CHAPTER 14: Equilibrium
• Reactions are reversible
• A + B ó C + D ( forward)
• C + D ó A + B (reverse)
• Initially there is only A and B so only the forward reaction is possible
• As C and D build up, the reverse reaction speeds up while the forward reaction slows down.
• Eventually the rates are equal
• Eventually you reach a point where the reverse reaction is going as fast as the forward reaction.
• This is dynamic equilibrium.
• Dynamic equilibrium: The rate of the forward reaction is equal to the rate of the reverse reaction.
• The concentration of products and reactants stays the same, but the reactions are still running.
What is at Equilibrium?
• Rates are equal
• Concentrations are not.
• Rates are determined by concentrations
• The concentrations do not change at equilibrium.
MEASURING EQUILIBRUIM
• At equilibrium the concentrations of products and reactants are constant.
• Equilibrium position: How much product and reactant there are at equilibrium.
• We can write a constant that will tell us where the equilibrium position is.
• Keq: Equilibrium Constant
WRITING THE EQUILIBRIUM EXPRESSION
• For any reaction
aA + bB ó cC + dD
Keq = [C]c[D]d / [A]a[B]b
• Write the equilibrium expressions for the following reactions.
• 3H2(g) + N2(g) ó 2NH3(g)
• 2H2O(g) ó 2H2(g) + O2(g)
The units for Keq
• Based on the molarities of reactants and products at equilibrium
• Generally omit the units
Keq is CONSTANT
• At any temperature.
• Temperature affects rate.
• Equilibrium position is a set of concentrations at equilibrium.
• Does not depend on starting concentrations but does on temperature.
What does Keq tell us?
• If Keq > 1 Products are favored
• If Keq < 1 Reactants are favored
CALCULATING EQUILIBRIUM
Nitrogen reacts with hydrogen to produce ammonia. At equilibrium the [N2] = 0.921 M, [H2] =.763 M and [NH3] = 0.157 M. What is the equilibrium constant?
• First write the balanced equation.
• 3H2(g) + N2(g) ó 2NH3(g)
• Write the equilibrium expression
• Plug concentrations given into the expression
A gaseous mixture contains 0.30 mol CO, 0.10 mol H2, and 0.020 mol H2O, plus an unknown amount of CH4, in each liter. What is the [CH4] in this mixture? Keq = 3.92
CO(g) + 3 H2(g) óCH4(g) + H2O(g)
• Write equilibrium expression
• Plug in concentrations given into the equilibrium expression
• Given the starting concentrations and one equilibrium concentration
• Use stoichiometry to figure out other concentrations and Keq
• Learn to create a table of initial and final conditions (ICE Box).
• I: Initial
• C: Change
• E: Equilibrium
Examples:
1. When 4.0 mol HI was put into a 5.0 L flask, it gave an equilibrium mixture containing 0.442 mol I2. What is the Keq?
2 HI (g) ó H2 (g) + I2 (g)
• Calculate the molarity of HI and I2
• Do you have a way to solve for “x” ?
• Given the value of I2 at equilibrium.
• This is the value of “x”.
• Use this value to calculate equilibrium concentrations for H2 and HI
• Then plug theses values into the equilibrium expression.
2. When 0.10 mol H2S was put into a 10.0 L flask, it gave an equilibrium mixture containing 0.0285 mol H2. What is the Keq?
H2S (g) ó H2 (g) + S (g)
• Calculate the molarity of H2S and H2
• Given the value of H2 at equilibrium.
• Calculate “x” and [H2S]eq and [S]
Equilibrium Involving Gases
• Write equilibrium expression in terms of partial pressure
• Value of Kp not necessarily equal to the value of Keq
Relationship between Keq and Kp
• Kp = Keq (RT)Δn
• Δn = ∑(products coefficients) – ∑(reactants coefficients)
• R = 0.08206 L atm / mol K
Examples:
2 NO (g) + O2 (g) ó 2 NO2 (g) Keq = 4.67 x 10 13 at 298 K.
Find Kp.
Homogeneous Equilibria
1. Reactants and products are in the same phase
2. Usually in the gas phase
Heterogeneous Equilibria
1. Reactants and products are in different phases
2. In the Keq expression do not write pure solids or liquids in the expression.
3. The concentration of a pure liquid or solid is a constant at a particular temperature.
Heterogeneous Equilibria
• H2(g) + I2(s) ó 2HI(g)
• Write the Keq expression.
REACTION QUOTIENT
• Tells you the direction the reaction will go to reach equilibrium
• Calculated the same as the equilibrium constant, but for a system not at equilibrium
The Reaction Quotient
• Q = [Products]coefficient
[Reactants] coefficient
• Compare value to equilibrium constant
What Q tells us
• If Q<K
- Not enough products
- Shift to right
• If Q>K
- Too many products
- Shift to left
• If Q=K system is at equilibrium
Examples:
1. If the [Co] and [Cl2] are each 0.15 M and the [CoCl2] = 1.1 x 10 -3 M, is the reaction at equilibrium? If not in which direction will it proceed?
CoCl2 (g) ó Co (g) + Cl2 (g) Keq = 170
2. At 740 ºC, Keq = 0.0060 for the reaction: CaCO3 (s) ó CaO (s) + CO2 (g). Find Q and predict how the reaction will proceed if [CO2] = 0.0004 M.
Le Chatelier’s Principle
• If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.
Changing Concentration
• If you add reactants shifts to the right
• If you add products shifts to the left
• If you remove reactants shifts to the left
• If you remove products shifts to the right
Changing Temperature
• Doesn’t just change the equilibrium position, changes the equilibrium constant.
• The direction of the shift depends on whether it is exo- or endothermic
Exothermic
• Think of heat as a product
• Releases heat
• Raising temperature push toward reactants.
• Shifts to left
Endothermic
• Think of heat as a reactant
• Raising temperature push toward products.
• Shifts to right.
Change Pressure
• Shrink Volume by increasing pressure
• Shift to the side with fewer gas molecules
• Increase volume by decreasing pressure
• Shift to the side with more gas molecules
• CCl4 (g) è C (s) + 2 Cl2 (g)
• Decrease Volume
Shift to left
Le Chatelier’s Principle
• Can you increase the amount of product in each of the following equations by increasing the pressure? Explain.
• Shrink volume increase pressure shift to side with less gas molecules
CO2(g) + H2(g) ó CO (g) + H2O (g)
No because there is the same number of gas molecules on each side
4 CuO (s) ó 2 Cu2O (s) + O2 (g)
No favor reactant side since it has fewer gas molecules
2 SO2 (g) + O2 (g) ó 2 SO3 (g)
Yes, product side has less gas molecules