Equilibrium
Closed system - a system in which no materials enter or leave. Constant numbers of atoms of each type.
Equilibrium - A state in which the properties (color, pressure) of a closed system at a constant temperature do not change.
Examples of equilibrium previously encountered -
Vapor pressure over a liquid in a closed
container at a constant temperature
Saturated solution with solute present undissolved
Dynamic nature of equilibrium - Although no visible change in properties is occurring at equilibrium, at the molecular level processes are occurring.
Vapor pressure -
molecules are evaporating and condensing
rate of evaporation = rate of condensation
Saturated solution -
molecules leave the solute crystal at the same rate they return to it
Physical Equilibrium
Vapor Pressure in a closed container at constant Temperature
I. Liquid is added to empty container and sealed. Initially,
rate of evaporation > rate of condensation.
II. Rate of condensation is increasing.
III. Equilibrium exists.
rate of evaporation = rate of condensation
Saturated solution
I. Iodine crystals are added to a mixture of water and alcohol. Iodine begins dissolving.
II. Iodine continues to dissolve, and the color of the solution darkens.
III. Equilibrium exists. No observable changes in color occur.
rate of dissolving = rate of crystallization
Chemical Equilibrium
Consider the reaction of
H2(g) + I2(g) <-----> 2HI(g)
H2 and HI are colorless. I2 is a purple. The progress of the reaction can be followed by graphing the color intensity with time.
Starting with only H2(g) and I2(g) the color intensity decreases as reactants are used and products are formed. When equilibrium is reached, there is no further change in the color intensity of the system.
Approach to equilibrium
The same state of equilibrium is reached (concentration of reactants and products) whether we start with 1.0M H2 and 1.0 M I2 and no HI or with 2.0 M HI and no H2(g) or I2(g).
At equilibrium, the
rate of the forward reaction =
rate of the reverse reaction
We can show the system is in equilibrium by writing
2HI(g) <-----> H2(g) + I2(g).
Once equilibrium is reached it is maintained as long as the conditions remain constant.
Equilibrium Constant
From the lab we find that for the reaction
Fe3+(aq) + SCN-(aq) <-----> FeSCN2+(aq),
the expression
[FeSCN2+] = 140
[Fe3+][SCN-]
remains constant when equilibrium is reached.
We call this expression the equilibrium constant, Kc.
For this reaction, no matter what concentrations we start with, at equilibrium the concentrations will satisfy the equilibrium constant.
We can write equilibrium constant expressions for any reaction.
Consider
Example 1
A(g) + B(g) <-----> C(g)
Kc = [C]
[A][B]
Example 2
A(g) + A(g) <-----> C(g)
Kc = [C] = [C]
[A][A] [A]2
In general, Kc is the equilibrium constant for a reaction. The concentrations are in units of molarity .
The equilibrium constant has a certain value for a given reaction at a constant temperature.
For the general reaction,
nA(g) + mB(g) <-----> pC(g) + qD(g)
the equilibrium expression is
Kc= [C]p[D]q
[A]n[B]m
The concentrations of pure solids and liquids are not written in the equilibrium constant since they remain constant at a constant temperature.
For example, the equilibrium constant expression for the reaction
CaCO3(s) <-----> CaO(s) + CO2(g)
is written
Kc = [CO2].
The concentrations of the solids CaCO3 and CO2 are included in the value of Kc.
Note - Although solids and liquids do not appear in the equilibrium constant expression, they must be present for equilibrium to be established.
Some Equilibrium Constants
Ag+(aq) + 2NH3(aq) <-----> Ag(NH3)2+(aq)
K = [Ag(NH3)2]= 1.7 x 107 at 25o C
[Ag+][NH3]2
N2O4(g) <----> 2NO2(g)
K = [NO2]2= 8.3 x 10-1 at 55o C
[N2O4]
2HI(g) <----> H2(g) + I2(g)
K = [H2][I2]= 1.84 x 10-2 at 423o C
[HI]2
CH3COOH(aq) <----> H+(aq) + CH3COO-(aq)
K = [H+][CH3COO-]= 1.8 x 10-5 at 25o C
[CH3COOH]
Facts about K
K can only be determined by experiments.
Each substances must be present at equilibrium.
A large value of K (>1) means products are favored at equilibrium.
A small value of K (<1) means that reactants are favored at equilibrium.
Relationship between Kc and Kp
For Kc quantities are expressed in molarity units (mol/L).
For reactions involving gases, the quantities of reactants and products can be expressed in pressure units of atm. When expressed this way the equilibrium constant is denoted as Kp.
For the reaction,
nA(g) + mB(g) <----> pC(g)
Kp = PCp
PAnPBm
where pressure is measured in atm.
Mathematically, Kc and Kp can be related.
Kp = (RT)nKc
where n = number of moles of product -
number of moles of reactants
and R = 0.0821 L-atm/mol-K
Example -
2HI(g) <----> H2(g) + I2(g) at 25o C
Kc = 0.016
Since the number of moles of product and reactant are equal, n = 0; therefore,
Kp= Kc = 0.016
Example -
The Haber process is a method for synthesizing ammonia, a major constituent of fertilizers, by the following reaction
N2(g) + 3H2(g) <----> 2NH3(g)
Kc for this reaction at 472o C is 0.105. Calculate Kp for this reaction.
Kp = (RT)-2Kc
= (0.0821x(472+273))-2(0.105)
= 2.81 x 10-5
Finding K from values for other reactions
Example 1 -
A(g) + 2B(g) <----> 2C(g)
Kf = [C]2
[A][B]2
For the reverse reaction
2C(g) <----> A(g) + 2B(g)
Kr = [A][B]2= 1
[C]2 Kf
Opposite reactions have Ks that are inverses of each other.
Example 2 -
A(g) + B(g) <----> 2C(g) + D(g)K1 = [C]2[D]
[A][B]
2C(g) + D(g) <----> 3E(g) + F(g) K2 = [E]3[F]
[C]2[D]
For the overall reaction
A(g) + B(g) <----> 3E(g) + F(g)
K = [E]3[F] = [E]3[F] .[C]2[D]= K1xK2
[A][B] [C]2[D][A][B]
In general, when an overall reaction is found by adding reactions, the equilibrium constant for the overall reaction is the product of the individual equilibrium constants.
Example 3 -
A(g) + B(g) <----> 2C(g)K1
A(g) + B(g) <----> 2C(g)K1
2A(g) +2B(g) <----> 4C(g) K = K1xK1 = K12
When creating a new equation by multiplying all the species coefficients by n, the resulting K is found by raising the original K to the power n.
Equilibrium Calculations
The following examples relate to the reaction:
2HI(g) <----> H2(g) + I2(g)
at 520o C in a sealed container.
Example 1 -
Starting with an initial concentration of 0.200 M HI, it is found that at equilibrium the concentration of HI is 0.160 M. Calculate Kc for the reaction.
Note: -2[HI] = [H2] = [I2] from the equation coefficients.
HIH2I2
Initial Conc 0.20000
change-2xxx
.040+.020+.020
Equilibrium Conc .200-2x=.160.020.020
2x = .040
Kc = [H2][I2] = (0.020)(0.020) =0.016
[HI]2 (0.160)2
Example 2 -
Starting with 0.100 M H2 and 0.100 M I2, at equilibrium the concentration of H2 is 0.020 M. Find the other concentrations at equilibrium and calculate Kc.
HI H2 I2
Initial Conc00.1000.100
change+2x-x-x
Equilibrium Conc.160 .100-x =.020.020
x = .080
Kc = (0.020)(0.020) = 0.016
(0.160)2
Note - Starting with equivalent amounts of reactants or products the same equilibrium state is reached.
Example 3 -
Initial concentrations of all species is 0.100 M.
Find the equilibrium concentrations.
Since K<1, the equilibrium will favor the reactants. H2 and I2 will react to form more HI.
Assume x M of H2 and I2 react.
2HI(g) <----> H2(g) + I2(g)
+2x-x-x
HIH2I2
Initial conc 0.1000.1000.100
change +2x-x-x
Equilibrium conc 0.100+2x0.100-x0.100-x
Using the equilibrium constant,
(0.100-x)(0.100-x) = 0.016
(0.100+2x)2
0.0100 - 0.200x + x2= 0.016
0.0100 +0.400x + 4x2
Cross multiplying
0.0100 - 0.200x + x2 = 0.00016 + 0.0064x + 0.064x2
Rearranging
0.936x2 - 0.2064x + 0.00984 = 0
Using the quadratic equation
x = 0.070 or 0.53 the second root is impossible
[HI] = 0.100 + 0.140 = 0.240 M
[H2] = [I2] = 0.100 - 0.070 = 0.030 M