Chemistry 2000 (Spring 2014)
Problem Set #6: Free Energy, Temperature and Equilibrium
Solutions
Comment on Answers in Text
Equilibrium constant expressions should initially be written in terms of activities (not
concentrations or pressures) and should include activities of solids and liquids as well.
e.g.
The text lists the reaction quotient for 15.13(a) Qc  CO 2 .
First, we want to work with Q not Qc.
aCaO (s )  aCO 2 ( g )
The reaction quotient should be written as Q 
.
aCaCO
3 (s
)
Note that, when the values for the activities are filled in, this expression will simplify to:
Q 
aCaO (s )  aCO
aCaCO (s )
3
2 (g )

 pCO 2 ( g ) 

 1bar 
1
1

pCO ( g )
1bar
2
Since pressure and volume are inversely related (just as concentration and volume are
inversely related), this expression gives the same overall answer as the version in the text
– that a decrease in pressure will result in a decrease in Q and therefore promote reaction
in the forward direction.
So, why use Q instead of Qc (or K instead of Kc)? Q can be used in r G  r G   RT ln Q and
K can be used in r G   RT ln K . Neither Qc nor Kc can be used in either of those formulas,
so Qc and Kc are not very useful!
Reaction Quotient Expressions and/or Equilibrium Constant Expressions for Asterisked
Questions
Note that font limitations in Word prevent use of proper equilibrium arrows for the balanced
chemical equations in this document. You should be using proper equilibrium arrows.
15.12
(a)
4 NO(g) + O2(g) → 2 N2O3(g)
Q 
a

2
N 2O 3 ( g )
a   a
a 
Q 
a   a 

a
 a
Q 
a   a
4
NO( g )
O2 ( g )
6
(b)
2 SF6(g) + 4 SO3(g) → 6 SO2F2(g)
SO 2F 2 ( g )
2
4
SF6 ( g )
SO 3 ( g )
2
(c)
2 SClF5(g) + H2(g) → S2F10(g) + 2 HCl(g)
S 2F10 ( g )
HCl ( g )
2
SClF5 ( g )
H 2( g )
15.14
(a)
2 NO2Cl(g) → 2 NO2(g) + Cl2(g)
a
Q 
 a
a 
a   a
Q 
a 
a   a 
Q 
a   a 
2
NO2 ( g )
Cl 2 ( g )
2
NO2Cl ( g )
2
(b)
2 POCl3(g) → 2 PCl3(g) + O2(g)
PCl 3 ( g )
O2 ( g )
2
POCl 3 ( g )
2
(c)
4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g)
6
N 2( g )
H 2O ( g )
4
3
NH 3 ( g )
O2 ( g )
15.18
While the numerical values of all activities of pure solids and liquids are 1, they should still be
written in the initial reaction quotient expression.
(a)
(b)
2 Na2O2(s) + 2 CO2(g) → 2 Na2CO3(s) + O2(g)
H2O(l) → H2O(g)
Q 
Q 
a
a
 a
  a 
2
Na2CO 3 ( s )
O2 ( g )
2
Na2O 2 ( s )
2
CO 2 ( g )
aH O g
2 ( )
aH O l
2 ( )
(c)
NH4Cl(s) → NH3(g) + HCl(g)
Q 
aNH
 aHCl ( g )
3( g )
aNH Cl s
4
( )
15.20
While the numerical values of all activities of pure solids and liquids are 1, they should still be
written in the initial reaction quotient expression.
aNa2CO 3( s )  aCO 2 ( g )  aH 2O( g )
Q 
(a)
2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
2
a
(b)
(c)
SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g)
H2SO4(l) + SO3(g) → H2S2O7(l)
Q 
Q 
NaHCO3 ( s )


aSn s  aH O g
( )
aSnO
2(s )

2 ( )
 aH 2 ( g )


2
aH S O
2 2 7 (l )
aH SO
2
4 (l )
 aSO3( g )
2
Answers to Questions in Silberberg (only those w/out answers at the back of the book)
Note that font limitations in Word prevent use of proper equilibrium arrows for the balanced
chemical equations in this document. You should be using proper equilibrium arrows.
15.3
While there is no net change in reactant or product concentration, both the forward
reaction and the reverse reaction still proceed at equilibrium. When the system is at
equilibrium, the rates of the forward and reverse reactions are the same.
15.4
K is very large for a reaction that goes essentially to completion. If a reaction is almost
complete, the activities of the products will be relatively high and the activities of the
reactants will be very low. Since K is found by dividing activities of products by
activities of reactants, this will give a large value for K.
15.6
The value of Q is not constant. The value of Q depends on the activities of products and
reactants in the system and changes as they change. Adding more of one reactant, for
example, would decrease the value for Q but would have no impact on the value for K.
15.10 Yes. Q for the reverse reaction is the inverse of Q for the forward reaction:
1
Qreverse 
Qforward
This is because the product(s) of the forward reaction are the reactant(s) of the reverse
reaction, and the reactant(s) of the forward reaction are the product(s) of the forward
reaction.
a 
2
e.g.
N2(g) + O2(g) → 2 NO(g)
2 NO(g) → N2(g) + O2(g)
Q 
Q 
NO( g )
aN
2( g )
aN
2( g )
 aO2 ( g )
 aO2 ( g )
a 
2
NO( g )
15.34
(a)
Three red spheres indicates 0.0300 mol of D in 1.00 L, so cD = 0.0300 mol/L.
Three blue spheres indicates 0.0300 mol of E in 1.00 L, so cE = 0.0300 mol/L.
  
K 

a    
aE aq
(
1 mol
L
)
2
D ( aq )
(b)
0.0300mol
L
cE
cD
2
1 mol
L
1 mol
L
0.0300mol
L
1 mol
L
  0.0300  33.3
 0.000900
2
Scene B:
Three red spheres indicates 0.0300 mol of D in 0.500 L, so cD = 0.0600 mol/L.
Three blue spheres indicates 0.0300 mol of E in 0.500 L, so cE = 0.0600 mol/L.
  
Q 

a    
aE aq
(
)
2
D ( aq )
0.0600mol
L
cE
1 mol
L
cD
1 mol
L
2
1 mol
L
0.0600mol
L
1 mol
L
  0.0600  16.7
 0.003600
2
Q < K so the reaction will proceed in the forward direction.
Scene C:
Three red spheres indicates 0.0300 mol of D in 0.500 L, so cD = 0.0600 mol/L.
Six blue spheres indicates 0.0600 mol of E in 0.500 L, so cE = 0.120 mol/L.
  
Q 

a    
aE aq
(
)
2
D ( aq )
0.1200mol
L
cE
1 mol
L
cD
1 mol
L
2
1 mol
L
0.0600mol
L
1 mol
L
  0.1200  33.3
 0.003600
2
Q = K so the system is at equilibrium.
15.39 See high school lecture notes and/or text
15.57 When discussing Le Châtelier’s principle, a “disturbance” is any change to the system.
This could mean addition or removal of a quantity of reactant or product, a change in the
pressure, a change in the volume, a change in the temperature, etc.
15.60 The word “constant” in the term “equilibrium constant” implies that the value of the
equilibrium constant (K) does not change. This is true as long as the temperature does
not change:
o K does not change when reactant concentration(s) change.
o K does not change when product concentration(s) change.
o K does not change when pressure changes.
o K does not change when volume changes.
15.61
(a)
An increase in reactant concentration increases the rate of the forward reaction since
there are more reactant molecules. It does not affect the rate of the reverse reaction
*until* the concentration of product molecules is increased (which is inevitable since the
forward reaction is now faster than the reverse reaction). As more reactant molecules are
converted into product molecules, the rate of the forward reaction slows down while the
rate of the reverse reaction speeds up – until the two rates are again equal and equilibrium
is again reached. Thus, the equilibrium position has changed but the equilibrium constant
has not changed.
(b)
A decrease in volume increases the proximity of the gas particles. The frequency of
collisions between gas molecules therefore increases – and, as a result, the frequency of
productive collisions between gas particles also increases. (Not every collision results in
a reaction! Only productive collisions result in reactions.) This will increase the rates of
both the forward and reverse reactions (if both involve gas particles reacting), but it will
have a greater impact on the rate of the reaction with more gas particles reacting. Thus,
the equilibrium position will shift toward the side with fewer gas particles.
(d)
Changing the temperature of the system changes K since changing temperature changes
ΔrG r G   r H   Tr S  , and ΔrG is related to the equilibrium constant
r G   RT ln K  .
If a reaction is endothermic (ΔH > 0), it must have a positive entropy change (ΔS > 0) in
order for it to be spontaneous (ΔG < 0) since r G  r H  Tr S . Increasing the
temperature increases the contribution from the TΔS term and, since ΔS > 0, this makes
ΔG more negative. When ΔG is more negative, the forward reaction is more strongly
favoured.
(c)
Thus, an endothermic reaction will have a larger equilibrium constant at higher
temperatures.
The reverse of an endothermic reaction is an exothermic reaction. The same argument
used in the answer to part (d) can be used to demonstrate that the reverse (exothermic)
reaction will have a smaller equilibrium constant at higher temperatures since Kforward =
1/Kreverse.
Note that it is possible to have a reaction that is exothermic (ΔH < 0) and has a positive
entropy change (ΔS > 0). Such a reaction is only favoured in the forward direction and
the concept of equilibrium would not usually be considered in that case since the reverse
reaction would never be favoured.
15.62 Scene A is the best representation of the system at equilibrium. The activity of a pure
solid is always 1. So, as long as there is *some* XY(s) and *some* Y(s), the amounts of
each are not relevant to the equilibrium position.
15.79
(a)
Yes
(b)
No
(c)
Yes
(d)
No
The forward reaction is exothermic so lowering the temperature slows the rate of
the forward reaction less than it slows the rate of the reverse reaction. Thus, the
equilibrium position shifts toward making more products (including CaCO3).
Increasing volume decreases the pressure. The only reactant/product affected by
this change is the CO2 as it is the only gas. Collisions involving CO2 molecules
will be less frequent and therefore the rate of the forward reaction will slow. (For
this system, the rate of the reverse reaction will not change.) Thus, the
equilibrium position shifts toward making more reactants.
Increasing the partial pressure of CO2 will have the opposite effect of the change
in part (b). Collisions involving CO2 molecules will be more frequent and
therefore the rate of the forward reaction will increase. (For this system, the rate
of the reverse reaction will not change.) Thus, the equilibrium position shifts
toward making more products (including CaCO3).
Since the activity of CaCO3(s) is equal to 1 regardless of the quantity (as long as
there is some), removing CaCO3 will have no effect on the equilibrium position.
15.110
(a)
At point A, decreasing the temperature (while maintaining the same pressure) promotes
conversion of graphite to diamond. Thus, at this pressure, the formation of diamond from
graphite is exothermic (ΔH < 0).
(b)
Since increasing the pressure promotes conversion of graphite to diamond, diamond must
be denser than graphite.
18.65 See pages 764-765 of text
18.66
(a)
If all reactants and products are in their standard states, x represents ΔrG.
(b)
B represents point 1, a mixture containing 100% reactants. (A looks very similar but,
because the two reactants are separated, preventing any reaction, the system is at
equilibrium in scene A. It is not at equilibrium in scene B because the reactants are then
able to react.)
(c)
C represents point 2. At equilibrium, there is a mixture of reactants and products. The
graph shows that the reaction has proceeded more than halfway so there are more product
molecules than reactant molecules; this rules out D.
18.102
(a)
D depicts the change in Gsystem for the formation of HI. Gsystem has its minimum value
(for the system under the given set of conditions) at equilibrium. As long as the value for
K is between 0 and ∞, the system will contain a mixture of reactants and products at
equilibrium. Only graph D has a minimum value for Gsystem consistent with such a
mixture.
(b)
To answer this question, we must assume that the melting point of water/ice is still lower
than 1 C at 1 bar. This is probable since 1 bar = 0.987 atm. (Recall that normal
melting points are measured at 1 atm – not 1 bar.)
A depicts the change in Gsystem for ice melting at 1 C and 1 bar. As long as the
temperature is maintained above the melting point of the ice, the system will not reach
equilibrium until after all the ice has melted (and the water has warmed to 1 C). As
such, the minimum value for Gsystem occurs when the system has reached 100% products
(or proceeded 100%).
11.12 Recall that not all particles in a sample have the same energy; there is a MaxwellBoltzmann distribution of energies. So, both the solid and liquid will contain particles
with higher and with lower energies than the average.
Also, recall that all particles (except those at 0 K) are constantly in motion – even those
in a solid are rotating and vibrating.
Finally, it is important to recognize that intermolecular forces are important in both the
solid and liquid phases.
The two simultaneous processes are the melting of solid benzene and the freezing of
liquid benzene.
Some of the particles at the surface of the solid will acquire a high enough energy to
overcome some of the intermolecular forces holding them in the solid lattice. These
particles therefore enter the less ordered liquid phase.
Some of the particles in the liquid phase lose some of their energy to other particles as
they collide with them. This leaves them as lower-energy particles and they no longer
have enough energy to overcome the intermolecular forces ordering the lattice. They are
therefore incorporated into the solid phase.
11.15 Diagram C represents the situation at point 1 (solid and liquid phases in equilibrium)
Diagram A represents the situation at point 2 (liquid and gas phases in equilibrium)
Diagram D represents the situation at point 3 (solid and gas phases in equilibrium)
11.16
(a)
No, the pressure does not change.
The initial pressure was the equilibrium vapour pressure for that substance at that
temperature. If there is still liquid left after the vessels are connected (increasing the
volume) and the system allowed to re-establish equilibrium then the final pressure is still
the equilibrium vapour pressure for that substance at that temperature.
(b)
The activity of the liquid is 1 in both cases, and the equilibrium constant for this process
does not change. As such, the activity of the gas must be the same at equilibrium.
Yes, the pressure changes; the final pressure is lower than the initial pressure.
Since the liquid is removed before the vessels are connected (increasing the volume),
there is no more to vapourize and therefore the equilibrium vapour pressure cannot be
reached. Since we are instead comparing two samples of gas with different volumes, the
new pressure can be calculated using Boyle’s Law (p1V1 = p2V2) or the ideal gas law.
11.17 We can measure the relative densities of each substance in the solid and liquid phases.
An easy way to do so would be to partially freeze/melt each sample to give a mixture of
solid and liquid.
Substance A will have a solid that is more dense than the liquid, so the solid will sink.
Substance B will have a liquid that is more dense than the solid, so the solid will float.
11.29
(a)
This is not the easiest graph to read given the non-linear axes and lack of gridlines.
The answers to ii. and iv. are relatively clear cut. Use the triple point and critical point
to let you know where -57 C and +31 C lie on the x-axis. -120 C is lower than -57 C
while +40 C is higher than +31 C. -40 C and +20 C must lie between those two
points and appear like they may appear in the green (liquid) region, but that is open to
interpretation.
i.
At 20 C and 2026 kPa, there may be liquid CO2 in the cylinder.
ii.
At 40 C and 2026 kPa, there is no liquid CO2 in the cylinder. (It is gas.)
iii.
At -40 C and 2026 kPa, there may be liquid CO2 in the cylinder.
iv.
At -120 C and 2026 kPa, there is no liquid CO2 in the cylinder. (It is solid.)
(b)
Atmospheric pressure (101.3 kPa) is lower than the pressure at the triple point, so the dry
ice chunks will not melt. They will instead sublime as they warm to room temperature.
(c)
If a container is nearly filled with dry ice then sealed and warmed to room temperature, it
will initially sublime (assuming the initial pressure was atmospheric pressure); however,
as more particles enter the gas phase, the pressure will increase. It is likely that this will
elevate the pressure above the pressure at the triple point so, as long as the container
remains sealed, it is likely that there will be a point at which some solid CO2 melts.
(d)
No. The liquid phase cannot form at a temperature (or pressure) below the triple point.
11.112
(a)
I, II, III and V
(b)
IV
(c)
V to IV then IV to liquid then liquid to I
(d)
There are triple points for each of the following phase combinations:
 I, II and liquid
 II, IV and liquid
 IV, V and liquid
 II, III and IV
 III, IV and V
11.124 Since liquid gallium has a higher density than solid gallium, the slope of the solid-liquid
line for gallium must be negative (like that of water). As such, the temperature at the
triple point will be higher than the normal melting point.
11.125
(a)
The question gives the volume of the bottle (4.7 L) and the temperature (11 C).
The total mass of ethanol will only be relevant to part (a) of this question if all the ethanol
vaporizes (i.e. if more than 0.33g ethanol is required to reach the equilibrium vapour
pressure). Otherwise, that information will be used later in the question.
To find the mass of C2H5OH(g), it is necessary to find the equilibrium constant for the
evaporation of C2H5OH. That can be used to find the activity of C2H5OH(g) at
equilibrium (then the equilibrium vapour pressure then the moles then the mass).
Step 1: Write a balanced chemical equation for this process.
C 2H 5OH(l )  C 2H 5OH( g )
Step 2: Find ΔrG at 25 C.
r G   f G C 2H 5OH ( g )   f G C 2H 5OH (l ) 
kJ
kJ
   174.8 mol

  168.6 mol
kJ
r G   6.2 mol
Step 3: Find K at 25 C.
r G   RT ln K
K e
K e
 r G 
RT
kJ 
 6.2 mol
1000J
8.314462molJ K 298.15K  1kJ
K  e 2.5  0.082
Step 4: Find K at -11 C. This requires finding ΔrH first.
K1 = 0.082
T1 = 25 C = 298.15 K
K2 = ???
r H   f H C 2H 5OH ( g )   f H C 2H 5OH (l ) 
T2 = -11 C = 262.15 K
kJ
kJ
   277.63 mol

  235.1 mol
kJ
kJ
r H   42.53 mol
 42.5 mol
K   H o
ln  2   r
R
 K1 
1
1
  
 T1 T 2 
kJ
J
 K  42.5 mol
1000
 
1
1
1kJ
ln  2  


J
 K 1  8.314462mol K   298.15K  262.15K
K2
 e 2.36
K1
K 2  K 1  e 2.36  0.082  e 2.36  0.0078

  2.36
 
Step 5: Find activity of C2H5OH(g) at equilibrium at -11 C.
K 
aC H OH g
2
5
( )
aC H OH l
2
5
( )
aC H OH g  K  aC H OH l  0.00781  0.0078
2
5
( )
2
5
( )
Step 6: Find equilibrium vapour pressure of C2H5OH(g) at -11 C.
aC H OH g 
2
5
( )
pC H OH
1bar
2
5
pC H OH  aC H OH l  1bar  0.00781bar   0.0078bar
2
5
2
5
( )
Step 7: Find moles of C2H5OH(g) at equilibrium at -11 C.
Pressure is not high, so it is reasonable to assume ideal gas behaviour.
pV  nRT
nC H OH 
2
5

pV
RT
0.0078bar 4.7L   1m 3  105 Pa
Pa m
8.314462mol

 1000L 1bar
K 262.15K
3
nC H OH  0.0017mol
2
5
Step 8: Find mass of C2H5OH(g) at equilibrium at -11 C.
mC 2H 5OH  nC 2H 5OH  M C 2H 5OH
 0.0017mol 
46.0688g
1mol
mC H OH  0.077g
2
5
Since there was 0.33 g C2H5OH(l) placed in the container, 0.077 g was available to vaporize.
Therefore, the vapour contains 0.077 g C2H5OH(g).
(b)
Repeat steps 4 to 8 of part (a), but for a temperature of 20. C instead of -11 C
Step 4: Find K at 20 C.
K1 = 0.082
T1 = 25 C = 298.15 K
K2 = ???
T2 = 20 C = 293.15 K
ΔrH = 42.53 kJ/mol (3 sig. fig.)
 K 2  r H o
ln   
R
 K1 
1
1
  
 T1 T 2 
kJ
J
 K  42.5 mol
1000
 
1
1
1kJ
ln  2  


J
 K 1  8.314462mol K   298.15K  293.15K
K2
 e 0.293
K1
K 2  K 1  e 0.293  0.082  e 0.293  0.061

  0.293
 
Step 5: Find activity of C2H5OH(g) at equilibrium at 20 C.
K 
aC H OH g
2
5
( )
aC H OH l
2
5
( )
aC H OH g  K  aC H OH l  0.0611  0.061
2
5
( )
2
5
( )
Step 6: Find equilibrium vapour pressure of C2H5OH(g) at 20 C.
aC H OH g 
2
5
( )
pC H OH
1bar
2
5
pC H OH  aC H OH l  1bar  0.0611bar   0.061bar
2
5
2
5
( )
Step 7: Find moles of C2H5OH(g) at equilibrium at 20 C.
Pressure is not high, so it is reasonable to assume ideal gas behaviour.
pV  nRT
nC H OH 
2
5

pV
RT
0.061bar 4.7L 
1m 3 105 Pa


Pa m
8.314462mol

 1000L 1bar
K 293.15K
3
nC H OH  0.012mol
2
5
Step 8: Find mass of C2H5OH(g) at equilibrium at 20 C.
mC 2H 5OH  nC 2H 5OH  M C 2H 5OH
 0.012mol 
46.0688g
1mol
mC H OH  0.54g
2
5
Since there was 0.33 g C2H5OH(l) placed in the container, 0.54 g was not available to vaporize.
Therefore, all of the ethanol vaporizes.
11.129
(a)
Find the equilibrium constants at each temperature (from the equilibrium vapour
pressures) then use the two pairs of K and T values to find the standard enthalpy of
vaporization for CH3OCH3.
Step 1: Write a balanced chemical equation for this process.
CH 3OCH 3(l )  CH 3OCH 3( g )
Step 2: Find equilibrium constants at each temperature from equilibrium vapour pressures
K 
aC H OH g
2
( )
aC H OH l
2
K 
5
5
( )
aC H OH g
2
5
( )
aC H OH l
2
5
( )


pC 2H 5OH( g )


1bar
1
pC 2H 5OH( g )
1bar
1
  101.3kPa  1000Pa  1bar
1bar
1kPa
105 Pa
  53.3kPa  1000Pa  1bar
1bar
1kPa
105 Pa
 1.013
 0.533
at -23.7 C
at -37.8 C
Step 3: Find ΔrH (aka ΔvapH)
K1 = 1.013
T1 = -23.7 C = 249.45 K
K   H o  1
1
  
ln  2   r
R  T1 T 2 
 K1 
r H o

1
1
 0.533 

ln 


J
 1.013  8.314462mol K   249.45K  235.35K
 0.642  - 2.889 10-5 mol
r H o
J

r H o 
(b)

 0.642
- 2.889 10-5
mol
J

 2.22  104
T2 = -37.8 C = 235.35 K
K2 = 0.533
J
mol


 
kJ
 22.2 mol
r H   f H CH 3OCH 3( g )   f H CH 3OCH 3(l ) 
f H CH 3OCH 3(l )   f H CH 3OCH 3( g )   r H 
kJ
kJ
  22.2 mol

  185.4 mol
kJ
f H CH 3OCH 3(l )   207.6 mol
17.73 AgCl and AgBr both dissociate into a 1 : 1 mixture of cation and anion.
As such, whichever species has a smaller Ksp value will be less soluble in water since
 c Ag (aq )  c X (aq ) 
a Ag (aq )  a X (aq )  1molL  1molL  c Ag (aq )  c X (aq )

K sp 


2
1
a AgX ( s )
1 mol
L2
Ag2CrO4 dissociates into a 2 : 1 mixture of cations and anion. As such,
a   a

2
K sp
Ag (aq )
CrO42(aq )
a Ag CrO
2
4(s )
2


 c Ag (aq )   c CrO42(aq ) 
 1 mol   1mol  c  2 c 2
L
  L   Ag (aq ) CrO4 (aq )

3
1
1 mol
L3
Since the activity of the Ag+ cations is squared, a direct comparison between the Ksp of
Ag2CrO4 and AgCl (or AgBr) is not appropriate.
Additional Practice Problems
1.
The reaction 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) comes to equilibrium at 400oC in a 2L
flask. Analysis of the equilibrium mixture shows that it contains 60g of Cl2, 12g of H2O,
20g of HCl and 8g of O2. Calculate the equilibrium constant.
Since quantities of all reactants and products in an equilibrium mixture are provided, the
equilibrium constant can be calculated from their activities.
a  a

a  a
2
K
Cl 2 ( g )
H 2O ( g )
4
HCl ( g )
O2 ( g )


2
Recall that, for gases, activities are calculated from pressures (in bar).
Sample activity calculation:
Step 1: Find moles of the gas
1mol
nCl 2  60g 
 0.85mol
70.9054g
Step 2: Find pressure of the gas (in bar)
pV  nRT
nCl RT
V
Pa m
0.85mol 8.314462mol

  1000L  1bar
K 400  273.15 K

2L 
1m 3 105 Pa
pCl 
2
2
3
pCl  23.68bar  2  102 bar
2
Step 3: Find activity of the gas
p
23.68bar
aCl 2  Cl 2 
 23.68  2  102
1bar
1bar
While each activity only has 1 sig. fig. (due to the volume only having 1 sig. fig.), make
sure you carry through all the digits in your calculator when calculating K. Of course, K
will only be accurate to 1 sig. fig. as well, so the final answer should be rounded.
If your calculator doesn’t have enough different spaces in memory to carry through all
digits, a good rule of thumb is to carry through at least 2-3 extra digits until you get to
the final answer.
Once all four activities have been calculated, use them to calculate the equilibrium
constant.
a  a

a  a
2
K
Cl 2 ( g )
H 2O ( g )
4
HCl ( g )
O2 ( g )
  23.68 18.64
 15.35 6.996
2
2
4
2
 0.5
2.
The equilibrium constant for the reaction H2(g) + I2(s) → 2HI(g) is 0.352 at 25oC. Suppose
than an excess of solid iodine is placed in a rigid flask with 0.400 bar of hydrogen gas
and 0.300 bar of hydrogen iodide. In what direction will the reaction proceed?
The easiest approach to this question is to calculate Q and compare it to K.
 If Q is smaller than K, the forward reaction will proceed.
 If Q is larger than K, the reverse reaction will proceed.
 If Q is equal to K, the system is already at equilibrium so there is no net reaction.
Q is calculated from activities of reactants and products.
Q
a

0.300bar 2

  0.225
1bar

 0.400
bar
aH (g ) aI (s )   1bar 1
2
HI ( g )
2
2
Q < K therefore the reaction will proceed forward (as written).
3.
Calculate the equilibrium constant for the reaction CO(g) + Cl2(g) → COCl2(g) at 25oC.
Since quantities of all reactants and products in an equilibrium mixture are not provided,
the equilibrium constant must be calculated from thermodynamic data. Standard free
energies of formation can be found in the appendix of any first year chemistry textbook.
(Appendix B in Silberberg)
Step 1: Calculate standard free energy change for the reaction
r G   f G COCl 2( g )   f G CO ( g )   f G Cl 2( g ) 
kJ
kJ
kJ
   137.2 mol
  0 mol

  206 mol
kJ
kJ
r G   68.8 mol
 69 mol
Step 2: Calculate equilibrium constant for the reaction
 r G   RT ln K
K e
K e
 r G 
RT
kJ 
 69 mol
1000J
8.314462molJ K 298.15K  1kJ
K  e 28  1.1  1012
4.
Using the solubility product of barium sulfate and the standard free energies of formation
of solid barium sulfate and of the aqueous sulfate ion, calculate the standard free energy
of formation of the aqueous Ba2+ ion and compare the value obtained to that given in the
data tables in your textbook.
Standard free energies of formation and solubility products can be found in the appendix
of any first year chemistry textbook. (Appendices B and C in Silberberg)
K sp (BaSO 4 )  1.1  1010
kJ
f G BaSO 4 (s )   1353 mol


kJ
f G  SO 42(aq )  741.99 mol
The reaction described by the solubility product of BaSO4 is:
BaSO 4(s )  Ba (2aq )  SO42(aq )
The standard free energy change for this reaction can be calculated from its equilibrium
constant (i.e. Ksp):
J
kJ
r G   RT ln K  8.314462molJ K 298.15K ln 1.1  1010  56844 mol
 56.84 mol


The standard free energy of formation of the aqueous Ba2+ ion can then be calculated
from the standard free energy change for the reaction and all the other relevant standard
free energies of formation.
r G   f G  Ba (2aq )  f G  SO 42(aq )  f G BaSO 4 (s ) 










f G  Ba (2aq )  r G   f G BaSO 4 (s )   f G  SO 42(aq )

kJ
kJ
kJ
   1353 mol
   741.99 mol

 56.84 mol
kJ
f G  Ba (2aq )  554 mol
Compare to fG(Ba2+(aq)) = -560.7 kJ/mol (Appendix B of Silberberg)
While these answers do not match perfectly, they are consistent to two significant figures.
5.
What is the vapour pressure of a solution made by dissolving 50g of ammonium sulphate
in 300g of water at 40oC? The vapour pressure of pure water at 40oC is 7373 Pa.
While this question can be solved ‘from scratch’ using activities and equilibrium constant
expressions, it’s easier to approach using Raoult’s law*:
pH 2O  X H 2O  pHo 2O
Step 1: Calculate the moles of each species in solution
1mol
nH 2O  300g 
 16.6526mol  16.7mol
18.0152g
1mol
n NH 4 2 SO 4  50g 
 0.38mol
132.140g
nNH  n NH

4
4 2 SO 4
nSO  n NH
2
4
4 2 SO 4

2molNH 4
1mol NH 4 2 SO4

1molSO 42
1mol NH 4 2 SO 4
 0.76molNH 4
 0.38molSO 42
Step 2: Calculate the mole fraction of water in the solution
nH 2O
16.7mol
X H 2O 

 0.936
nH 2O  nNH   nSO 2 16.7mol  0.76mol  0.38mol
4
4
Step 3: Calculate the vapour pressure of water above the solution
pH 2O  X H 2O  pHo 2O  0.9367373Pa   6902Pa  6.90  103 Pa  6.90kPa
*Raoult’s law is derived from activities and equilibrium constant expressions.
6.
In lab, we often rinse wet glassware with acetone to remove the water then use a stream
of air to evaporate off the acetone. The structure of acetone and vapour pressure curves
for acetone and water are shown below. (1 atm = 1.01325 bar = 760 mmHg)
Vapour Pressure Curves for Acetone and Water
Equilibrium Vapour Pressure (mmHg)
800
700
600
500
Acetone
Water
400
300
200
P20'C(acetone) = ~178 mmHg
100
0
-80
-60
-40
-20
P20'C(H2O) = 17 mmHg
(a)
0
20
40
o
Temperature ( C)
60
80
100
120
Tobp(acetone) = 56.5oC Tobp(H2O) = 100oC
What type(s) of intermolecular forces are responsible for water’s solubility in acetone?
dipole-dipole forces (including hydrogen bonding between H of water and O of acetone
as well as hydrogen bonding between H of one water molecule and O of another water
molecule)
(b)
What are the normal boiling points of acetone and water? see graph above; normal
boiling point is the vapour pressure at 1 atm (760 mmHg); interpolation on the graph
gives the normal boiling point of acetone as ~56 ˚C and the normal boiling point of water
as ~100 ˚C.
(c)
What are the vapour pressures of acetone and water at room temperature (20 ˚C)? see
graph above; interpolation gives a vapour pressure of ~17 mmHg = 0.023 bar for water
and a vapour pressure of ~178 mmHg = 0.237 bar for acetone.
(d)
Briefly, justify the relative boiling points and vapour pressures of water and acetone.
Water molecules can hydrogen bond with each other. Acetone has no hydrogen atoms
bonded to N, O or F so acetone molecules cannot hydrogen bond with each other. As
such, the intermolecular forces between water molecules are stronger than the
intermolecular forces between acetone molecules. If the intermolecular forces are
stronger, fewer molecules will have enough energy to escape the liquid phase. Thus, the
vapour pressure of water is lower.
(e)
Why does blowing a stream of air over acetone-wet glassware accelerate evaporation?
The liquid acetone and acetone vapour exist in equilibrium. The stream of air reduces the
vapour pressure of acetone so that more acetone must evaporate to reach the equilibrium
vapour pressure. In other words, the air pushes the acetone vapour away, so more
acetone has to evaporate to restore the equilibrium.
(f)
Why is blowing a stream of air over water-wet glassware much less effective?
Water has a much lower equilibrium vapour pressure than acetone. As such, there are
fewer water vapour molecules to push away and fewer water molecules need to evaporate
to restore the equilibrium. This method would eventually work but it might take hours!
(g)
If you need dry glassware for the following week’s lab, is this method of drying
necessary? Why or why not?
No, it isn’t necessary. As long as the wet glassware is not stored in a sealed environment,
the water (or acetone) will slowly evaporate until the glassware is dry.