Equilibrium II
15.6 Using the Equilibrium Constant (Keq)
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.
#1
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?
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?
#2
#3
At 373 K, Kp = 0.416 for the equilibrium:
are equal, what is the equilibrium pressure of Br2?
2 NOBr  2 NO + Br2. If the pressures of NOBr and NO
#4
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?
Predicting the Direction of Reaction
For a general reaction: aA + bB cC + dD
We define Q, the reaction quotient, as:
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
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?
(a)
[COCl2] = 2.00  10-3 M [CO] = 3.3  10-6 M [Cl2] = 6.62  10-6 M
(b)
[COCl2] = 4.50  10-2 M [CO] = 1.1  10-7 M [Cl2] = 2.25  10-6 M
(c)
[COCl2] = 0.0100 M [CO] = [Cl2] = 1.48  10-6 M
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).
- The system must consume the H2 and produce products until equilibrium is re-established
- [H2] and [N2] will decrease and [NH3] increase until Q = Keq.
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 (Keq)
• 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
Consider the endothermic reaction:
Co(H2O)62+(aq) + 4Cl–(aq)  CoCl42–(aq) + 6H2O(l)
•
•
•
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).
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
#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
#8 For the reaction,
PCl5 (g)  PCl3 (g) + Cl2 (g) ΔHrxn = +111 kJ.
Fill in the following table:
Change to reaction
Reaction shifts
Change in K
Add PCl5
Remove Cl2
_______
______
Add Ar
______
Decrease P
______
Increase T
______
Add a catalyst
______
Decrease V of container
19.7
Free Energy and the Equilibrium Constant
Gibbs Free Energy: ΔG
If ΔG is (+) the reaction is nonspontaneous
If ΔG is (–) the reaction is spontaneous
If ΔG = 0 the system is at equilibrium
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:
At equilibrium, Q = Keq and ΔG = 0
0 = ΔGo + RT ln Keq
ΔGo = – RT ln Keq
From the above we can conclude:
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)
#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.
#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)
20.6
Effect of Concentration on Cell EMF
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:
•
The Nernst equation can be simplified by collecting all the constants together and using a temperature of
298K:
(n is the # of e transferred)
#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]
Cell EMF and Chemical Equilibrium
A system is at equilibrium when ΔG = 0 and ΔE = 0.
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.
#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?
#15 Using the standard state reduction potentials, calculate the equilibrium constant at 298 K:
(a) Zn(s) + Sn+2(aq)  Zn+2(aq) + Sn(s)
HW#2: 15 (46, 48, 50, 52, 56, 62, 64) 19(78, 80, 84) 20(52, 58)