1. Chemical Equilibrium
4/24/2017

2. The Concept of Equilibrium
Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2:
N2O4(g)  2NO2(g).
At some time, the color stops changing and we have a mixture of N2O4 and NO2.
Chemical equilibrium is the point at which the concentrations of all species are constant.

3. The Concept of Equilibrium

5. The Concept of Equilibrium
At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4:
The double arrow implies the process is dynamic.
Consider
Forward reaction: A  B Rate = kf[A]
Reverse reaction: B  A Rate = kr[B]
At equilibrium kf[A] = kr[B].
NO2 – equilibrium

6. The Concept of Equilibrium

7. The Equilibrium Constant
For a general reaction in the gas phase
the equilibrium constant expression is
where Keq is the equilibrium constant.

8. The Equilibrium Constant
For a general reaction
the equilibrium constant expression for everything in solution is
where Keq is the equilibrium constant.

9. The Equilibrium Constant

11. The Equilibrium Constant
The Magnitude of Equilibrium Constants

12. The Equilibrium Constant
The Direction of the Chemical Equation and Keq
In the reverse direction:

13. The Equilibrium Constant
Other Ways to Manipulate Chemical Equations and Keq Values
The reaction
has
which is the square of the equilibrium constant for

14. The Equilibrium Constant
Other Ways to Manipulate Chemical Equations and Keq Values
Equilibrium constant for the reverse direction is the inverse of that for the forward direction.
When a reaction is multiplied by a number, the equilibrium constant is raised to that power.
The equilibrium constant for a reaction which is the sum of other reactions is the product of the equilibrium constants for the individual reactions.

16. The Equilibrium Constant
Given at 700K:
H2(g) + I2(g)  2 HI(g) Keq = 54.0
N2(g) H2(g)  2 NH3(g) Keq’ = 1.04x10-4
Determine Keq’’ at 700K for:
2 NH3(g) + 3 I2(g)  6 HI(g) + N2(g)
Answer: 1.51x109
Unit(s) of the Equilibrium Constant
Dimensionless because:
Partial Pressures divided by Reference Pressure of 1 atm.
Molarities divided by Reference Concentration of 1 Molar.

17. Answer: 1.51x109 Given at 700K: H2(g) + I2(g)  2 HI(g) Keq = 54.0
N2(g) H2(g)  2 NH3(g) Keq’ = 1.04x10-4
Determine Keq’’ at 700K for:
2 NH3(g) + 3 I2(g)  6 HI(g) + N2(g)
Answer: 1.51x109

18. Heterogeneous Equilibria
When all reactants and products are in one phase, the equilibrium is homogeneous.
If one or more reactants or products are in a different phase, the equilibrium is heterogeneous.
Consider:
experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

19. Heterogeneous Equilibria
The concentration of a solid or pure liquid is its density divided by molar mass.
Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant.
For the decomposition of CaCO3:

20. Heterogeneous Equilibria
We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.
The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.
Examples:

22. Calculating Equilibrium Constants
Proceed as follows:
Tabulate initial and equilibrium concentrations (or partial pressures) given.
If an initial and equilibrium concentration is given for a species, calculate the change in concentration.
Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species.
Deduce the equilibrium concentrations of all species.
Usually, the initial concentration of products is zero. (This is not always the case.)

23. Calculating Equilibrium Constants
A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC. At equilibrium, the partial pressures of the gases were determined to be 7.38 atm H2, 2.46 atm N2 and atm NH3. Calculate Keq .
Enough ammonia is dissolved in 5.00 L of water at 25oC to produce a solution that is M in ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of OH- is 4.64x10-4 M. Calculate Keq at 25oC for the reaction.
Gaseous sulfur trioxide decomposes to gases of sulfur dioxide and oxygen at high temperature in a sealed container. Initially the vessel is charged at 1000 K with SO3(g) at a partial pressure of atm. At equilibrium, the SO3 partial pressure is atm. Calculate the value of Keq at 1000 K.

24. A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC
A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC. At equilibrium, the partial pressures of the gases were determined to be 7.38 atm H2, 2.46 atm N2 and atm NH3. Calculate Keq .
Answer: 2.79x10-5

25. Enough ammonia is dissolved in 5
Enough ammonia is dissolved in 5.00 L of water at 25oC to produce a solution that is M in ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of OH- is 4.64x10-4 M. Calculate Keq at 25oC for the reaction.
Answer: 1.80x10-5

26. Gaseous sulfur trioxide decomposes to gases of sulfur dioxide and oxygen at high temperature in a sealed container. Initially the vessel is charged at 1000 K with SO3(g) at a partial pressure of atm. At equilibrium, the SO3 partial pressure is atm. Calculate the value of Keq at 1000 K.
Answer:

27. Applications of Equilibrium Constants
Predicting the Direction of Reaction
We define Q, the reaction quotient, for a general reaction as
Q = K only at equilibrium.

28. Applications of Equilibrium Constants
Predicting the Direction of Reaction
If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K).
If Q < K then the forward reaction must occur to reach equilibrium.

30. Applications of Equilibrium Constants
Calculating Equilibrium Constants
The same steps used to calculate equilibrium constants are used.
Generally, we do not have a number for the change in concentration line.
Therefore, we need to assume that x mol/L of a species is produced (or used).
The equilibrium concentrations are given as algebraic expressions.

31. Applications of Equilibrium Constants
Predicting the Direction of Reaction
At 1000 K the value of Keq for the reaction 2SO3(g)  2SO2(g) + O2(g) is Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are 0.16 atm for SO3, 0.41 atm for SO2 , and 2.5 atm for O2 .
At 448oC, Keq is 51 for: H2(g) + I2(g)  2HI(g) . Predict how the reaction will proceed to reach equilibrium at 448oC if we start with 2.0x10-2 mol HI , 1.0x10-2 mol H2 , and 3.0x10-2 mol I2 in a 2.00-L container.
[Hint: Calculate partial pressures using Ideal Gas Law. Calculate Q.]
[Answer: Q= Reaction proceeds from left to right (i.e. more products)]

33. Applications of Equilibrium Constants
Calculating Equilibrium Concentrations
At 500 K the reaction: PCl5(g)  PCl3(g) + Cl2(g) , has Keq = In an equilibrium mixture at 500 K, the partial pressure of PCl5 is atm and that of PCl3 is atm. What is the partial pressure of Cl2 in the equilibrium mixture?
A L flask is filled with mol of H2 and mol of I2 at 448oC. The value of the Keq for the reaction: H2(g) + I2(g)  2HI(g) , at 448oC is What are the partial pressures of H2 , I2 , and HI in the flask at equilibrium?
[Hint: Solve for initial partial pressures using Ideal Gas law. Set up initial/change/equil’m table and define x as change in partial pressures of reactants. Solve for x in Keq expression using quadratic formulation.] [Answer: x = 55.3 atm; H2 = 3.90 atm, I2 = 63.1 atm, HI = 111 atm.]

34. 2. A 1. 000-L flask is filled with 1. 000 mol of H2 and 2
2. A L flask is filled with mol of H2 and mol of I2 at 448oC. The value of the Keq for the reaction: H2(g) + I2(g)  2HI(g) , at 448oC is What are the partial pressures of H2 , I2 , and HI in the flask at equilibrium?
[Hint: Solve for initial partial pressures using Ideal Gas law. Set up initial/change/equil’m table and define x as change in partial pressures of reactants. Solve for x in Keq expression using quadratic formulation.] [Answer: x = 55.3 atm; H2 = 3.90 atm, I2 = 63.1 atm, HI = 111 atm.]

35. Le Châtelier’s Principle
Consider the production of ammonia
As the pressure increases, the amount of ammonia present at equilibrium increases.
As the temperature decreases, the amount of ammonia at equilibrium increases.
Can this be predicted?
Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

37. Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Consider the Haber process
If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).
The system must consume the H2 and produce products until a new equilibrium is established.
So, [H2] and [N2] will decrease and [NH3] increases.

39. Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Adding a reactant or product shifts the equilibrium away from the increase.
Removing a reactant or product shifts the equilibrium towards the decrease.
To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).
We illustrate the concept with the industrial preparation of ammonia.

41. Le Châtelier’s Principle
Effects of Volume and Pressure Changes
As volume is decreased pressure increases.
Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.
That is, the system shifts to remove gases and decrease pressure.
An increase in pressure favors the direction that has fewer moles of gas.
In a reaction with the same number of product and reactant moles of gas, pressure has no effect.

42. Le Châtelier’s Principle
Effects of Volume and Pressure Changes
An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.
The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.
The system moves to reduce the number moles of gas (i.e. backward reaction is favored).

43. Le Châtelier’s Principle
Effect of Temperature Changes
The equilibrium constant is temperature dependent.
For an endothermic reaction, H > 0 and heat can be considered as a reactant.
For an exothermic reaction, H < 0 and heat can be considered as a product.

44. Le Châtelier’s Principle
Effect of Temperature Changes
Adding heat (i.e. heating the vessel) favors away from the increase:
if H > 0, adding heat favors the forward reaction,
if H < 0, adding heat favors the reverse reaction.
Removing heat (i.e. cooling the vessel), favors towards the decrease:
if H > 0, cooling favors the reverse reaction,
if H < 0, cooling favors the forward reaction.

45. Le Châtelier’s Principle
Effect of Temperature Changes
Consider
for which DH > 0.
Co(H2O)62+ is pale pink and CoCl42- is blue.
If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue.
Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl42-.
If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink.
Since H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H2O)62+.

47. Le Châtelier’s Principle
The Effect of Catalysis
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 affect the composition of the equilibrium mixture.

49. Le Châtelier’s Principle Summary
If a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.
Adding a reactant or product shifts the equilibrium away from the increase.
Upon pressure increase, reaction will proceed towards the side with the fewest number of gaseous moles.
Adding heat (i.e. heating the vessel) favors away from the increase:
if ∆H > 0, adding heat favors the forward reaction,
if ∆H < 0, adding heat favors the reverse reaction.
A catalyst does not affect the composition of the equilibrium mixture.

50. Chemical Equilibrium