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 –3891.0 kJ/mol. When 1.731 g of compound A (molar mass = 101.61 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 7.441 °C. Using this data, what is the heat capacity (calorimeter constant) of the calorimeter?

Suppose a 3.021 g sample of a second compound, compound B, was combusted in the same calorimeter, and the temperature rose from 25.01 °C to 29.38 °C. What is the heat of combustion per gram of compound B?

Explanation / Answer

Solution-

Mass of compound A=1.731g

Moles of compound A=Mass/molar mass=1.731/101.61

=0.01703 mol

So the Heat released in reaction=Moles of compound*heat of combustion

                                =0.01703*3891.0 =66.26 kJ

Let heat capacity of calorimeter be C.

C*change in temperature=Heat released in reaction

C*7.441=66.26

Calorimeter constant, C=8.90 kJ/C

b)

Let heat of combustion be h

Mass*h=Calorimeter constant*Change in temperature

3.021*h=8.90*(29.38-25.01)

h=12.87 kJ/g

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