122.05 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 7485 Using this data, what is the heat capacity (calorimeter constant) of the calorimeter? Number Suppose a 3.103 g sample of a second compound, compound B, was combusted in the same calorimeter, and the temperature rose from 23.55°C to 28.80 cc. What is the heat of combustion per gram of compound B? Number 0 kJ / g

Explanation / Answer

Given that;

Heat of combustion = -3359.0 KJ/ mole

Amount of compound A = 1.409 g

Molar mass of A = 122.05 g / mole

First calculate the number of moles of compound A:

Number of mole = amount in g/ molar mass

= 1.409 g/ 122.05

= 0.0115 mole

energy released = -3359 kJ/mole * 0.0115 mole

= - 38.6285 kJ
negative sign indicates the energy release only
Heat capacity of the calorimeter = total amount of energy/ temperature change

= 38.6285 kJ / 7.485 deg C = 5.161 kJ/deg C

Part 2

Total amount of energy = Heat capacity of the calorimeter * temperature change

= 5.161 kJ/deg C * (28.80 -23.55) deg C

= 27.095 kJ

Combustion per gram = total energy / compound din g

= 27.095 kJ/ 3.103 g

= 8.73 kJ/ g

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