Equilibrium II
15.6 – Using Keq
15.7 – Le Chậtelier’s Principle
19.7
20.5
Calculating Equilibrium
Concentrations
• The same steps used to calculate equilibrium
constants are used to calculate equilibrium
concentrations.
• Generally, we do not have a number for the
change in concentration.
• We need to assume that x mol/L of a species is
produced (or used).
• The equilibrium concentrations are given as
algebraic expressions.
At 2000oC, the equilibrium constant for the reaction: 2 NO (g)  N2 (g) +
O2 (g) is Kc = 2.40  103. If the initial concentration of NO is 0.200 M, what
are the equilibrium concentrations of NO, N2 and O2?
2 NO(g)

N2(g)
+
O2(g)
init
0.200M
0
0
change
- 2x
x
x
equil. 0.200 – 2x
x
x
#1
[N 2 ][O 2 ]
3
=
2.4
x
10
[NO]2
(x)(x)
3
=
2.4
x
10
(0.200 - 2x)2
x
= 2.4 x 103
0.200 - 2x
x
 49.0
0.200 - 2x
Kc =
x = 49.0(0.200 - 2x)
x = 9.80 - 98.0x
99.0x = 9.80
x = 0.0990 M = [N 2 ] = [O 2 ]
[NO] = 0.200 - 2(.0990) = 0.002 M
#2 At 100oC, Kc = 0.078 for the reaction:
SO2Cl2  SO2 + Cl2.
In an equilibrium mixture of the three gases, the concentrations
of SO2Cl2 and SO2 are 0.108 M and 0.052 M, respectively.
What is the partial pressure of Cl2 in the equilibrium mixture?
[SO 2 ][Cl 2 ]
Kc =
[SOCl 2 ]
(0.052)[Cl 2 ]
.078 =
 [Cl 2 ]  0.16 M
(0.108)
PV = nRT
-->
P=
nRT
V
-->
P = MRT
PCl2 = (0.16)(0.0821)(373) = 5.0 atm
#3 At 373 K, Kp = 0.416 for the equilibrium:
2 NOBr  2 NO + Br2. If the pressures of NOBr and NO
are equal, what is the equilibrium pressure of Br2?
Kp =
2
(PNO
)(PBr2 )
2
NOBr
P
when PNO = PNOBr , they cancel
K p = PBr2 = 0.416 atm
#4
(b)
For the reaction: H2 + I2  2 HI Kc = 55.3 at 700 K. In a 2.00 L flask
containing an equilibrium mixture of the three gases, there are 0.056 g H2
and 4.36 g I2. What is the mass of HI in the flask?
[HI]2
Kc =
= 55.3
[H 2 ][I 2 ]
.056 g
2.02 g / mol
 .014 M
2.00 L
[H 2 ] 
4.36 g
[I2 ] 
254 g / mol
 .00858M
2.00 L
[HI]2
= 55.3
(.014)(.00858)
[HI]=0.081 M
(0.081mol / L)( 2.00 L)(128 g / mol )  21gHI
Applications of Equilibrium
Constants - Predicting the Direction
of Reaction
For a general reaction: aA + bB cC + dD
We define Q, the reaction quotient, as:

C D
Q
a
b
A  B
c
d
Where [A], [B], [C], and [D] are molarities (for
substances in solution) or partial pressures (for
gases) at any given time.
Q is a K expression with non-equilibrium
concentration values.
• Q = Keq only at equilibrium.
• If Q < Keq then the forward reaction must occur
to reach equilibrium.
– Reactants are consumed, products are formed.
– Q increases until it equals Keq.
• If Q > Keq then the reverse reaction must occur
to reach equilibrium.
– Products are consumed, reactants are formed.
– Q decreases until it equals Keq.
#5
(a)
(b)
(c)
(a)
(b)
(c)
At 100oC, the equilibrium constant for the reaction:
COCl2  CO + Cl2
has the value Kc = 2.19  10-10. Are the following mixtures of COCl2, CO, and Cl2 at
100oC at equilibrium?
[COCl2] = 2.00  10-3 M [CO] = 3.3  10-6 M [Cl2] = 6.62  10-6 M
[COCl2] = 4.50  10-2 M [CO] = 1.1  10-7 M [Cl2] = 2.25  10-6 M
[COCl2] = 0.0100 M [CO] = [Cl2] = 1.48  10-6 M
[CO][Cl 2 ]
Kc =
= 2.19 x 10-10
[COCl 2 ]
(3.3 x 10-6 )(6.62 x 10-6 )
Q=
= 1.1 x 10-8
-3
(2.00 x 10 )
Q > K, reaction shifts left to attain equilibrium
(1.1 x 10-7 )(2.25 x 10-6 )
Q=
= 5.5 x 10-12
-2
(4.50 x 10 )
Q < K, reaction shifts right to attain equilibrium
(1.48 x 10-6 )2
Q=
= 2.19 x 10-10
(0.0100)
Q = K, reaction is at equilibrium
15.7 Le Châtelier’s Principle
Le Châtelier’s principle: If a system at equilibrium
is disturbed by a change in temperature, a change in
pressure, or a change in the concentration of one or
more components, the system will shift its
equilibrium position in such a way as to counteract
the effects of the disturbance.
Change in Reactant or Product Concentration
• If a chemical system is at equilibrium and we add or
remove a product or reactant, the reaction will shift so as
to reestablish equilibrium.
• For example, consider the Haber process:
N2(g) + 3H2(g)
2NH3(g)
If H2 is added while the system is at equilibrium, Q < Keq
The system must respond to counteract the added H2 (by
Le Châtelier’s principle).
• That is, the system must consume the H2 and produce
products until a new equilibrium is established.
• Therefore, [H2] and [N2] will decrease and [NH3]
increase until Q = Keq.
Effects of Volume and Pressure Changes
Effects of Volume and Pressure Changes
• If the equilibrium involves gaseous products or reactants,
the concentration of these species will be changed if we
change the volume of the container.
• For example, if we decrease the volume of the container,
the partial pressures of each gaseous species will
increase.
• Le Châtelier’s principle predicts that if pressure is
increased, the system will shift to counteract the
increase.
– the system shifts to remove gases and decrease pressure.
.
An increase in pressure favors the direction
that has fewer moles of gas.
A decrease in pressure favors the direction
that has more moles of gas.
In a reaction with the same number of moles
of gas in the products and reactants,
changing the pressure has no effect on the
equilibrium.
No change will occur if we increase the total
gas pressure by the addition of a gas that is
not involved in the reaction because the
partial pressures of the gases will stay
constant.
Effect of Temperature Changes
Temperature affects the equilibrium constant:
• For an endothermic reaction, heat can be
considered a reactant.
– Adding heat causes a shift to the right
– Removing heat causes a shift to the left
• For an exothermic reaction, heat can be
considered a product.
– Adding heat causes a shift to the left
– Removing heat causes a shift to the right
• At room temperature, an equilibrium mixture (light
purple)
• The mixture is placed in a beaker of warm water.
– The mixture turns deep blue - shift toward products
- CoCl42–
• The mixture is placed in ice water.
– The mixture turns bright pink - shift toward
reactants - Co(H2O)62+.
The Effect of Catalysts
• A catalyst lowers the activation energy
barrier for the reaction.
• Therefore, a catalyst will decrease the time
taken to reach equilibrium.
• A catalyst DOES NOT effect the
composition of the equilibrium mixture
#6 Consider the following equilibrium, for which ΔH < 0:
2 SO2(g) + O2(g)  2 SO3(g). + heat
How will each of the following changes affect an equilibrium mixture of the
three gases?
(a) O2(g) is added to the system
(b) the reaction mixture is heated
(c) the volume of the reaction vessel is doubled
(d) a catalyst is added to the mixture
(e) the total pressure of the system is increased by adding a noble gas
(f) SO3(g) is removed from the system
(a) shifts right
(b) shifts left
(c) shifts left – increasing volume decreases pressure  favors the
side with more moles of gas
(d) no effect
(e) no effect
(f) shifts right
#7
How do the following changes affect the value of the equilibrium
constant for a gas-phase exothermic reaction:
(a) removal of a reactant or product
(b) decrease in the volume
(c) decrease in the temperature
(d) addition of a catalyst
(a) no effect
(b) no effect
(c) Temperature affects Keq:
exothermic: inverse relationship
endothermic: direct relationship
increase Keq
(d) no effect
#8 For the reaction,
PCl5 (g)  PCl3 (g) + Cl2 (g) ΔHrxn = +111 kJ.
Fill in the following table:
Change to reaction
Add PCl5
Reaction shifts
right
Change in K
none
right
none
no shift
none
Decrease P
right
none
Increase T
right
increases
no shift
none
left
none
Remove Cl2
Add Ar
Add a catalyst
Decrease V of container
Review!
• Gibbs Free Energy: ΔG
– Combines enthalpy and entropy to tell us whether a
reaction will be spontaneous or not
• If ΔG is (+) the reaction is nonspontaneous
• If ΔG is (–) the reaction is spontaneous
• If ΔG = 0 the system is at equilibrium
19.7 Free Energy and the Equilibrium Constant
• ΔGo applies under standard conditions, but most
reactions don’t happen under standard conditions!
• ΔG and Q apply to any conditions.
• It is useful to determine whether substances will react
under specific conditions:
ΔG = ΔGo + RT lnQ
On equation sheet!
• At equilibrium, Q = Keq and ΔG = 0
0 = ΔGo + RT ln Keq
On equation sheet!
ΔGo = – RT ln Keq
ΔGo = – RT ln Keq
From the above we can conclude that at 298K:
• If ΔGo is negative, then Keq is greater than 1
(-) ΔG means spontaneous rxn, favors products
• If ΔGo = 0, then Keq = 1
• If ΔGo is positive, then Keq is less than 1
(+) ΔG means nonspontaneous rxn, favors reactants
#9 Explain qualitatively how ΔG changes for each of the following
reactions as the partial pressure of O2 is increased
(a)
2 CO(g) + O2(g)  2 CO2(g)
(b)
2 H2O2(l)  2 H2O(g) + O2(g)
ΔG = ΔGo + RT lnQ
(a) ΔG becomes smaller (or more negative)
(b) ΔG becomes larger (or more positive)
#10 Consider the reaction: 2 NO2(g)  N2O4(g).
(a) Using data from Appendix C, calculate ΔGo at 298 K.
(b) Calculate ΔG at 298 K if the partial pressures of NO2 and
N2O4 are 0.40 atm and 1.60 atm respectively.
(a) ΔGo = ΔGo N2O4 (g) - 2 ΔGo NO2(g)
= 98.28
- 2(51.84)
= - 5.40 kJ
 PN O
(b) G = G + RT ln  22 4
 PNO

2




 1.60 
G = - 5.40 kJ + (.008314 x 298)[ln 
]
2 
 0.40 
G = 0.3 kJ
#11 Use data from Appendix C to calculate Kp at 298 K for each of the following reactions:
(a) H2(g) + I2(g)  2 HI(g)
(a) G = 2G HI(g) - G H 2 (g) - G I 2 (g)
= 2(1.30) - 0 - 19.37 = - 16.77 kJ
G 0   RT ln Kp
16.77  (.008314)(298) ln Kp
K p  870
20.6 Effect of Concentration on Cell EMF
Review:
E is positive – spontaneous
E is negative - nonspontaneous
E =0 - equilibrium
• A voltaic cell is functional until E = 0 at which
point equilibrium has been reached. (The cell is
then “dead.”)
• The point at which E = 0 is determined by the
concentrations of the species involved in the
redox reaction.
The Nernst Equation
• We can calculate the cell potential under nonstandard conditions.
• Recall that: ΔG = ΔGo + RT lnQ and ΔG = - nFE
• We can substitute in our expression for the free energy change:
- nFE = - nFEo + RT lnQ
• Rearranging, we get the Nernst equation:
E  E RT lnQ
nF
On AP Equation Sheet!
R=8.31
• The Nernst equation can be simplified by collecting all the constants
together and using a temperature of 298 K:
0.0592
E = E° log Q
n
(n is the # of e transferred)
On AP Equation Sheet!
#12 At 298K we have the reaction:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
[Cu2+] = 5.0 M and [Zn2+] = 0.050M:
Q = [Zn+2] / [Cu+2]
0.0592
E = E° log Q
n
(a) Cu+2(aq) + 2e  Cu(s)
Zn(s)  Zn+2(aq) + 2e
Eored = +.34 V
Eoox = + 0.763V
Eo = + 1.10 V
0.0592
0.050
Ecell = 1.10V log
= 1.16V
2
5.0
Cell EMF and Chemical Equilibrium
A system is at equilibrium when Q=K and ΔE = 0.
0.0592
E = E° log Q
n
0.0592
0 = E° log K
n
nE°
log K =
0.0592
On AP Equation Sheet!
Thus, if we know the Eo, we can calculate the
equilibrium constant.
#13
What is the effect on the emf of the cell which has the overall
cell reaction:
Zn(s) + 2 H+1(aq)  Zn+2(aq) + H2(g)
for each of the following changes?
(a) the pressure of the H2 is increased in the cathode compartment
(b) zinc nitrate is added to the anode compartment
(c) sodium hydroxide is added to the cathode compartment,
decreasing [H+1]
(d) the surface area of the anode is doubled. E  E RT lnQ
nF
(a) Q increases, E decreases
(b) [Zn+2] increases, Q increases, E decreases
(c) [H+1] decreases, Q increases, E decreases
(d) No effect – does not appear in the Nernst equation
#14 A voltaic cell is constructed that uses the following reaction and
operates at 298 K:
Zn(s) + Ni+2(aq)  Zn+2(aq) + Ni(s)
(a) What is the emf of this cell under standard conditions?
(b) What is the emf of this cell when [Ni+2] = 3.00 M and
[Zn+2] = 0.100 M?
(a) Ni+2(aq) + 2e  Ni(s)
Zn(s)  Zn+2(aq) + 2e
Ni+2(aq) + Zn(s)  Ni(s) + Zn+2(aq)
Eored = - 0.28 V
Eoox = + 0.763V
Eo = + 0.48 V
0.0592
[Zn +2 ]
(b) E = E log
n
[Ni +2 ]
0.0592
[0.100]
E = 0.48 V log
= 0.53 V
2
[3.00]
#15 Using the standard state reduction potentials listed in Appendix E,
calculate the equilibrium constant for the following reaction at 298 K:
(a) Zn(s) + Sn+2(aq)  Zn+2(aq) + Sn(s)
n E
log K =
0.0592
(a) E = Eox + Ered = 0.763 + (- 0.136) = 0.627 V ;
n E
2(0.627)
log K =
=
0.0592
0.0592
K = 1.5 x 1021
n=2