Dr. Rabih O. Al-Kaysi

Ext: 47247Email: kaysir@ksau-hs.edu.sa

General Chemistry Course # 111, two creditsSecond Semester 2009

King Saud bin Abdulaziz University for Health Science

Textbook: Principles of Modern Chemistry by David W. Oxtoby, H. Pat Gillis, and Alan Campion (6 edition; 2007)

Lecture 15Chemical Equilibrium

• Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2:

N2O4(g) 2NO2(g).

• After some time, the color stops changing and we have a mixture of N2O4 and NO2.

• Chemical equilibrium:

1) is the point at which the concentrations of

reactants and products are constant 2) Chemical equilibrium occurs when the reaction

forward and reverse reaction have equal rates

k forward = k reverse

Equilibrium

• Using the collision model: N2O4(g) 2NO2(g).

• At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g) N2O4(g)) does not occur.• Only the production of NO2 will occur.

• As the amount of NO2 increases, there is a chance that two NO2 molecules will collide constructively to form N2O4.

• Thus, the forward chemical reaction has an opposing reaction that will increase with the increase in product formation.

Equilibrium: Collision Model

• The point at which the rate of decomposition: Rateforward

N2O4(g) 2NO2(g)

equals the rate of dimerization: Ratereverse

2NO2(g) N2O4(g).

is dynamic equilibrium. (Rate forward = Rate reverse )

• The equilibrium is dynamic because the reaction has not stopped: Only the opposing rates are equal.

• A healthy human body is in a state of dynamic equilibrium. You loose that equilibrium when you get sick, infected or dead!

Dynamic 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], which implies

• The mixture at equilibrium is called an equilibrium mixture.

Expressing Equilibrium Reactions

N2O4(g) 2NO2(g)

A B

• No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.

N2 + 3H2 2NH3

The Equilibrium Constant: Keq

The Equilibrium Constant: Keq

• For a general reaction in solution where all the species are in solution

the equilibrium constant expression for everything in solution is

where Keq is the equilibrium constant, A, B, C, and D are the reactants and products respectively, and a, b, c, and d are the stoichiometric coefficients.

aA + bB cC + dD

ba

dc

eqKBA

DC

Eg: Decomposition of Ozone O3 to oxygen O2 is an equilibrium process

2O3 3O2 Keq = [O2]3 / [O3]2

Properties of Keq

• In solution Keq is based on the Molarities of reactants and products at equilibrium.

• We generally omit the units of the equilibrium constant.

• Note that the equilibrium constant expression has products over reactants.• K>>1 implies products are favored, and equilibrium lies

to the right.

• K<<1 implies reactants are favored, and equilibrium lies to the left.

• The same equilibrium is established not matter how the reaction is begun.

Forward Equilibrium Direction• An equilibrium can be approached from any

direction.• Example:

• Has

• K >1 implies an equilibrium favoring product or NO2 formation.

N2O4(g) 2NO2(g)

46.6

42

2

ON

2NO P

PKeq Keq = [NO2]2 / [N2O4] = 6.46

• In the reverse direction:

• Thus,

Reverse Equilibrium Direction

46.61

155.02NO

ON

2

42 P

PKeq

2NO2(g) N2O4(g)

K >1 implies an equilibrium Not favoring product or N2O4 formation.

• Homogeneous equilibrium: When all reactants and products are in one phase. Eg: solution, or gas phase all in gas phase

• Heterogeneous equilibrium: When one or more reactants or products are in a different phase, the is.

• Consider:

– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

Heterogeneous and Homogeneous Equilibria

CaCO3(s) CaO(s) + CO2(g)

N2 + 3H2 2NH3

• The concentration (Molarity) of a solid or pure liquid is its (density / molar mass) X 1000. Eg: The concentration of pure water [H2O] = (1/18) X 1000 = 55.6 M

The Molarity of Hg liquid = 67.8 M. • Because neither density nor molar mass is a variable, the concentrations

of solids and pure liquids are constant no matter how much is present

• For the decomposition of CaCO3:

• Thus, if a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium expression for the reaction.

Heterogeneous Equilibria

]CO[constant]CO[]CaCO[

CaO][22

3eqK

CaCO3(s) CaO(s) + CO2(g)

Assumptions• 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.

• This applies to solids or pure liquids with constant concentration

Heterogeneous Equilibria

• Eg:• In one of their experiments, Haber and co-workers introduced a mixture of hydrogen and

nitrogen into a reaction vessel and allowed the system to attain chemical equilibrium at 472 oC. The equilibrium mixture of gases was analyzed and found to contain 0.1207 M H2, 0.0402 M N2, and 0.00272 M NH3. From these data, calculate the equilibrium constant, Keq, for

• N2(g) + 3H2(g) 2NH3(g)

• Eg:• Gaseous Hydrogen iodide is placed in a closed container at 425 oC, where it partially

decomposes to hydrogen and iodine: 2HI (g) H2(g) + I2(g). At equilibrium, it is found that [HI] = 3.35*10-3M; [H2] = 4.79*10-4M; [I2] = 4.79*10-4M. What is the value of Keq at this temperature.

• A mixture of 0.100 mole of NO, 0.050 mole of H2, and 0.050 mole of H2O is placed in a 1.00-L vessel. The following equilibrium is established:

• 2NO(g) + 2H2(g) N2(g) + 2H2O(g)

• (a) Calculate the Keq for the reaction.

• We define Q, the reaction quotient, for a general reaction

as

• Q = K only at equilibrium.

Predicting Direction of Reaction

aA + bB cC + dD

ba

dc

PP

PPQ

BA

DC

Eg:

• At 448 oC the equilibrium constant, Keq, for the reaction: H2(g) + I2(g) 2HI(g) is 50.5. Predict how the reaction will proceed to reach equilibrium at 448 oC if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2, and 3.0 x 10-2 mol of I2 in a 2.0-L container.

• 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.

• If Q = K then the reaction is at equilibrium

Reaction Quotient

• 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.

Calculating Equilibrium Concentrations

Eg:

• A 1.00-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 oC. The value of the equilibrium constant, Keq, for the reaction:

H2(g) + I2(g) 2HI(g) at 448 oC is 50.5. What are the concentration of HI, H2, and I2 in the flask at equilibrium.

• 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.

Introducing Le Châtelier’s Principle

N2(g) + 3H2(g) 2NH3(g)

• 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.

Change in Reactant or Product Concentrations

N2(g) + 3H2(g) 2NH3(g)

• 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).

Application of Le Châtelier’s Principle

• 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.

Effect of Temperature

• 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 effect the composition of the equilibrium mixture.

The Effect of Catalysis