Chemistry 112
Third Hour Exam
Name:____________
1. (12 points) Hurricane Wilma had a record low pressure of 882 millibar. While this isn’t
a unit we usually use in chemistry, the conversion factor is listed on you table of conversion
factors. Calculate the pressure of Hurricane Wilma in Pascal, atm, and torr.
882 millibar x (10-3 Bar/millibar) x (105Pa/1 bar) = 8.82x104 Pa
8.82x104 Pa x (1atm/101325 Pa) = .870 atm
.870 atm x (760 torr/1 atm) = 662 torr
2. (13 points) I have 28 grams of N2 gas at STP. What is the volume of this gas?
28 g N2 x (1 mole/28.02g) = 1.00 mole N2
At STP 1 mole of gas has a volume of 22.42L, so this would be 22.42 L of N2
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3. (12 points) The van der Walls equation for a non-ideal gas is:
In this equation what are a and b, and what physical properties of the gas are they
correcting for?
(First state what a corrects for, then state what b corrects for.)
a is a correction term that corrects for the fact that gas molecules do exert some force on each
other when they collide.
b is a correction term that corrects for the fact that gas molecules to have a volume that must
be taken into account.
. (12 points) Define the following terms
Enthalpy H=E + PV, or a measure of energy that includes internal energy plus a
term that includes the pressure and volume of the system.
qv Heat flow measured when the volume of a system is constant.
Specific heat capacity The amount of heat it takes to raise the temperature of oen
gram of a material by one oC.
Constant volume calorimeter A ‘Bomb’ calorimeter. A calorimeter that is made with
a heavy steel enclosure so the volume of the system cannot change.
Intensive property A property of a system like pressure or temperature that does not
depend on the amount of material in the system.
Enthalpy of formation The amount of heat released when a compound is created
from its elements. Both the compound and the elements are assumed to be in theri
standard states.
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5. (13 points) A biology experiment requires the preparation of a water bath at 37.0oC. The
temperature of the cold tap water is 22.0oC, and the temperature of the hot tap water is
55.0oC. If a student starts with 110.0 g of cold tap water, how much hot water must be
added to reach 37.0oC. (The specific heat capacity of water is 4.183 J/oC@g)
Heat lost by hot water = heat gained by cold water
Technically one is positive and one is negative, but I will define both as positive sothe equation
works out.
Heat lost by hot water
(55-37) x 4.183 x X =
Heat gained by cold water
(37-22) x 4.183 x 110g
Note: specific heat capacity of water falls out of this equation, so you don’t really need it!
(55-37)=(37-22)110
18 X = 15(110)
X = 15(110)/18
= 91.7g
6. (13 points) Given the following data
Fe2O3(s) + 3CO(g) 6 2Fe(s) + 3CO2(g)
3Fe2O3(s) + CO(g) 6 2 Fe3O4(s) + CO2(g)
Fe3O4(s) + CO(g) 6 3FeO(s) + CO2(g)
)Ho = -25 kJ
)Ho = -39 kJ
)Ho = +16 kJ
Calculate )Ho for the reaction
FeO(s) + CO(g) 6 Fe(s) + CO2(g)
I focused on the Fe product first and took
Fe2O3(s) + 3CO(g) 6 2Fe(s) + 3CO2(g)
)Ho = -25 kJ
But coefficient for Fe is wrong so multiply by ½:
½ Fe2O3(s) + 3/2 CO(g) 6 1Fe(s) + 3/2 CO2(g)
)Ho = -25/2 kJ
Next I got rid of the Fe2O3(s) reactant using:
3Fe2O3(s) + CO(g) 6 2 Fe3O4(s) + CO2(g) )Ho = -39 kJ
To do this I need to reverse the equation
2 Fe3O4(s) + CO2(g) 63Fe2O3(s) + CO(g) )Ho = +39 kJ
And multiply by 1/6
2/6 Fe3O4(s) + 1/6 CO2(g) 63/6 Fe2O3(s) + 1/6 CO(g) )Ho = +39/6 kJ
Now I need to get rid of the Fe3O4(s) and change it to FeO using the reverse of the last equation:
3FeO(s) + CO2(g) 6Fe3O4(s) + CO(g)
)Ho = -16 kJ
and multiply by 2/6 = 1/3
3/3FeO(s) + 1/3 CO2(g) 61/3 Fe3O4(s) + 1/3 CO(g)
)Ho = -16/3 kJ
4
Let’s add all these up and see what we have:
½ Fe2O3(s) + 3/2 CO(g) 6 1Fe(s) + 3/2 CO2(g)
2/6 Fe3O4(s) + 1/6 CO2(g) 63/6 Fe2O3(s) + 1/6 CO(g)
3/3FeO(s) + 1/3 CO2(g) 61/3 Fe3O4(s) + 1/3 CO(g)
)Ho = -25/2 kJ
)Ho = +39/6 kJ
)Ho = -16/3 kJ
½ Fe2O3(s) + 3/2 CO(g)+2/6 Fe3O4(s) + 1/6 CO2(g) +3/3FeO(s) + 1/3 CO2(g)
61Fe(s) + 3/2 CO2(g) + 3/6 Fe2O3(s) + 1/6 CO(g) +1/3 Fe3O4(s) + 1/3 CO(g)
½ Fe2O3(s) + 3/2 CO(g)+1/3 Fe3O4(s) + 1/2 CO2(g) +1FeO(s)
61Fe(s) + 3/2 CO2(g) + 1/2 Fe2O3(s) + 1/2 CO(g) +1/3 Fe3O4(s)
Cancelling common terms:
1 CO(g) +1FeO(s)
61Fe(s) + 1 CO2(g)
Which is what I wanted, so adding up the )H’s
-25/2 + 39/6 -16/3 = -12.5 + 6.5 -5.333 = -11.33
7. (12 points) Calculate the frequency and energy associated with a photon of UV light at
a wavelength of 220 nm.
c=8<
c/8=<
3x108/220x10-9 = 1.36x1015 sec-1
E=h<
E=6.626x10–34J@sec x 1.36x1015 sec-1 = 9.04x10-19J
8. (13 points) Explain why the spectrum of hydrogen gas only contains a few distinct
frequencies of light, instead of a continuous spectrum of light.
The spectrum of any atom is caused by electrons in the molecule falling from high energy orbitals
into lower energy orbitals. Since quantum theory restricts the energy of individual orbitals to certain
distinct levels, the transitions are also restricted to certain distinct levels, so you only see light at
frequencies that correspond to these specific changes of energies. The real question then becomes,
why to you see continuous spectra for some elements and compounds? The answer to that question
is a bit beyond the scope of this class.