Chemistry 152 Discussion Section. Have the students work through as many probles as
possible. What they do not finish be sure to go over with them. Problems like these will
be on the exam.
For several problems the following standard state properties are useful:
-1274.45
0
-393.51
-285.84
-276.98
218.16
29.36
37.13
75.30
111.96
1) The respiratory system uses oxygen to degrade glucose to carbon dioxide and water.
The net equation is
a) Calculate the enthalpy of reaction at 310K. Assume all heat capacities
are constant between 298K and 310K.
Solution: At T=298K
Then
Finally:
b) Assume respiration occurs at constant pressure. Calculate the amount of
heat produced by respiration per mole of oxygen absorbed at 298K.
Solution:
c) An average person uses about 1.2 moles of oxygen per hour in the
course of respiring. Calculate the amount of heat produced by respiration
per kilogram per hour by an average person. Assume a weight of 70 kg.
Solution:
d) Most of the heat produced by the body is lost. But suppose that a person
were clothed in a thermally insulated suit such that all the heat from
respiration were retained by the body. After 10 hours of respiration, how
much would the body temperature of the person described in part c rise?
Assume a heat capacity of 4.18kJ kg-1 K-1.
Solution:
2) Certain yeast can degrade glucose into ethanol and carbon dioxide in a process called
alcoholic fermentation according to the equation:
a) Calculate the enthalpy of reaction. You can find appropriate heats of
formation in your text appendices.
Solution:
b) Calculate the amount of heat produced at constant pressure and at T=298K per
mole of glucose fermented.
Solution:
c) Calculate the work done per mole glucose fermented as a result of the CO2 gas
expanding against a constant external pressure of 1 atm, and at T=298K.
Assume carbon dioxide behaves ideally and that the molar volume of an ideal
gas at T=298K and P=1 atm is V=24.45Lmol-1. Give your answer in Joules.
Solution:
3)
d) Calculate the energy change ΔE when one mole of glucose ferments at
T=298K and 1 atm.
Solution:
A super-heated liquid is a liquid that exists above its normal boiling temperature.
When a superheated liquid vaporizes it does so irreversibly. The heat that is
absorbed is not the usual heat of vaporization ΔHvap, which is applicable only at
the normal boiling point. With these remarks in mind, calculate ΔHsys for the
vaporization of one mole of superheated liquid water at a temperature of 383 K
and a pressure of 1 atm. At the normal boiling point T=373 K the enthalpy of
vaporization of water is ΔHvap=40,700 J/mole. For liquid water the molar heat
capacity at constant pressure CP=75 J mole-1 K-1. For water vapor CP=33.5 J mole1 -1
K . Hint: Describe a three step path. For each step calculate ΔH. Assume all
heat capacites are constant from 373-383 K.
Solution: You can solve this by setting up a thermodynamic cycle:
4) 18 grams of water ice at 230 K are placed in thermal contact with a large heating
block maintained at a temperature of 350 K. Calculate the enthalpy change of the water
ΔH, when 18 grams of water ice at 230 K are converted to liquid water at 350K. For ice
Cp=37.7 J/K. For liquid water Cp=75 J/K. Assume the heat capacity odof ice is constant
between 230K and 273K. Assume the heat cvapacity of liquid water is constant between
273k and 350K. For water ΔHfusion=6.01 kJ/mole at T=273K. The external pressure is 1
atm.
.
Solution: