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Chapter 13Chapter 13
What Is It?What Is It?
The state where concentrations of all reactants and products remain constant with time.
At the molecularlevel, the reaction
continues. Macroscopically,
the reaction appears static.
Why Is It?Why Is It?
• Reaction rate depends on concentration.
• Collision Theory says more collisions = faster reactions.
• As the reaction progresses, the concentration of products begins to increase.
Nothing appears to change because the rates of reaction are so naturally slow. A catalyst is used to begin the process for
commercial production of ammonia. Haber applied Le Chatelier’s Principle to maximize the forward reaction.
Equilibrium Position is Equilibrium Position is Determined By:Determined By:
• Initial Concentrations
• Energies of reactants and products
• Organization (disorder) of reactants and products
N2(g) + 3H2(g) 2NH3(g)
Le Chatelier’s PrincipleLe Chatelier’s Principle
When a stress is applied to a system at equilibrium, the
reaction shifts to relieve the stress.
What stresses out a Reaction?What stresses out a Reaction?
• Heat (heat is measured in kJ or kcal)
• Pressure (related to # of moles on each side of the reaction)
• Concentration (how much of each component is added)
…When any of these increase or decrease from original conditions, that is stress.
The Equilibrium ConstantThe Equilibrium Constant
• The equilibrium constant is represented by K
xw
zy
BA
DCK
][][
][][
Where: wA + xB yC + zD
Try Me!Try Me!
Write the equilibrium expression for the following reaction:
4 NH3 + 7O2 4NO2 + 6H2O
Calculate the value of the equilibrium constant if the concentrations of the reactants and products are as follows:
[NH3] = 3.1 x 10-2 mol/L
[O2] = 5.4 x 10-2 mol/L
[NO2] =3.1 x 10-2 mol/L
[H2O] = 4.7 x 10-2 mol/L
Number of equilibrium positions available Number of values for K
Equilibrium PositionEquilibrium Position
• A set of concentrations that indicate whether products or reactants dominate while the rxn is at equilibrium.
Equilibrium and PressuresEquilibrium and Pressures
• Pressure and concentration are interchangeable
K Kpvs
In General:
Kp = K(RT)nFigure out the relationship between K
and Kp for the Haber Process.Do Now!Do Now!
nn = = productsproducts - -reactantsreactants
PV = nRTP = (n/V)RT
n/V = concentration
Heterogeneous EquilibriaHeterogeneous Equilibria
• Pure liquids and pure solids are not included in the equilibrium expression for a reaction.
• Concentrations of PURE liquids and solids cannot change.
Applications of the Equilibrium Applications of the Equilibrium ConstantConstant
• The Equilibrium constant gives information that will allow us to…
– Decide how likely it is that the reaction will occur.– Determine if a reaction is at equilibrium given a set of
concentrations.– Determine which direction the reaction must shift in order
to reach an equilibrium position.
If K is greater than 1The reaction is muchMore likely to occur
Spontaneously(the equilibrium lies farthest to the right)
If K is greater than 1The reaction is muchMore likely to occur
Spontaneously(the equilibrium lies farthest to the right)
If K is very smallThe reaction is not
likely to be spontaneous
(the equilibrium is near the reactants)
If K is very smallThe reaction is not
likely to be spontaneous
(the equilibrium is near the reactants)
Spontaneous DoesNot
MeanFast!
Spontaneous DoesNot
MeanFast!
The Reaction QuotientThe Reaction Quotient
• Obtained by replacing initial concentrations into the concentrations of the equilibrium expression.
• Three possible cases:
Q > KShift Left
Q = KNo Shift
Q < KShift Right
Try Me!!
• For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0 x 10-2. Predict the direction in which the system will shift to reach equilibrium:
[NH3]o = 1.0 x 10-3 M
[N2]o = 1.0 x 10-5 M
[H2]o = 2.0 x 10-3 M
Q = 1.3 x 107
Shift to the Left
Try Me Again and Again!
• Same problem, different conditions:[NH3]o = 2.0 x 10-4 M
[N2]o = 1.5 x 10-5 M
[H2]o = 3.54 x 10-1 M
[NH3]o = 1.0 x 10-4 M
[N2]o = 5.0 M
[H2]o = 1.0 x 10-2 M
Q = 2.0 x 10-3
Shift to the Right
Q = 6.01 x 10-2
No Shift
Solving for Concentrations and Solving for Concentrations and PressuresPressures
• Several Types of problems and Solving methods.
• Plug and Chug• ICE• ICE with Stoichiometry• ICE with the quadratic equation
Type 1: Plug and ChugType 1: Plug and Chug
Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2.
N2O4(g) 2NO2(g)
Type 2: Initial, Type 2: Initial, Change,EquilibriumChange,Equilibrium
• At a certain temperature a 1.00L flask initially contained 0.298 mol PCl3(g) and 8.70 x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00 x10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the reaction
PCl5(g) PCl3(g) + Cl2(g)
Calculate the concentrations of all species and the value of k.
Type 3: ICE with StoichiometryType 3: ICE with Stoichiometry
• Hydrofluoric acid vapor decomposes into its elemental components hydrogen and fluorine. At 700K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.00 mol of each component is mixed in a 1.00L flask.
Type 4: ICE with QuadraticType 4: ICE with Quadratic
• A 1.00L flask is filled with 1.0 mol H2 gas and 2.0 mol I2 gas at 448oC. The value of the equilibrium constant Kc for the reaction
H2 + I2 2HI
at 448oC is 50.5. What are the equilibrium concentrations of H2, I2 and HI in mol/L?