Chemical Equilibrium
A state in which the concentration of all reactants and products
remains constant.
Chemical Equilibrium
• As a reaction progresses and reactants are used up forward
reaction slows down.
• As reaction progresses products build up and reverse reaction
starts to pick up speed.
• When slowing forward reaction and accelerating reverse
reaction rates become equal equilibrium is reached. Reaction
appears to go nowhere. Goes in both (forward and reverse)
directions simultaneously. Some molecules are undergoing
forward reaction while others are undergoing reverse
reaction.
• At equilibrium chemistry is occurring as fast as ever but
concentrations of reactants and products no longer change
with time.
At equilibrium both of the following reactions are occurring at the
same rate
N2O4(g) ! NO2(g) + O2(g)
NO2(g) + O2(g) ! N2O4(g)
The Equilibrium Constant (Kc)
For the reaction jA+ kB <=> lC + mD
Coefficients in the chemical equation become subscripts in the
equilibrium expression
Kc = [C]l[D]m
[A]j[B]k
Reactants or products that are solids or liquids do not appear in
the equation. Gases or aqueous substances appear
N2O4(g) <=> 2NO2(g)
Kc = [NO2]2
[N2O4]
AgCl(s) ! Ag+(aq) + Cl-(aq)
Kc = [Ag+][Cl-]
(AgCl is a solid so it is not present)
In a gaseous reactions the reactants and products can be expressed
in partial pressures
From the gaseous reactions above the Kp expression in
Kp = P2NO2
PN2O4
The relationship between Kc and Kp is:
Kp = Kc(RT)Δ n
Δn is the change in number of moles of gas as reactants become
products (Kc = Kp when the number of moles of gas remain the
same)
Write the equilibrium constant expressions, Kc and Kp for the
following reaction:
CaCO3(s) ! CaO(s) + CO2(g)
An equilibrium mixture of H2, I2, and HI gases at 425 oC is
determined to consist of 4.5647 x 10-3 mol/L of H2, 7.378 x 10-4
mol/L of I2 and 1.3544 x 10-2 mol/L of HI. What is the equilibrium
constant for the system at this temperature?
Working with the equilibrium expression
The equilibrium expression for the reverse reaction is the
reciprocal of that of the forward reaction.
Give the equilibrium constant for the reverse reactions for the
examples above.
A large equilibrium constant (greater than 1) favors the
forward reaction. A small equilibrium constant (less than 1)
favors the reverse reaction.
For the gas phase reaction
H2(g) + I2(g) = 2 HI(g)
Kc = 50.3 at 731 K. Equal amounts (0.100 M each) are
introduced to a container, and then the temperature is
raised to 731 K.
Is the system at equilibrium? If not, what direction is the
reaction?
Solution
When equal amounts are present,
[HI]2
(0.1)2
Qc = ---------- = -------- = 1 < Kc
(50.3)
[H2] [I2]
(0.1)2
There is a tendency to increase [HI] to reach equilibrium at
731 K.
Exercise 1 Writing Equilibrium Expressions
Write the equilibrium expression for the following reaction:
4 NH3(g) + 7 O2(g) !!4 NO2(g) + 6 H2O(g)
Exercise 2 Equilibrium Expressions for Heterogeneous Equilibria
Write the expressions for K and Kp for the following processes:
a. The decomposition of solid phosphorus pentachloride to
liquid phosphorus trichloride and chlorine gas.
b. Deep blue solid copper(II) sulfate pentahydrate is heated to
drive off water vapor to form white solid
copper(II) sulfate.
Exercise 3 Calculating the Values of K
The following equilibrium concentrations were observed for the
Haber process at 127°C:
[NH3] = 31 × 10−2 mol/L
[N2] = 8.5 × 10−1 mol/L
[H 2] = 3.1 × 10−3 mol/L
a. Calculate the value of K at 127°C for this reaction.
b. Calculate the value of the equilibrium constant at 127°C for the
reaction:
2 NH3(g) <=>!N2(g) + 3 H2(g)
c. Calculate the value of the equilibrium constant at 127°C for the
reaction given by the equation:
1/2N2(g) + 3/2H2(g) <=> NH3(g)
Exercise 4 Equilibrium Positions
The following results were collected for two experiments involving
the reaction at 600°C between gaseous sulfur dioxide and oxygen
to form gaseous sulfur trioxide:
Show that the equilibrium constant is the same in both cases.
Exercise 5 Calculating Values of Kp
The reaction for the formation of nitrosyl chloride
2 NO(g) + Cl2(g) <=> 2 NOCl(g)
was studied at 25°C. The pressures at equilibrium were found to be
PNOCl = 1.2 atm
PNO = 5.0 × 10-2 atm
PCl2 = 3.0×10-1 atm
Calculate the value of Kp for this reaction at 25°C.
Exercise 6 Calculating K from Kp
Using the value of Kp obtained in Sample Exercise 13.4, calculate
the value of K at 25° C for the reaction:
2 NO(g) + Cl2(g) <=> 2 NOCl(g)