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Unit 4: Equilibrium, Acids & Bases Part 1: Equilibrium. Equilibrium Constant Expressions Calculating the Value of an Equilibrium Constant Applications of Equilibrium Constants LeChatelier’s Principle. Chemical Equilibrium. - PowerPoint PPT Presentation
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Unit 4: Equilibrium, Acids & Bases
Part 1: Equilibrium
Equilibrium Constant ExpressionsCalculating the Value of an
Equilibrium ConstantApplications of Equilibrium
ConstantsLeChatelier’s Principle
Chemical Equilibrium
One of the challenges that industrial chemists face is to maximize the yield of product obtained in a reaction.
Some reactions do not go to completion.The reaction stops short of the
theoretical yield.
Unreacted starting materials are still present, but no additional product is formed.
Chemical Equilibrium
Time
Mo
lar
Co
nc
en
tra
tio
n
Hydrogen
Nitrogen
Ammonia
Consider the reaction to produce ammonia:
N2 (g) + 3 H2 (g) 2 NH3 (g)
Chemical Equilibrium
After a period of time, the composition of the reaction mixture stays the same even though most of the reactants are still present.
Although it is not apparent, chemical reactions are still occurring within the reaction mixture.
N2 (g) + 3 H2 (g) 2 NH3 (g)
2 NH3 (g) N2 (g) + 3 H2 (g)
Chemical Equilibrium
The reaction has reached chemical equilibrium and is best represented by the equation:
N2 (g) + 3 H2 (g) 2 NH3 (g)
The double arrow indicates that the reaction is an equilibrium reaction.The reaction occurs in both
directions simultaneously.
Chemical Equilibrium
Chemical equilibrium:A state of dynamic balance in which
the opposing reactions are occurring at equal rates
Rate of forward reaction (R P) = rate of reverse reaction (P R)
Consider a simple system at equilibrium:Forward: A B Rate = kf[A]Reverse: B A Rate = kr[B]
At equilibrium, the rate for the forward reaction equals the rate of the reverse reaction.
kf[A] = kr[B]
Rearranging: [B] = kf = a constant[A] kr
Chemical Equilibrium
Chemical Equilibrium
At chemical equilibrium, the concentrations of the reactants and products do not change.ratio of products over reactants is
constant
Note: This does not mean that the concentrations of the reactants and products are identical to each other.
Chemical Equilibrium
For a general, balanced equilibrium reaction:
a A + b B d D + e E
the equilibrium condition is expressed by the equation:
Kc = [D]d [E]e
[A]a [B]b
where Kc = equilibrium constant obtainedwhen concentrations areexpressed in molarity
Equilibrium-constant expression
Chemical Equilibrium
The equilibrium constant, Kc, is the numerical value obtained when the actual equilibrium concentrations (in M) of reactants and products are substituted into the equilibrium constant expression.Kc is unitless.
The subscript c indicates that all concentrations used to calculate the value of Kc were expressed in M .
The equilibrium constant expression for the following reaction is:
Kc = [Ag(NH3)2+
]
[Ag+] [NH3]2
Ag+ (aq) + 2 NH3 (aq) Ag(NH3)2+
(aq)
Chemical Equilibrium
Chemical Equilibrium
Some equilibrium reactions involve reactants and products that are all in the same phase.Homogeneous equilibriumExample: N2 (g) + 3 H2 (g) 2 NH3
(g)
Some equilibrium reactions involve reactants and/or products that are in different phasesheterogeneous equilibriumExample: Ag+ (aq) + Cl- (aq) AgCl
(s)
Chemical Equilibrium
An example of a heterogeneous equilibrium:
CO2 (g) + H2 (g) CO (g) + H2O (l)
If a solid or liquid is involved in a heterogeneous equilibrium, its concentration is constant and is not included in the equilibrium constant expression.
For this example:
Kc = [CO]
[CO2][H2]
Example: Write the equilibrium constant expression, Kc, for the following reactions:
Cd2+ (aq) + 4 Br- (aq) CdBr42- (aq)
CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g)
Chemical Equilibrium
Example: Write the equilibrium constant expression, Kc, for the following reactions:
Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq)
Ti (s) + 2 Cl2 (g) TiCl4 (l)
Chemical Equilibrium
Chemical Equilibrium
When all reactants and products in a chemical equilibrium are gases, the equilibrium constant expression can also be written in terms of the partial pressures of the gases.
Kp = the equilibrium constant in terms of partial pressures of reactants and products
Partial pressure: the pressure exerted by a particular
gas in a mixture of gases
Chemical Equilibrium
For the general chemical equation:
a A (g) + b B (g) d D (g) + e E (g)
the equilibrium constant expression is:
Kp = (PD)d (PE)e
(PA)a (PB)b
where Kp = equilibrium constant in terms of
pressure
PD = partial pressure of D in atm.
Chemical Equilibrium
The numerical values of Kc and Kp are different for most reactions.
Kp = Kc (RT)n
where R = 0.0821 atm.Lmol.K
T = temperature in K n = change in # of moles = # mol products - # mol reactants
Chemical Equilibrium
Example: Write the equilibrium constant expression, Kp, for the following reaction:
CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g)
Chemical Equilibrium
Example: Write the equilibrium constant expression, Kp, for the following equilibrium:
CO2 (g) + H2 (g) CO (g) + H2O (l)
Chemical Equilibrium
Example: Write the equilibrium constant expression, Kp, for the following reaction:
Ti (s) + 2 Cl2 (g) TiCl4 (l)
Chemical Equilibrium
The solubility-product constant (Ksp) describes the equilibrium that is established during the dissolution of an ionic compound in water as a saturated solution is formed. In a saturated solution, the undissolved
solid and its hydrated ions are in equilibrium.
For the dissolution of CaF2:
CaF2 (s) Ca2+ (aq) + 2 F- (aq)
Ksp = [Ca2+] [F-]2
Chemical Equilibrium
Dissolution:The process of dissolving a
substance in a solvent
Notes:The expression for Ksp excludes
solids (just like other heterogeneous equilibria)
The value for Ksp is calculated using the concentrations (in M) of the ions.
Chemical Equilibrium
Example: Write the solubility product constant expression for the dissolution of silver chromate.
Magnitude of Equilibrium Constants The magnitude of Kc, Kp, and Ksp varies
widely depending on the reaction.
N2 (g) + O2 (g) 2 NO (g)
Kc = [NO]2 = 1 x 10-30
[N2] [O2]
CO (g) + Cl2 (g) COCl2 (g)
Kc = [COCl2] = 4.57 x 109
[CO] [Cl2]
Magnitude of Equilibrium Constants When Kc (or Kp or Ksp) < 1, more
reactants than products are present at equilibrium.
N2 (g) + O2 (g) 2 NO (g)
Kc = [NO]2 = 1 x 10-30
[N2] [O2]
Equilibrium lies to the left.Reactants are favored.
When Kc (or Kp or Ksp) is > 1, more products than reactants are present at equilibrium.
CO (g) + Cl2 (g) COCl2 (g)
Kc = [COCl2] = 4.57 x 109
[CO] [Cl2]
Equilibrium lies to the right.Products are favored.
Magnitude of Equilibrium Constants
Magnitude of Equilibrium ConstantsExample: Are reactants or products favored in the following reaction?
H2 (g) + I2 (g) 2 HI (g) Kc = 50.5
Magnitude of Equilibrium Constants Equilibrium can be approached from
either direction.N2O4 (g) 2 NO2 (g)
If N2O4 (g) is placed in a reactor at 100oC, N2O4 will decompose to form NO2 (g).
If NO2(g) is placed in a reactor at 100oC, NO2 will react to form N2O4.
Magnitude of Equilibrium Constants For an equilibrium reaction, the
direction that we write the chemical equation is arbitrary. Influences the way we write the
equilibrium constant expression and the value of the equilibrium constant.
Magnitude of Equilibrium Constants For the reaction,
N2O4 (g) 2 NO2 (g)
the equilibrium constant expression is:Kc = [NO2]2 = 0.212 at 100oC
[N2O4]
For the reaction,
2 NO2 (g) N2O4 (g)
the equilibrium constant expression is:Kc = [N2O4] = 4.72 at 100oC
[NO2]2
Magnitude of Equilibrium Constants The equilibrium constant expression
and the value of the equilibrium constant for a reaction written in one direction is the reciprocal of the one written in the opposite direction.
A B Kc = [B] [A]
B A Kc = [A][B]
Kc (forward) = 1
Kc (reverse)
Magnitude of Equilibrium ConstantsExample: Given that Kc for the formation of phosgene is 4.57 x 109, what is the value of Kc for the decomposition of phosgene?
COCl2 (g) CO (g) + Cl2 (g) Kc = ?
CO (g) + Cl2 (g) COCl2 (g) Kc = 4.57 x 109
Calculating Equilibrium Constants In order to calculate the value of an
equilibrium constant, you must know eitherconcentrations of reactants and
products at equilibrium (for Kc or Ksp)
partial pressures of reactants and products at equilibrium (for Kp)
Calculating Equilibrium Constants
On the exam, you must be able to calculate the value of equilibrium constant when given any of the following: The equilibrium concentrations or partial
pressures of reactants and products
The equilibrium # moles (or grams) of reactants and products and the volume of the container
The initial quantity of reactant(s) present and the quantity of one reactant (or product) at equilibrium.
Calculating Equilibrium ConstantsExample: PCl5 is prepared at 450 K according to the following reaction. What is the value of Kp if the partial pressure of the three gases at equilibrium are: PPCl3 = 0.124 atm, PCl2 = 0.157 atm, and PPCl5 = 1.30 atm?
PCl3 (g) + Cl2 (g) PCl5 (g)
Calculating Equilibrium Constants Write the expression for Kp
Substitute the pressure of each reactant or product:
Calculating Equilibrium ConstantsExample: At equilibrium, a reaction mixture contains 0.0360 mol H2, 0.0570 mol N2, 0.414 mol H2O, and 0.186 mol NO in a 3.00 liter reactor. Calculate the value of Kc for the following reaction.
2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)First, write the expression for Kc
Next, calculate all concentrations:[H2] =
[N2] =
[H2O] =
[NO] =Finally, plug concentrations into expression for Kc
Calculating Equilibrium Constants
Kc = 654
Calculating Equilibrium ConstantsExample: A mixture of 0.678 mol of H2 and 0.440 mole of Br2 is heated in a 2.00-L reactor at 700 K. At equilibrium, 0.283 mol of H2 are present in the reactor. What are the equilibrium concentrations of H2, Br2, and HBr? Calculate Kc for the reaction.
H2 (g) + Br2 (g) 2 HBr (g)Write the expression for Kc:
Calculating Equilibrium Constants Determine the initial concentrations of
the reactants and products as well as the equilibrium concentration of the reactant (H2) given in the problem.
[H2]initial =
[Br2]initial =
[HBr]initial =
[H2]equil =
Calculating Equilibrium Constants
Initial 0.339 M 0.220 M 0.000 M
Change
Equil. 0.1415 M
H2 (g) Br2 (g) 2 HBr
Set up an “ICE” table showing initial conc., change in concentration, equilibrium conc. of all reactants and products.
Calculating Equilibrium Constants
Calculating Equilibrium Constants Finally, use the equilibrium
concentrations of the reactants and products to determine the value of Kc.
Applications of Equilibrium Constants The magnitude of Kc, Kp, or Ksp indicates
the extent to which a reaction will proceed. Products favored (Kc >> 1)Reactants favored (Kc << 1)
Kc can also be used to predict the direction a reaction mixture must
go to reach equilibrium
equilibrium concentrations of reactants and products
Applications of Equilibrium Constants In order to use Kc to predict the
direction in which a reaction mixture must go in order to reach equilibrium, we must calculate the reaction quotient (Q).The value obtained when the
concentrations (or partial pressures) of reactants and products under conditions that are not necessarily equilibrium conditions are substituted into the equilibrium constant expression.
Applications of Equilibrium Constants The value of the reaction quotient can
be compared to the value of Kc or Kp in order to determine the direction the reaction must proceed to reach equilibrium. If Q = K
reaction mixture is at equilibrium If Q < K
reaction must proceed toward products (toward the right)
If Q > K reaction must proceed toward
reactants (toward the left)
Applications of Equilibrium ConstantsExample: At 1000 K, the value of Kc for the reaction 2 SO3 (g) 2 SO2 (g) + O2 (g) is 4.08 x 10-3. Calculate the reaction quotient and predict the direction in which the reaction will proceed to reach equilibrium if the initial concentrations of reactants and products are [SO3] = 2 x 10-3 M, [SO2] = 5 x 10-3 M, and [O2] = 3 x 10-2 M.
Applications of Equilibrium Constants Kc =
Q =