Exam 3 Review SOLUTION
Ch. 5 – Gas Laws
1. A sample of carbon dioxide has a pressure of 0.975 atm at –10.0°C in a 0.500 L flask. The sample is
transferred completely to a 0.750 L flask, and allowed to come to room temperature (25.0°C). What
is the pressure inside this new flask?
P1 = 0.975 atm
V1 = 0.500 L
T1 = -10.0°C = 263.2K
P1 V1
T1
=
P2 = ? atm
V2 = 0.750 L
T2 = 250.°C = 298.2K
P2 V2
T2
P2 =
P1 V1 T2
T1 V2
=
(0.975 atm ) (0.500 L ) ( 298.2 K ) = 0.736 atm
( 263.2 K ) (0.750L )
2. A sample of an organic compound weighs 0.495 g. This sample is a gas collected in a flask that has
an exact volume of 127 mL. At 98°C, the pressure of the gas is 754 mmHg in the flask. Calculate
the molecular weight of the organic compound in the flask.
P = 754 mmHg x
V = 127 mL x
1 atm
= 0.992 atm
760 mmHg
1L
= 0.127 L
1000 L
T = 98°C + 273 = 371 K
PV = nRT
n=
(
)(
)
0.992 atm 0.127 L
PV
=
= 4.14 x 10-3 mol
RT
"
L ! atm %
$# 0.08206 mol ! K '& 371 K
Molar Mass =
(
)
#g
0.495 g
=
= 119 g/mol
# mol
4.14 x 10-3 mol
n = ? moles
3. Chlorine gas can be produced from the electrolysis of sea water, according to the following balanced
equation: [12 points]
2 NaCl (aq)  2 Na0 (s) + Cl2 (g)
How many moles of chlorine gas can be produced from 350.0 ml of sea water that is 2.50 M in
NaCl?
1 mol Cl2
1L
2.50 mol NaCl
x
x
= 0.438 mol Cl2
1000 mL
L
2 mol NaCl
I lost the portion of the problem that then relates moles of Cl2 to a volume!
350.0 mL x
4. Two flasks, one containing 1.500 L of He at 0.825 atm and the other containing 2.35 L of N2 at
0.135 atm, are connected by a stopcock. The entire system is being held at a constant 25.0°C. What
is the final pressure of the system, when the stopcock is opened and the gases are allowed to mix?
What is the mole fraction of He in this system?
PHe = 0.825 atm
PN = 0.135 atm
VHe = 1.500 L
VN = 2.35 L
2
2
T = 25.0 °C = 298.2 K
Vtotal = 1.500 L + 2.35 L = 3.85 L
We need to find new pressures of He and N 2 at the total volume, using P1 V1 = P2 V2 :
Pnew,He =
(0.825 atm ) (1.500 L ) = 0.321 atm
Ptotal = Pnew,He
! He =
Pnew,N =
(0.135 atm ) ( 2.35 L ) = 0.0824 atm
2
3.85 L
+ Pnew,N = 0.321 atm + 0.0824 atm = 0.403 atm
3.85 L
2
P
mol He
0.321 atm
= He =
= 0.797
mol total
Ptotal
0.403 atm
5. Suppose you have two rigid container of equal volume, cylinders A and B. Cylinder A contains 1.00
kg carbon monoxide, CO, and cylinder B contains 1.00 kg methane, CH4. Which cylinder has a
greater pressure at 25°C? [8 points]
In order to compare pressures, we simply need to figure out how many moles of each gas there is!
1000 g
1 mol
1000 g
1 mol
x
= 35.7 mol CO
1.00 kg CH 4 x
x
= 62.3 mol CH 4
1 kg
28.01 g
1 kg
16.04 g
Moles and pressure are directly related to one another. Since there are more moles of methane, cylinder
B has a greater pressure.
1.00 kg CO x
Ch. 6 – Thermochemistry
6. Consider the thermochemistry of CaCl2 dissolving in water, as it is commonly available for purchase
as a de-icer in winter. [25 points]
CaCl2 (s) 
Ca2+ (aq) +
2 Cl- (aq)
-795.0 kJ/mol
-542.96 kJ/mol
-167.46 kJ/mol
a) Determine the enthalpy of this reaction from the enthalpies of formation.
!H rxn = " ni!H f (products) - " ni!H f (reactants)
(
)(
) (
)(
)
(
)(
)
= #$ 1 mol Ca 2+ -542.96 kJ/mol + 2 mol Cl- -167.46 kJ/mol %& - #$ 1 mol CaCl2 -795.0 kJ/mol %&
= -82.9 kJ/mol
b) Determine the enthalpy change if 3.25 kg of CaCl2 is used to de-ice a sidewalk.
1000 g
mol
x
= 29.3 mol CaCl2
1 kg
110.986 g
-82.9 kJ
29.3 mol x
= -2430 kJ
mol
3.25 kg x
c) If the same 3.25 kg of CaCl2 is added to 100.0 mL of water at 4.00°C, what is the final temperature
of the water? The density of water at this temperature is 1.00 g/mL.
This is a heat transfer problem, where q1 = - q2. We use the heat calculated in part b for q1, and the
heat of water will be q2. The mass of the water is found from the volume and density.
-2430 kJ x
'!
*
1000 J
1.00 g $ !
J $
= - )# 100.0 mL x
4.184
T
4.00°C
,
1 kJ
ml &% #"
g¡C &% f
("
+
3
Tf = 5.81 x 10 °C
(
)
7. The heat of formation of propene gas, C3H6, is, by definition, the enthalpy change for the reaction:
a) 3/2 C2 (g) + 3 H2 (g)  C3H6 (g)
b) 3 C (g) + 6 H (g)  C3H6 (g)
c) C3H8 (g)  C3H6 (g) + H2 (g)
d) 3 C (s) + 3 H2 (g)  C3H6 (g)
e) C3H4 (g) + H2 (g)  C3H6 (g)
A heat of formation is defined as making one mole of a compound from their elements in their
standard states.
8. The enthalpy change for the thermochemical equation
2HF(g)  H2(g) + F2(g) This is 2x and reversed from the formation reation!
is 542.2 kJ. The enthalpy of formation of HF(g) in kilojoules per mole must be
a) –542.2.
b) –271.1.
c) –1084.
d) 271.1.
e) 542.2
9. Calculate the standard enthalpy of formation of carbon disulfide from its elements given that
C (s) + O2 (g)  CO2 (g)
∆H = –393.5 kJ
SO2 (g)  S (s) + O2 (g)
∆H = 296.8 kJ
CS2 (l) + 3 O2 (g)  CO2 (g) + 2 SO2(g)
∆H = –1076.8 kJ
a) 386.5 kJ
b) –1173.5 kJ
c) –1767.1 kJ
d) 89.7 kJ
e) 1276.9 kJ
Target reaction: C (s) + 2 S (s)  CS2 (l)
10. This question is concerned with the heat change when ammonia is formed from its elements:
N2(g) + 3H2(g)  2NH3(g)
∆H = –92 kJ
Therefore, 92 kJ is the quantity of heat that is
a) lost to the surroundings when 1 mol of hydrogen is consumed.
b) lost to the surroundings when 1 mol of ammonia is formed.
c) lost to the surroundings when 2 mol of ammonia are formed.
d) gained from the surroundings when 1 mol of ammonia is formed.
e) gained from the surroundings when 2 mol of ammonia are formed.
11. How much heat is gained when 400 g of copper is warmed from 22.6 to 81.0°C? The specific heat
of copper is 0.385 J/g•°C.
a) –9000 J
b) 3500 J
c) 8990 J
d) 8993.6 J
e) 9000 J
12. A reaction is performed in 100.0 mL of water in an insulated calorimeter. If the reaction is
exothermic, what happens to the temperature of the water?
a) The temperature of the water goes up.
b) The temperature of the water stays the same.
c) The temperature of the water goes down.
13. The standard heat of formation of lead nitrate corresponds to the reaction:
a) Pb(OH)2 (s) + 2 HNO3 (aq)  Pb(NO3)2 (aq) + 2 H2O (l)
b) Pb2+ (aq) + 2 NO3- (aq)  Pb(NO3)2 (s)
c) Pb2+ (aq) + 2 NO3- (aq)  Pb(NO3)2 (aq)
d) Pb (g) + N2 (g) + 3O2 (g)  Pb(NO3)2 (g)
e) Pb (s) + N2 (g) + 3 O2 (g)  Pb(NO3)2 (s)
14. For the reaction of hydrogen and oxygen
H2 (g) + 1/2 O2 (g)  H2O (l)
a)
b)
c)
d)
ΔH° = -285.8 kJ
all of the following statements are true except
if the equation is multiplied by 2, the change in enthalpy is –571.6 kJ.
the value –571.6 pertains to one mole of liquid water.
per mole of O2, the change in enthalpy is –571.6 kJ.
if the equation is reversed, the change in enthalpy is +285.8 kJ.
Ch. 7 – Energy and Light
15. Determine the energy associated with violet light, at 5.10 x 102 nm, in both J/photon and kJ/mol.
1m
5.10 x 102 nm x -9
= 5.10 x 10-7 m
10 nm
6.626 x 10-34 J " s 2.998 x 108 m " s-1
hc
E=
=
= 3.90 x 10-19 J/photon
-7
!
5.10 x 10 m
-19
3.90 x 10 J
1 kJ
6.022 x 1023 photons
kJ
x
x
= 235
photon
1000 J
mol
mol
(
)(
)
16. Calculate the energy of an n=4 electron in Cl16+, in kJ/mol.
a) –3.936 x 10-17 J
E4 = -
b) 23698 kJ/mol
Z2
Rh = -
17 2
R h = -3.9376 x 10-17 J
photon
n
4
-17
23
-3.9376 x 10 J
1 kJ
6.022 x 10 photons
x
x
= -23700 kJ/mol
photon
1000 J
1 mol
c) –23698 kJ/mol
d) -5576 kJ/mol
2
2
17. Determine the energy of an electron in the n = 4 state in Fe25+, in kJ/mol.
Z2
262
E 4 = - 2 R h = - 2 R h = -9.21 x 10-17 J
photon
n
4
-9.21 x 10-17 J
1 kJ
6.022 x 1023 photons
x
x
= -55500 kJ/mol
photon
1000 J
1 mol
18. Calculate the energy, in J, of a photon of wavelength of 22.5 nm.
a) 1.49 x 10-41 J
b) 1.49 x 10-32 J
c) 8.83 x 10-27 J
d) 2.95 x 10-26 J
e) 8.83 x 10-18 J
22.5 nm x
E=
hc
=
!
1m
9
= 2.25 x 10-8 m
10 nm
6.626 x 10-34 J " s 2.998 x 108 m " s-1
(
)(
-8
2.25 x 10 m
) = 8.83 x 10
-18
J/photon