Chemical Kinetics & Equilibrium Chang, Chapters 13 & 14 Bylikin Chapters 6 and 7

Chapter 13/14 Outline Chapter 13 –Kinetics –Collision Theory –Reaction Rate –Catalysts –Macroscopic and Microscopic factors affecting rate –Interpreting Data Chapter 14 Concept of Equilibrium –Equilibrium Constant –Equilibrium Expression Relationship between Reaction Rate and Equilibrium Constant Le Châtelier’s Principle (factors affecting equilibrium)

Chemical Kinetics The study of the rates of chemical reactions Kinetics tells us the rates of reactions under specific conditions Kinetics studies the macroscopic constraints to understand changes on the molecular level

Collision Theory There are 3 Requirements that must be met in order for a reaction to occur: –Collision –Orientation –Energy

Reaction Rate Reaction Rate=The change in the amount of reactants with respect to time. –Consider A  Q –Evaluation of the rate at a point yields the instaneous rate of change (d y /d x ) –Rate = -  [A] =  Q  T  T

Stoichiometry and Rates Consider the following reaction: 2A  Q Two moles of A will disappear for each mole of B that forms Rate = -1  [A] =  [Q] 2  t  t

Observations of Rate Reaction rate can be determined by monitoring any of the physical properties that can be related to a changing molecular species –Example: Oxidation of Crystal Violet We can measure the change in Abs of light at 590 nm –Example: Decomposition of H 2 O 2 We can measure the formation of O 2(g) –Example: Hydrolysis of Ether We can measure ion conductivity

Catalysts A catalysts = a substance that increases the rate of a chemical reaction without being consumed E a with a catalyst is less than E a without There are heterogeneous catalysts, homogeneous catalysts, and biological catalysts (enzymes)

Macroscopic Observations and Microscopic Effects Macroscopic ConstraintMicroscopic Effect Concentration of ReagentsNumber of Collisions/Energy of Collisions TemperatureNumber of Collisions/Energy of Collisions Catalyst/InhibitorsEnergy of Collisions/Orientation Physical StateNumber of Collisions MediumEnergy of Collisions/Orientation

Interpreting Data Calculate the instantaneous rate of disappearance of C 4 H 9 Cl at t= 0 s (the initial rate) and again at t=600 s.

Interpreting Data Loss of Mass

Equilibrium Consider a Can of Soda H 2 CO 3(aq)   CO 2(aq) + H 2 O (l) CE CO 2(aq)   CO 2(g) PE –A Dynamic Equilibrium exists: Reactants and Products are continually exchanging, but there is no net change in the overall composition. –Chemical systems go to equilibrium because they are driven by thermodynamics (they go to the point of balance) –There is no indication of the TIMESCALE required to reach equilibrium

Equilibrium Constant Expressions Consider the following equilibrium 2NO 2(g)  N 2 O 4(g) ambercolorless –K eq is the equilibrium constant (unitless). –K eq = [N 2 O 4 ] = [Products] stoichiometric coefficient [NO 2 ] 2 [Reactants] stoichiometric coefficient –Pure solids and liquids are not included in K eq expressions because their concentrations are intensive properties and do not depend on amount –The K eq value gives insight on the extent of the reaction If the K eq is small, the reactants are favored If the K eq is large, the products are favored

Equilibrium Constant Expressions (cont) There are two different forms of K eq –K c is used when the equilibrium expression is written using concentrations of reactants and products in molarity –K P is used when the equilibrium expression is writing using partial pressures of reactants and products in atm 2NO 2(g)  N 2 O 4(g) –K c = [N 2 O 4 ] while K P = P N2O4 [NO 2 ] 2 P 2 NO2 –K P = K c RT  n

Equilibrium Expression and Equilibrium Constant Practice Write the equilibrium expression, K c for each of the following reactions: –Cu 2+ (aq) + 4NH 3(aq)   Cu(NH 3 ) 4 2+ (aq) –Ag + (aq) + Cl - (aq)   AgCl (s) –H 2 O 2(g)   H 2(g) + O 2(g) –HOCl (aq) + H 2 O (l)   H 3 O + (aq) + OCl - (aq) Which of the silver halides is least soluble? –AgCl, K eq = 1.6x –AgBr, K eq = 3.3x –AgI, K eq = 1.5x10 -16

Relationship between Reaction Rate and Equilibrium Constant A  k r k f  B Forward Rate = k f [A]Reverse Rate = k r [B] In an equilibrium, the forward reaction rate = reverse reaction rate, k f [A] = k r [B] [B] = k r = K eq [A] k f K eq = the ratio of the rates and is constant at a given temperature Regardless of the reaction mechanism, K eq can ALWAYS be written as

Reaction Quotient (Q) Reaction Quotients are calculated by substituting initial concentrations into the equilibrium constant (instead of equilibrium concentrations) By comparing the reaction quotient (Q) to the equilibrium constant (K eq ), you can determine which direction the reaction will proceed

Reaction Quotient (cont) PCl 5(g)   PCl 3(g) + Cl 2(g) K P 250 o C –KP = P PCl3 ∙P Cl2 = 1.05 P CL5 –Initial Conditions: PCl 5 = atm PCl 3 = atm PCl 2 = atm –Q = (0.223)(0.111) = (0.177) –Q<K, predict that PCl 5 will decrease and PCl 3 and Cl 2 will increase

Le Châtelier’s Principle A change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or counteract the effect of the change. Conditions that cause change: –Pressure (volume – in gasses) –Temperature (change magnitude of k) –Concentration of Reagents

Le Châtelier’s Principle - Concentration of Reagent 2CrO H +   H 2 O + Cr 2 O 7 2- K c = 1x10 14 (a) Add HCl then the equilibrium with shift __________. What happens to K c ? (b)Add NaOH the equilibrium will shift __________. What happens to K c ?

Le Châtelier's Principle - Temperature 2CrO H +   H 2 O + Cr 2 O 7 2-  H o rxn = kJmol -1 ( c) As the temperature increases the equilibrium will shift __________. What happens to K c ?

Le Châtelier's Principle – Pressure Change N 2(g) + 3H 2(g)  2NH 3(g)  H Ө = kJ (d) If the volume of the container is reduced by ½ the equilibrium will shift __________. What happens to K c ? (e) As the temperature decreases the equilibrium will shift __________. What happens to K c ?

Changes that Have NO Effect Increasing the pressure of a system with equal moles of gas on either side of the equilibrium Addition of an inert gas at a constant volume Addition of a catalyst (thereby lowering E a )