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 -3377.0 kJ/mol. When 1.609 g of compound A (molar mass = 103.91 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 7.713 degree C. Using this data, what is the heat capacity (calorimeter constant) of the calorimeter? Suppose a 3.575 g sample of a second compound, compound B, was combusted in the same calorimeter, and the temperature rose from 25.69 degree C to 31.58 degree C. What is the heat of combustion per gram of compound B?

Explanation / Answer

1)

number of mol of compound burnt,

n = mass / molar mass

= 1.609/103.91

= 0.0155 mol

heat released,

q = n*delta H

= 0.0155 mol * (-3377.0 KJ/mol)

= -52.29 KJ

This heat is absorbed by calorimeter

use

|q| = C*delta T

52.29 KJ = C * 7.713 oC

C = 6.780 KJ/oC

Answer: 6.780 KJ/oC

2)

Q = C*delta T

= 6.780 KJ/oC * (31.58 - 25.69) oC

= 39.93 KJ

This is heat released by substance

heat of combustion = -39.93 KJ/3.575 g

= -11.17 KJ/g

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