Chemistry 114
First Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. According to Wikipedia the ‘air’ pressure on Mars is 0.6 kPa. Convert his pressure to:
A. (4 points) Atm
B. (4 points) mmHg
C.(4 points) psi
2. (12 points) The steel liquid propane tank you hook to your barbeque contains 20 lbs, or
9.1 kg of C3H8. What is the volume of propane gas that I can release from this tank in
Spearfish, when the pressure is .88 atm and the temperature is 35oC.
1
3.In the following 4 situations, give me the equation you would use to solve the problem,
but DO NOT SOLVE! (Hint: the answer is NOT PV=nRT)
A. (3 points) The initial pressure is 5 atm and the volume is 6 liters, what is the final volume
if the pressure is changed to 7 atm?
Variables: P&V Constants:n,R,T
P1V1=nRT=P2V2
B. (3 points) The initial pressure is 500 psi when the temperature is 300K, what is the final
pressure when the temperature changed to 500K?
Variables:P&T
PV=nRT;
P1/T1=nR/V=P2/T2
Constants: n,R,V
C. (3 points) If you have .5 moles of a gas at 50 kPa pressure, what is the new pressure
if you add .2 more moles of gas?
Variables: n & P
PV=nRT
P1/n1=RT/V=P2/n2
Constants: T,V,R
D. (3 points) If you have 15 moles of a gas at 500 torr and 500K , what is the new pressure
if you remove 5 moles of gas and increase the temperature to 700K?
Variables: n,P,&T Constants: R,V
PV=nRT
P1/n1T1=R/V=P2/n2T2
4. (12 points) What is the density of Br2 gas at STP?
Quick and Dirty:
At STP (1 atm & 273K) 1 mole of gas occupies 22.42 liters
1 mole of Br2 gas has a mass of 79.9g x 2 = 159.8g
Density = 159.8g/22.42l = 7.13 g/l
Long way:
P/RT ×molar mass = density
Density = 1atm/(.08206l@atm/K@mol × 273K) ×159.8g = 7.13g/mol
2
5. Define the following terms (2 points each)
Heat of formation - Energy gained or lost when a compound in its standard state if
forms from its elements in their standard state.
Molar Bond energy - The energy required to break 1 mole of a given type of
chemical bond.
Hess’s Law - If you add two or more chemical reactions together to create a new
overall chemical reaction, you can find the ÄH of that new, overall reaction by adding the
ÄH’s of the individual reactions together.
Heat Capacity - The amount of heat required to raise the temperature of an object
1 C (or 1K).
o
Specific Heat Capacity - The amount of heat required to raise the temperature of
one gram of material by 1oC (or 1K).
Bomb Calorimeter - A calorimeter that is made of steel so its volume does not
change. It is used to measure the ÄU of a reaction, since heat measured at a constant
volume equals ÄU.
6. If my math is right, a butane cigarette lights contains 4 g of butane which corresponds
to about .001 moles. This should correspond to 22.4 mLs of gas at STP.
A. (6 points) Calculate the work involved in expanding a gas from 0 to 22.4 mLs final
volume. (At STP)
W= -PÄV
= -1atm ×.0224L = -.0224l@atm
-.0224 l@atm x 101.3J/l@atm = -2.27J
B. (6 points) When burned, .001 moles of butane will release -2.87 kJ of heat
energy. What is ÄU for the process of burning .001 moles of butane?
ÄU=q+w
= -2870 J + (-2.3J)
= -2872J
3
7. (12 points) An earlier problem dealt with propane, C3H8. Use the following table of ÄHo f
values to determine ÄHrxn for the reaction C3H8(g) + 5O2(g) 63CO2(g) + 4H2O(g)
Substance
ÄHof(kJ/mol)
C3H8(g)
-103.8
CO2(g)
-393.5
H2O(l)
-285.8
H2O(g)
-241.8
ÄHrxn =3npÄHfp - 3nrÄHfr
Note: O2 is en elemental form, so ÄHfo is zero and I do not have it in the table.
=[3(-393.5) + 4(-241.8)]-[1(-103.8) + 5(0)]
= -1180.5 - 967.2 + 103.8
= -2043.9 kJ
8. (12 points) I am going to drop a 500 gram block of aluminum that is at 99oC into 500
mLs of water that is at 1oC. Assuming adiabatic conditions, what is the final temperature
of the system?
(The specific heat capacity of aluminum is .83 J/mol@K, the specific heat capacity of water
is 4.184 J/mol@K, and the density of water is 1g/mL)
Adiabatic means ÄH=0
so 0 = Heat lost when aluminum cools + Heat gained as water warms
0 = S.H.C.al ×mass Al ×ÄTAl + S.H.C.H2O ×mass H2O ×ÄTH2O
0= [0.83 ×500×(TF-99)] + [4.184×500×(TF - 1)]
-[0.83 ×500×(TF-99)] = [4.184×500×(TF - 1)]
-0.83 ×500×(TF-99) = 4.184×500×(TF - 1)
-415 (TF-99) = 2092 (TF - 1)
-415 TF + 41085 = 2092TF - 2092
41085+ 2092 = 2092TF + 415TF
43177 = 2507TF
TF = 431775/2507 = 17.2oC
4