AP Chemistry Chemical Equilibrium
Chemical Equilibrium ­ A Dynamic Experience!
Many reactions are reversible, and we end up with a mixture of reactants and products that are eventually in a state of Chemical (dynamic) Equilibrium.
Chemical Equilibrium ­ A Dynamic Experience!
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Place 1.000 mol CO and 3.000 mol H2 in a 10.00 L vessel at 1200 K
Initially, high concentrations of reactants and a high forward reaction rate. Over time though...
14.1 ­ 14.3
AP Chemistry Chemical Equilibrium
Chemical Equilibrium ­ A Dynamic Experience!
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
1.000 mol CO and 3.000 mol H2 in a 10.00 L vessel at 1200 K and reach equilibrium.
The vessel is cooled and the water vapor becomes liquid (0.387 mol of water).
What is the molar composition of the equilibrium mixture?
Moles
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Initial
Change
Equilibrium
Start with 4 mol, completion is 2 mol.
Equilibrium = intermediate = 3.226 mol
The Equilibrium Constant!
Look at our previous example, starting with
1.000 mol CO and 3.000 mol H2.
We run a second experiment starting with
2.000 mol CO and 3.000 mol H2.
Our equilibrium molar compositions are found to be:
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Exp. 1: .613 mol
1.839 mol
.387 mol
.387 mol
Exp. 2: 1.522 mol 1.566 mol
.478 mol
.478 mol
Conclusion?
14.1 ­ 14.3
AP Chemistry Chemical Equilibrium
The Equilibrium Constant (Kc)
aA + bB cC + dD
Kc =
[C]c [D]d
[A]a [B]b
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Kc =
[CH4] [H2O]
[CO] [H2]3
Only gases are in the Kc expression.
Liquid and solid concentrations are constant
The Equilibrium Constant (Kc)
N2 (g) + 3 H2 (g) 2 NH3 (g)
us
o
n
ge ria
o
m lib
Ho qui
E
Kc =
1/2 N2 (g) + 3/2 H2 (g) NH3 (g)
Kc =
3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g)
s
ou
n
e
og bria
r
te
li
He Equi
[H2]4
Kc =
[H2O]4
*equilibrium not affected by amounts of solids/liquids...just need some present!
14.1 ­ 14.3
AP Chemistry Chemical Equilibrium
When given the Equilibrium Constant (Kc) expression, you can write the balanced equation from it!
Write the balanced equation for the gas reaction with the following Equilibrium Constant expression:
Kc =
[CO2]3 [H2O]4
[C3H8] [O2]5
C3H8 + 5 O2 3 CO2 + 4 H2O
The Equilibrium Constant (Kc) expression can also be manipulated...
The Equilibrium Constant expression for a reaction is:
Kc =
[NH3]4 [O2]5
[NO]4 [H2O]6
What is the Equilibrium Constant expression when the reaction is halved and reversed?
[NO]2 [H2O]3
Kc =
[NH3]2 [O2]5/2
14.1 ­ 14.3
AP Chemistry Chemical Equilibrium
The Equilibrium Constant (Kc)
The "c" in Kc means the equilibrium constant is defined in terms of molar concentrations
The Law of Mass Action:
a relation that states that the values of Kc are constant for a reaction at a given temperature, no matter what equilibrium concentrations are substituted into the expression
regardless of starting amount of reactants
The Law of Mass Action:
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Exp. 1: .613 mol
1.839 mol
.387 mol
.387 mol
Exp. 2: 1.522 mol 1.566 mol
.478 mol
.478 mol
10.00 L vessel, so M = mol/10.00
Kc =
14.1 ­ 14.3
[CH4] [H2O]
[CO] [H2]3
AP Chemistry Chemical Equilibrium
2 HI (g) H2 (g) + I2 (g)
When 4.00 mol HI was placed in a 5.00 L vessel at 731 K, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc for the decomposition of HI at this temperature?
2 HI (g) H2 (g) + I2 (g)
Molarity
Initial
Change
Equilibrium
Kc =
[H2] [I2]
[HI]2
(.0884)2
=
=
2
(.623)
2 HI (g) H2 (g) + I2 (g) Kc = .0201
at 731 K What is the value of Kc for the
following reaction at 731 K?
H2 (g) + I2 (g) 2 HI (g) It's the reverse reaction, so the Kc expression would be the reciprocal of the forward reaction!
So, the new Kc value will be the reciprocal of the forward Kc value!
14.1 ­ 14.3
AP Chemistry Chemical Equilibrium
The Equilibrium Constant (Kc)
The Kinetics Argument!
N2O4 (g) 2 NO2 (g)
Forward: Ratef = kf [N2O4]
Reverse: Rater = kr [NO2]2
At equilibrium: Ratef = Rater
The Equilibrium Constant (Kp)
The equilibrium constant expression can be expressed in partial pressures for gases instead of molar concentrations as the concentration of a gas is proportional to its partial pressure...remember?
PV = nRT, so n/V = P/RT
(M) constant at
a given temp
So... Kp = Kc (RT) n
n = sum of gas products ­ sum of gas reactants
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Kp =
14.1 ­ 14.3
Kp =
AP Chemistry Chemical Equilibrium
The Sum of Reactions (Kc or Kp)
When 2 reactions add together to get an overall reaction, the K for the overall reaction is the product of the K's Kc = 3.92
CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
Kc = 3.3 x 104 CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g)
Kc = (3.92) (3.3 x 104) = 1.3 x 105
14.1 ­ 14.3