Date post: | 18-Nov-2014 |
Category: |
Documents |
Upload: | api-19824406 |
View: | 339 times |
Download: | 0 times |
Dr. Rabih O. Al-Kaysi
Ext: 47247Email: [email protected]
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