Chemical Kinetics & Equilibrium Chapter 16

Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why? For a reaction to occur, molecules must have: 1. sufficient energy to break old bonds. 2. proper orientation for collision to be effective.

Three possible collision orientations-- a) & b) produce reactions, while c) does not.

Factors Affecting Rate of Reaction Concentration: The higher the concentration of the reactants, the more likely an effective collision will occur. Temperature: An increase in temperature increases: 1. the energy of a collision. 2. the number of collisions.

Plot showing the number of collisions with a particular energy at T1& T2, where T2 > T1 -- Boltzman Distribution.

Activation Energy, Ea Activation energy for a given reaction is a constant and not temperature dependent. Activation energy represents the minimum energy for a reaction to occur.

a) The change in potential energy as a function of reaction progress. Ea is the activation energy and E is the net energy change -- exothermic. b) Molecular representation of the reaction.

Enthalpy -- H Enthalpy -- at constant pressure, the change in enthalpy equals the energy flow as heat. Exothermic -- H is negative (-). Endothermic -- H is positive (+).

Catalysis Catalyst: A substance that speeds up a reaction without being consumed Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

Energy plots for a catalyzed and an uncatalyzed pathway for an endothermic reaction.

Effect of a catalyst on the number of reaction-producing collisions. A greater fraction of collisions are effective for the catalyzed reaction.

Catalysis The breakdown of the ozone layer is illustrated by the following equation: Cl + O3 ---> ClO + O2 O + ClO ---> Cl + O2 Cl + O3 + O + ClO ---> ClO + O2 + Cl + O2 Cl Net Reaction: O + O3 ----> 2O2 What is the catalyst? The intermediate?

Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

Reactions That Appear to Run to Completion 1. Formation of a precipitate. 2. Formation of a gas. 3. Formation of a molecular substance such as water. These reactions appear to run to completion, but actually the equilibrium lies very far to the right. All reactions in closed vessels reach equilibrium.

Molecular representation of the reaction 2NO2(g) ----> N2O4(g). c) & d) represent equilibrium.

Figure 16. 9: (a) The initial equilibrium mixture of N2, H2 and NH3 Figure 16.9: (a) The initial equilibrium mixture of N2, H2 and NH3. (b) Addition of N2. (c) The new equilibrium.

2NO2(g) <----> N2O4(g) Chemical Equilibrium 2NO2(g) <----> N2O4(g) The forward reaction goes to the right. The reverse reaction goes to the left. At equilibrium the rate of the reverse reaction equals the rate of the forward reaction.

Concentration profile for the Haber Process which begins with only H2(g) & N2(g).

The Law of Mass Action For jA + kB  lC + mD The law of mass action (Cato Guldberg & Peter Waage) is represented by the equilibrium expression:

Equilibrium Expression K is the equilibrium constant. [C] is the concentration expressed in mol/L. K is temperature dependent.

Equilibrium Constant, K For an exothermic reaction, if the temperature increases, K decreases. For an endothermic reaction, if the temperature increases, K increases.

Writing Equilibrium Expressions 2O3(g) <---> 3O2(g)

Writing Equilibrium Expressions Write the equilibrium expression for the following: H2(g) + F2(g) <---> 2HF(g) N2(g) + 3H2(g) <---> 2NH3(g)

Equilibrium Expression Write the equilibrium expression for the following reaction: 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)

Table 16.1: Results of Three Experiments for the Reaction N2(g) 1 3H2(g) 2NH3(g) at 500 ºC

Equilibrium Position For a given reaction at a given temperature, there is only one equilibrium constant (K), but there are an infinite number of equilibrium positions. Where the equilibrium position lies is determined by the initial concentrations of the reactants and products. The initial concentrations do not affect the equilibrium constant.

Homogeneous Equilibria Homogeneous equilibria are equilibria in which all substances are in the same state. N2(g) + 3H2(g) <---> 2NH3(g) H2(g) + F2(g) <---> 2HF(g)

Heterogeneous Equilibria . . . are equilibria that involve more than one phase. CaCO3(s)  CaO(s) + CO2(g) K = [CO2] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. This does not apply to gases or solutions.

The position of the equilibrium CaCO3(s) ---> CaO(s) + CO2(g) does not depend upon the amounts of solid CaCO3 or CaO.

Figure 16.10: The reaction system CaCO3(s) --> CaO(s) + CO2(g)

Heterogeneous Equilibria Write equilibrium expressions for the following: 2HOH(l) <---> 2H2(g) + O2(g) 2HOH(g) <---> 2H2(g) + O2(g) K = [H2]2[O2]

Heterogeneous Equilibria Write equilibrium expressions for the following: PCl5(s) <---> PCl3(l) + Cl2(g) CuSO4. 5 HOH(s) <---> CuSO4(s) + 5HOH(g) K = [Cl2] K = [HOH]5

Le Châtelier’s Principle . . . If a system at equilibrium is subjected to a stress, the equilibrium will be displaced in such direction as to relieve the stress.

Le Chatelier’s Principle If a reactant or product is added to a system at equilibrium, the system will shift away from the added component. If a reactant or product is removed, the system will shift toward the removed component.

Effect of Changes in Concentration on Equilibrium N2(g) + 3H2(g) <---> 2NH3(g) The addition of 1.000 M N2 has the following effect: Equilibrium Position I Equilibrium Position II [N2] = 0.399M [N2] = 1.348 M [H2] = 1.197 M [H2] = 1.044 M [NH3] = 0.203 M [NH3] = 0.304 M

Effect of Change in Concentration on Equilibrium Position I Position II K = 0.0602 K = 0.0602

Changes in Concentration Predict the effect of the changes listed to this equilbrium: As4O6(s) + 6C(s) <---> As4(g) + 6CO(g) a) addition of carbon monoxide b) addition or removal of C(s) or As4O6(s) c) removal of As4(g) a) left b) none c) right

The Effect of Container Volume on Equilibrium If the size of a container is changed, the concentration of the gases change. A smaller container shifts the equilibrium to the right -- N2(g) + 3H2(g) ---> 2NH3(g). Four gaseous molecules produce two gaseous molecules. A larger container shifts to the left -- two gaseous molecules produce four gaseous molecules.

The system of N2, H2, and NH3 are initially at equilibrium. When the volume is decreased, the system shifts to the right -- toward fewer molecules.

The Effect of Container Volume on Equilibrium Predict the direction of shift for the following equilibrium systems when the volume is reduced: a) P4(s) + 6Cl2(g) <---> 4PCl3(l) b) PCl3(g) + Cl2(g) <---> PCl5(g) c) PCl3(g) + 3NH3(g) <---> P(NH2)3(g) + 3HCl(g) a) right b) right c) none

Equilibrium Constant, K For an exothermic reaction, if the temperature increases, K decreases. N2(g) + 3H2(g) <---> 2NH3(g) + 92 kJ For an endothermic reaction, if the temperature increases, K increases. CaCO3(s) + 556 kJ <---> CaO(s) + CO2(g)

Effect of Temperature on Equilbirum Predict how the equilibrium will shift as the temperature is increased. N2(g) + O2(g) <---> 2NO(g) (endothermic) 2SO2(g) + O2(g) <---> 2SO3(g) (exothermic) shift to the right shift to the left

Effects of Changes on the System 1. Concentration: The system will shift away from the added component. 2. Temperature: K will change depending upon the temperature [treat heat as a reactant (endothermic) and as a product (exothermic)].

Effects of Changes on the System (continued) 3. Pressure: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer gaseous molecules.

Figure 16.12: Shifting equilibrium by changing the temperature

Magnitude of K A K value much larger than 1 means that the equilibrium system contains mostly products -- equilibrium lies far to the right. A very small K value means the system contains mostly reactants -- equilibrium lies far to the left. The size of K and the time required to reach equilibrium are not directly related!

Calculating Equilibrium Concentrations Gaseous phosphorus pentachloride decomposes to chlorine gas and gaseous phosphorus trichloride. If K = 8.96 x 10-2 , and the equilbrium concentration of PCl5 is 6.70 x 10-3 M and that of PCl3 is 0.300 M, calculate the concentration of Cl2 at equilibrium. PCl5(g) <---> PCl3(g) + Cl2(g)

Calculating Equilibrium Concentrations PCl5(g) <---> PCl3(g) + Cl2(g) K = 8.96 x 10-2 [PCl5] = 6.70 x 10-3 M [PCl3] = 0.300 M [Cl2] = ? [Cl2] = 2.00 x 10-3 M

Solubility Allows us to flavor foods -- salt & sugar. Solubility of tooth enamel in acids. Allows use of toxic barium sulfate for intestinal x-rays.

Solubility Product For solids dissolving to form aqueous solutions. Bi2S3(s)  2Bi3+(aq) + 3S2(aq) Ksp = solubility product constant and Ksp = [Bi3+]2[S2]3 Why is Bi2S3(s) not included in the solubilty product expression?

Writing Solubility Product Expressions Write the balanced equation for dissolving each of the following solids in water. Also write the Ksp expression for each solid. a) PbCl2(s) b) Bi2S3(s) c) Ag2CrO4(s) PbCl2(s) <---> Pb2+(aq) + 2Cl-(aq) Ksp = [Pb2+][Cl-]2 Bi2S3(s) <---> 2Bi3+(aq) + 3S2-(aq) Ksp = [Bi3+]2[S2-]3 Ag2CrO4(s) <---> 2Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2[CrO42-]

Solubility Product Calculations Cupric iodate has a measured solubility of 3.3 x 10-3 mol/L. What is its solubility product? Cu(IO3)2(s) <---> Cu2+(aq) + 2 IO3-(aq) 3.3 x 10-3 M ---> 3.3 x 10-3 M + 6.6 x 10-3 M Ksp = [Cu2+][IO3-]2 Ksp = [3.3 x 10-3][6.6 x 10-3]2 Ksp = 1.4 x 10-7

Calculating Solubility from Ksp Values The Ksp value for solid AgI(s) is 1.5 x 10-16 at 25 oC. Calculate the solubility of AgI(s) in water at 25 oC. AgI(s) <---> Ag+(aq) + I-(aq) [Ag+] = x Ksp = 1.5 x 10-16 = [Ag+][I-] [I-] = x Ksp = 1.5 x 10-16 = x2 x = 1.5 x 10-16 x = 1.2 x 10-8 M

Calculating Solubility from Ksp Values The Ksp value for solid lead chromate is 2.0 x 10-16 at 25 oC. Calculate its solubility in water at 25 oC. PbCrO4(s) <---> Pb2+(aq) + CrO42-(aq) [Pb2+] = x Ksp = 2.0 x 10-16 = [Pb2+][CrO42- ] [CrO42-] = x Ksp = 2.0 x 10-16 = x2 x = 2.0 x 10-16 x = 1.4 x 10-8 M

Solubility Product Calculations Cu(IO3)2(s) <---> Cu2+(aq) + 2 IO3-(aq) Ksp = [Cu2+][IO3-]2 If solid cupric iodate is dissolved in HOH; double & square the iodate concentration. If mixing two solutions, one containingCu2+ and and the other IO3-, then use the concentration of iodate and only square it.