I. Equilibrium
A. Definition 1
Equilibrium is the state in a reaction
when the concentrations of reactants and products are not changing
(static).
1. Example:
Co(H2O)62+
+ 4 Cl- <--> CoCl42-
+ 6 H2O
pink
blue
Each reaction is constantly occurring,
but there is no net change in the amounts of each.
B. Definition 2
Equilibrium is the state in which the
forward rate of reaction equals the reverse rate of reaction (dynamic).
1. Example:
Rate = k1[Co(H2O)62+][Cl-]4
= k-1[CoCl42-][H2O]6
or:
( = 3.8 x 104 M-2 at 127°C)
C. Keq is independent of reaction
mechanism because it is dependent on overall stoichiometry
not on how the molecules react.
Example:
O2 + N2 <--> 2 NO (combustion
of nitrogen)
Kc = [NO]2/[O2][N2]
regardless of proposed mechanism.
Remember: Keq is a measure of where a system is when it is not
changing over time
not what a system is doing from moment to moment.
Keq is a state function, like G, H and S.
II. Measuring Kc and getting
a feel for how complete is a reaction.
A. Theoretical example: aA + bB
<--> cC + dD
| Kc | A&B | C&D |
| 0.001 | high | low |
| 0.10 | sl. lower | sl. higher |
| 10 | low | high |
| 100 | very low | very high |
III. Calculating Keq with pressures:
Kp
A. Kp = equilibrium constant of
the reaction expressed with the pressures of molecules.
Theoretical
example: aA + bB <--> cC + dD
B. Relationship of Kp and Kc
1. Use Ideal
Gas Law: PV = nRT
P = nRT/V = CRT (C = concentration = n/V)
Thus Kp differs from Kc by factor of (RT)n
2. Example:
determining (RT)n for specific reaction.
2 NO (g) + Cl2 (g)
<--> 2 NOCl (g) at 25°C
Equilibrium Conditions
PNOCl = 1.2 atm
PNO = 0.050 atm
PCl2 = 0.30 atm
Kp = (PNOCl)2/(PNO)2(PCl2)
= (1.2 atm)2/(0.050 atm)2(0.30 atm)
= 1.9 x 103 1/atm
Kc = (PNOCl/RT)2/(PNO/RT)2(PCl2/RT)
= KpRT
= (1.9 x 103 1/atm)(0.08206 atm/M•K)(298 K)
= 4.7 x 104 1/M
Kp = Kc(RT)Dn
[Dn = (Sgas product
coeff. - Sgas reactant coeff.)]
Kp = Kc when they are unitless!!
IV. Heterogeneous Equilibria
A. What is the concentration
or pressure of a solid??
What is the concentration of a pure liquid??
ANS.: They have no true 'concentrations', but are assumed to have unit
activities,
which is simplified to [ ] = constant .
(Physical Chemistry topic)
B. Example: production of lime from decomposition
of CaCO3
CaCO3 (s) <--> CaO (s) + CO2
(g)
Kc
= [CaO][CO2]/[CaCO3] = C2[CO2]/C1
KcC1/C2
= [CO2] ; Kc' = [CO2]
(units = M)
Simplifying assumption applied here:
[pure liquid] and [pure solid] = unitless constants
V. Return to LeChatlier's Principle
A. Definition:
1. A system
at equilibrium that is perturbed will shift (change forward or reverse
rates) to re-establish equilibrium.
=> shift to make Q = Keq once again
2. Analogy:
think of a rock in a ditch (equilibrium):
If you push the rock up one side, then let go (perturbation), the rock
will roll back to the bottom of the ditch.
B. Perturbations
1. A change
in concentration of one of the components of the system: 2 NOCl
--> 2 NO + Cl2 (g)
a. Add NOCl: system shifts to the right (more NO and Cl2)
b. Add Cl2: system shifts to the left (more NOCl)
c. Remove NO: system shifts to the right (more NO and Cl2)
2. A change
in the pressure of one of the components or the whole system (could change
volume): PV = nRT
a. Ways of changing pressure
1. change concentrations of any component
2. change the volume of the container
Reduce volume by half:
pressure increases
Q changes because there are different numbers of molecules of reactants
(2) and products (3).
Kp = PCl2PNO2/PNOCl2
(atm)
=> if the pressure doubles, then Q doubles also.
The system will need to reduce the pressure to get Q = Kp once
again. - -> How?
By shifting to the left and reducing the total number of molecules present
(decrease P).
3. A change
in the temperature of the system
a. Remember the Arrhenius equation: k = Ae-Ea/RT
=> this shows that the rate of a reaction is dependent on temperature.
b. Direct or inverse dependence of Keq on T?
=> Is heat consumed in the forward reaction, or released?
1. Heat is consumed: forward reaction is endothermic.
(As the temperature of the system increases, more heat is available for
the forward reaction,
thus making it more favorable; Keq increases.)
2. Heat is produced: forward reaction is exothermic.
(As the temperature of the system increases, more heat is available for
the reverse reaction,
thus making the forward reaction less favorable; Keq
decreases.)
c. Keq actually changes with a change in temperature for exothermic
and endothermic reactions.
d. Example: N2 + 3 H2 -->
2 NH3 + 92 kJ (g)
| T (K) : | Kc (M-2) : |
| 500 | 90 |
| 600 | 3 |
| 700 | 0.3 |
| 800 | 0.04 |
VI. Working Problems Involving
Kc and Kp
A. Determine equilibrium concentrations if
given Kc
H2
+ F2 <--> 2 HF (g) ;
Kc = 1.0 x 102
Initial Conditions:
1.0 L flask
2.0 moles H2
=> 2.0 M
2.0 moles F2
=> 2.0 M
0 moles HF =>
0 M
1. Determine the reaction quotient (Q)
- put
the initial conditions into the equilibrium expression
Q = [HF]2/[H2][F2]
= 02 M2 /2.0 M x 2.0 M = 0 (unitless)
Q < Kc \
reaction shifts to the right
2. Determine the shift (x)
H2 + F2
<--> 2 HF (g)
2.0 2.0
0 M initial
-x -x
+ 2x M change
2.0-x 2.0-x
2x M final
Kc = 1.0
x 102 = (2x)2/(2.0-x)2
10 = 2x/2.0-x
20-10x = 2x
x = 20/12 = 1.7 M
3. Determine the equlibrium
Final Conditions: 1.0 L flask
2.0 M - 1.7 M H2 => 0.3 M H2
2.0 M - 1.7 M F2 => 0.3 M F2
0 M + 2(1.7 M) HF => 3.4 M HF
Check:
plug values into Kc expression:
Kc = (3.4 M)2/(0.3 M)2 = 128 (pretty close)
B. Determine Kc if given initial
concentrations and one equilibrium concentration:
PCl5 <--> PCl3 + Cl2
(g)
Initial Conditions: 1.00 L flask
0.01305 moles PCl5 => 0.01305 M PCl5
0.447 moles PCl3 => 0.447 M PCl3
0 moles Cl2 => 0 M Cl2
Equilibrium Condition: 0.0030 M Cl2
1. Determine
other equilibrium concentrations.
PCl5 <--> PCl3 +
Cl2 (g)
0.01305 0.447
0
M initial
-0.0030 +0.0030
+0.0030 M change
0.01005 0.450
0.0030 M final
2. Calculate
Kc
Kc = [PCl3][Cl2]/[PCl5] = (0.450
M)(0.0030 M)/(0.01005 M)
= 0.134 M