Unit 4: Equilibrium, Acids & Bases Part 1: Equilibrium

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.


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)


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)


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: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



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.


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


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)



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)


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)


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)



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


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.


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


Example: Write the equilibrium constant expression, Kp, for the following reaction:

CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g)


Example: Write the equilibrium constant expression, Kp, for the following equilibrium:

CO2 (g) + H2 (g) CO (g) + H2O (l)


Example: Write the equilibrium constant expression, Kp, for the following reaction:

Ti (s) + 2 Cl2 (g) TiCl4 (l)


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


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.


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


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 =


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 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 =

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Unit 4: Equilibrium, Acids & Bases Part 1: Equilibrium