CHEM 341
PHYSICAL CHEMISTRY
EXAM 1
Do not open this exam until
told to do so. The exam consists
of 5 pages, including this one. Count
them to insure that they are all there.
Constants and
conversions:
R = 8.31 J K-1
mole-1 NA
= 6.02 x 1023 MWwater
= 18
CP (liquid
water) = 75.3 J K-1 mole-1 CP (ice) = 37.0 J K-1 mole-1
DHfus (water) = 6.01 kJ/mole 1 atm = 101325 Pa = 101325 J/m3
Page |
Score |
2 |
/24 |
3 |
/38 |
4 |
/19 |
5 |
/19 |
Total |
/100 |
Conceptual Problems
(each problem is worth 4 points).
1) Pressure can be thought of in terms of:
a) Force per unit area
b) Energy per unit volume
c)
a and b
d) none of the above
2) The absolute temperature of a system is a measure of:
a) The density of molecules
b) The heat entering the system
c)
The average kinetic
energy of the molecules
d) The work a system is capable of performing
3) If heat is transferred reversibly from the environment to a system and nothing else changes, the entropy change of the system is:
a)
Positive
b) Negative
c) Zero
d) Can’t tell from the data given
4) We say a system is adiabatic when:
a) No energy or matter can enter or leave the system
b) The volume of the system is fixed
c) Heat freely passes between the system and the surroundings
d)
No heat can enter or leave the system
5) An endothermic reaction is one in which:
a)
Heat is taken up by the reaction
b) Heat is released by the reaction
c) Heat freely passes between the system and the surroundings
d) No heat can enter or leave the system
6) For a spontaneous reaction, which of the following is true:
a) DSSys > 0
b) DHSys < 0
c)
DSSys > q/T
d) DG > 0
7) At constant temperature and pressure, the Gibbs energy change (DG) of the system is always proportional to:
a) -DSSys
b) -DSSur
c)
-DSTot
d) DSSys -DSSur
8) In the equation DG = DH - TDS (at constant pressure and temperature):
a) DG represents the Gibbs energy change of the system
b) DG represents the Gibbs energy change of the surroundings
c)
DG represents the Gibbs energy change of the whole
universe
d) DG represents the Gibbs energy change of the molecules in the reaction itself
Numerical Problems
(each problem is worth 14 points). All work must be shown for credit.
9)
What is the pressure of 0.1 moles of nitrogen
gas (assume it is ideal) in a 10 liter container at 25 C?
P = nRT/V = (0.1 moles)(8.31 J/(K
mole))(298K)/10x10-3 m3 = 24.8 kPa
The only hard part is getting the
units right. Remember that a joule is a
kg m2/s2 so you have to make sure your volume units are
in terms of meters.
10) What is the
molar heat capacity of a 2 mole sample of liquid that rose in temperature by 10
C when supplied with 560 J of heat?
q = nCpDT so Cp =
q/(nDT) = 560 J/(2 mole x 10 K) = 28 J/(K mole)
11) Calculate the
standard enthalpy of reaction for 2H2O2 (l) à 2H2O(l)
+ O2(g) at 50 C. The standard
enthalpy of the reaction at 25 C is –196.0 kJ/mole (this is per mole of O2). The molar heat capacity at constant pressure
of H2O2 (l) is 89.1 J K-1 mole-1,
for H2O(l) is 75.3 J K-1 mole-1 and for O2(g)
is 29.4 J K-1 mole-1
Cool 2 moles H2O2 to 25 C: DH = nCpDT = (2 mole) (89.1 J K-1 mole-1) (-25K) = -4455 J
Perform the reaction at 25 C: DH =
-196,000 J
Heat 2 moles H2O to 50 C:
DH = nCpDT = (2 mole) (75.3 J K-1 mole-1) (25K) = 3765 J
Heat 1 mole O2 to 50 C:
DH = nCpDT = (1 mole) (29.4 J K-1 mole-1) (25K) = 735 J
Total enthalpy = DH =
-195.9 kJ/mole (per mole molecular oxygen that is).
12) Consider the reversible, isothermal expansion of 6 moles of an ideal gas with an initial volume of 15 liters to 150 liters. The external pressure is 1 atm. The temperature is 25 C. Determine DU, w, q, DSSys, DSSur, DSTot.
Since it is a reversible, isothermal
expansion, we know that DSTot
= 0 = DSSys + DSSur
Since it is isothermal and an ideal
gas, we know that DU = 0 = w + q
w = -nRTln(Vf/Vi)
= -34200 J
q = -w = 34200 J
DSSys = nRln(Vf/Vi) = 115 J/K
DSSur = -DSSys = -115 J/K
Quasi Real World
Problem (12 points)
13) A large backyard
swimming pool has about 100,000 liters of water in it. In
To cool 100,000 liters by 5 C
requires:
q = nCpDT
n = (100,000 liters) x (1000
g/liter) / (18 g/mole) = 5.55 x 106 moles
Cp = 75.3 J/(K mole)
DT = 5K
so q = 2.09 x 109 J
How much water do we have to
evaporate to remove this much heat?
q = nDHvap
n = q/DHvap = (2.09 x 109
J)/(44,000 J/mole) = 47,500 moles of water to evaporate.
This is 856,000 grams or 856 liters
of water.