PROBLEM 5.58 CALORIMETRY
A) You are given a mass of benzoic acid and its heat of combustion. From that get
the q ( which is also the ∆H). Then use the ∆T to calculate C (heat capacity of
the calorimeter). REMEMBER – in a bomb calorimeter you do not show the
mass of the water as it never changes for the same calorimeter device,
Bubbies.
All combustions are exothermic, have a
negative sign for q
1.640 g HC2H5O2 x -26.38kJ
1g
=-
43.2632 kJ (for this 1.640 g sample)= q
qreaction = -CBOMB calorimeter∆T
The heat lost by
combustion of benzoic
acid (- ∆H) is the heat
gained by the
calorimeter.
- 43.2632 kJ
= -CBOMB (27.20-22.25)
- 43.2632 kJ
= -CBOMB (4.95 CO)
PROBLEM 5.58 CALORIMETRY (continued)
- 8.74004 kJ / Co
8.74 kJ / Co
= - CBOMB
= CBOMB
Rounded to three
significant figures,
this is the heat
capacity of the
calorimeter for the
other parts of this
problem.
B) Use the heat capacity calculated in part A) and the data from the organic
sample to calculate q. Note it is the same calorimeter and the same heat
capacity.
The answer must be in kJ per g, so:
qreaction = -CBOMB calorimeter∆T
40.9032kJ = x : x = 30.987=31.0kJ/g
1.320g
1g
qreaction = 8.74 kJ / Co (4.68)= 40.9032kJ for 1.320g