Kinetics applies to the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs. At equilibrium: rate forward = rate reverse A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.

CHEMICAL EQUILIBRIUM - RATES OF REACTION k F Reactants  products k B Chemical reactions are a dynamic process, that is, reactions involve both forward and reverse processes. Chemical Equilibrium is reached by a reaction mixture when the rates of forward and reverse reactions becomes equal. k F = k B NO net change appears obvious although the system is still in constant motion.

Reaction Dynamics Time Rate Rate Forward Rate Reverse Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant.

Reaching equilibrium on the macroscopic and molecular levels. N 2 O 4 ( g ) 2NO 2 ( g )

5 H 2 (g) + I 2 (g)  2 HI(g) at time 0, there are only reactants in the mixture, so only the forward reaction can take place [H 2 ] = 8, [I 2 ] = 8, [HI] = 0 at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place [H 2 ] = 6, [I 2 ] = 6, [HI] = 4

H 2 (g) + I 2 (g)  2 HI(g) at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated [H 2 ] = 4, [I 2 ] = 4, [HI] = 8 at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium

LAW OF MASS ACTION K  equilibrium constant aA + bB  cC + dD K = Products Reactants K = [C] c [D] d = Q reaction quotient [B] b [A] a Write the equilibrium equation for: a. H 2 O + H 2 O  H 3 O+ + OH - b. 4NH 3(g) (g)  2N 2(g) + 6H 2 O (g) c. N 2(g) + 3H 2(g)  2NH 3(g)

HOMOGENEOUS EQUILIBRIA H 2(g) + I 2(g)  2HI (g) Kp = Kc = ? 2O 3(g)  3O 2(g) Kp = Kc = ? HETEROGENEOUS EQUILIBRIA 2H 2 O (l)  H 3 O + (aq)_ + OH -(aq) C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (g) Kc = ? ?

HETEROGENEOUS EQUILIBRIA Substance in more than one phase 1.CaCO 3(s)  CaO (s) + CO 2(g) Kc = [CaO][CO 2 ] [CaCO 3 ] How do the [ ] of solid express? Pure solids & liquid have constant [conc. ] ?

The reaction quotient for a heterogeneous system. solids do not change their concentrations

Q. Each of the mixtures listed below was placed in a closed container and allowed to stand. Which of these mixtures is capable of attaining the equilibrium, expressed by 1 a)pure CaCO 3 b)CaO & P CO2 > Kp c)solid CaCO 3 & P CO2 > Kp d)CaCO 3 & CaO

LE CHATELIER’S PRINICIPLE “If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration of one of its components, that system will shift it’s equilibrium position as to ‘counteract’ the effect of the disturbance.” Equilibrium can be disturbed by: - adding or removing components - a change in pressure - a change in volume - a change in temperature

PREDICTING THE DIRECTION OF THE SHIFT I. CATALYST A catalyst increases the rate at which equilibrium is achieved but not the composition of the equilibrium mixture. II.THE REACTION QUOTIENT Q < K: the reaction shifts to the products Q > K: the reaction shifts to the reactants III.CHANGES IN VOLUME Reducing the volume of a gas at equilibrium causes the system to shift in the direction that reduces the number of moles of gas.

The effect of pressure (volume) on an equilibrium system. + lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas

15 Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. The Effect of Volume Changes on Equilibrium

Sample Problem SOLUTION: Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position PROBLEM:How would you change the volume of each of the following reactions to increase the yield of the products. (a) CaCO 3 ( s ) CaO( s ) + CO 2 ( g ) (b) S( s ) + 3F 2 ( g ) SF 6 ( g ) (c) Cl 2 ( g ) + I 2 ( g ) 2 I Cl( g ) PLAN:When gases are present a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa.

IV.CHANGES IN TEMPERATURE When heat is added to a system, the equilibrium shifts in the direction that absorbs heat. Cooling has the opposite effect and shifts the equilibrium towards the side which produces heat. Endothermic Reactions: heat + Reactants  Products An increase in temperature leads to a shift towards the products, a decrease leads to a shift towards the reactants. (K eq increases) Exothermic Reactions: Reactants  Products + heat An increase in temperature leads to a shift towards reactants. (K eq decreases)

The Effect of Temperature Changes on Equilibrium

Sample Problem SOLUTION: Predicting the Effect of a Change in Temperature on the Equilibrium Position PROBLEM:How does an increase in temperature affect the concentration of the underlined substance and K c for the following reactions? (a) CaO( s ) + H 2 O( l ) Ca(OH) 2 ( aq )  H 0 = -82kJ (b) CaCO 3 ( s ) CaO( s ) + CO 2 ( g )  H 0 = 178kJ (c) SO 2 ( g ) S( s ) + O 2 ( g )  H 0 = 297kJ PLAN:Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on K c.

Practice - Le Ch âtelier’s Principle The reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? –adding more O 2 to the container –condensing and removing SO 3 –compressing the gases –cooling the container –doubling the volume of the container –warming the mixture –adding the inert gas helium to the container –adding a catalyst to the mixture

Example 1: N 2 O 4(g)  2 NO 2(g)  H = 58 kJ In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium? A) add N 2 O 4 B) remove NO 2 C) increase the total pressure by adding N 2 D) increase the volume of the container E) decrease the temperature

Example 2: PCl 5(g)  PCl 3(g) + Cl 2(g)  H = 88 kJ In which direction will the equilibrium shift when each of the following changes are made to a system at equilibrium? A) add Cl 2 B) temperature is increased C) the volume of the reaction system is decreased D) PCl 5 is added E) a catalyst is added

Q - The Reaction Quotient At any time, t, the system can be sampled to determine the amounts of reactants and products present. A ratio of products to reactants, calculated in the same manner as K tells us whether the system has come to equilibrium (Q = K) or whether the reaction has to proceed further from reactants to products (Q K). We use the molar concentrations of the substances in the reaction. This is symbolized by using square brackets - [ ]. For a general reaction aA + bB cC + dD where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as Q = [C] c [D] d [A] a [B] b

Reaction direction and the relative sizes of Q and K. Equilibrium: no net change reactants products Reaction Progress reactants products Reaction Progress

The change in Q during the N 2 O 4 -NO 2 reaction.

26 Q, K, and the Direction of Reaction

Predicting the direction of reaction: Q > K forms more reactants  Q = K equilibrium Q < Kforms more products  Note:1. K f = 1 K r 2. K = K n If the balanced equation is multiplied by a factor then the K (& Q) is multiplied by the exponent.

The range of equilibrium constants small K large K intermediate K

DIRECTION OF REACTIONS AND K eg 1. The following reaction is a means of “fixing” nitrogen: N 2(g) + O 2(g)  2 NO (g) A. If the value for Q at 25°C is 1 x , describe the feasibility of this reaction for Nitrogen fixation. B. Write the equilibrium expression, K c C. Write the equilibrium expression for 2NO (g)  N 2(g) + O 2(g) D. Determine the K c for “C”

K as either K c or K p K c = the equilibrium constant using concentrations. K p = the equilibrium constant using pressure K p = K c (RT)  n gas  n is the difference between the number of moles of products and moles of reactants R = L atm / mol K

Deriving the Relationship between K p and K c

32 Deriving the Relationship Between K p and K c for aA(g) + bB(g)  cC(g) + dD(g) substituting

workshop on K c vs. K p K p = K c (RT)  n gas R = L atm/ mol K Write K p then K c for: 1. N 2(g) + 3H 2(g)  2NH 3(g) 2. N 2 O 4(g)  2NO 2(g) 3. Calculate K p if K c = for #1 4. 2SO 3(g)  2SO 2(g) + O 2(g) if K c = 4.07 x 10 -3, what is K p ?

Calculating Variations on Q and K aA + bB cC + dD Q c = [C] c [D] d [A] a [B] b cC + dD aA + bB Q ’ = 1/Q c aA + bB cC + dD n Q c ’ = (Q c ) n For a sequence of equilibria, K overall = K 1 x K 2 x K 3 x …

Key stages in the Haber process for synthesizing ammonia.

CALCULATING THE K eq 1. In one experiment, Haber introduced a mixture of H 2Haber & N 2 into a reaction vessel and allowed the system to attain chemical equilibrium at 472°C. The equilibrium mixture of gases were analyzed and found to contain M H 2, M N 2, and M NH 3. Calculate K eq. 2. Nitryl Chloride, NO 2 Cl, is in equilibrium in a closed container with NO 2 and Cl 2. 2 NO 2 Cl(g)  2 NO 2 (g) + Cl 2 (g) Calculate K eq if [NO 2 Cl] = M [NO 2 ] = M & [Cl 2 ] = M

3. For the Haber process N 2(g) + 3H 2(g)  2NH 3(g) Kp = 1.45 x at 500°C If an equilibrium mixture of the three gas with partial pressures of atm for H 2 and atm for N 2, what is the partial pressure of NH 3 ?

Predicting Direction 1. Predicting the direction of reaction Q = reaction quotient at equil Q = KQ > K  species on Rt (prod) (no net Rx) react to form left K = [Equil] Q = [Non Equil] Q < K  forms more products Goal: Calculate Q to determine state of Rx, equil, more product or more reaction

I.C.E. & Predicting Direction 1. Sulfur Trioxide decomposes at High temperature in a sealed container. 2 SO 3(g)  2 SO 2(g) + O 2 (g) Initially the vessel is filled at 1000K with SO 3(g) at a concentration of 6.09 x M. At equilibrium, the [SO 3 ] is 2.44 x M. Calculate K c. 2. Calculate the value for Q and predict the direction in which the reaction will proceed towards equilibrium if the initial concentrations are: [SO 3 ] = 2.0 x M [SO 2 ] = 5.0 x M [O 2 ] = 3.0 x M

Workshop ICE & Predicting Directions 1. A mixture of 5.0 x mol of H 2 and 1.0 x mol of I 2 is placed in a 5.0L container at 448°C and allowed to come to equilibrium. Analysis of this equilibrium mixture shows that the [HI] is 1.87 x M. Calculate Kc:H 2(g) + I 2(g)  2HI (g) 2. At 448°C the equilibrium constant Kc for the reaction below is 50.5 H 2(g) + I 2(g)  2HI (g) Predict how the reaction will proceed to reach equilibrium if the initial amount of HI is 2.0 x mol, H 2 is 1.0 x mol, and I 2 is 3.0 x mol in a 2.00 L container.

Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures first decide which direction the reaction will proceed –compare Q to K define the changes of all materials in terms of x –use the coefficient from the chemical equation for the coefficient of x –the x change is + for materials on the side the reaction is proceeding toward –the x change is  for materials on the side the reaction is proceeding away from solve for x –for 2 nd order equations, take square roots of both sides or use the quadratic formula –may be able to simplify and approximate answer for very large or small equilibrium constants

1. For the Haber process N 2(g) + 3H 2(g)  2NH 3(g) Kp = 1.45 x at 500°C If an equilibrium mixture of the three gas started with partial pressures of atm for H 2 and atm for N 2, what is the partial pressure of NH 3 ? 2. A 1.00 L flask is filled with 1.00 mol of H 2 and 2.00 mol I 2 at 448°C K c is What are the equilibrium concentrations of H 2, I 2 & HI?

3.The equilibrium constant for the Haber process at 472°C is Kc = A 2.00 L flask is filled with mol of ammonia and is then allowed to reach equilibrium at 472°C. What are the equilibrium concentrations? N 2(g) + 3 H 2(g)  2 NH 3(g) 4. For the reaction PCl 5(g)  PCl 3(g) + Cl 2(g) at a certain temperature Kc equals 450. What will happen when 0.10 mol of PCl 5, 1.0 mol of PCl 3, and l.5 mol of Cl 2 are added to a 2.0-L container and the system is brought to the temperature at which Kc=450. What are the equilibrium concentrations?

EFFECT OF VARIOUS DISTURBANCES ON AN EQUILIBRIUM SYSTEM DISTURBANCE NET DIRECTION OF REACTION EFFECT ON VALUE OF K Concentration Increase (reactant)Toward formation of productNone Decrease (reactant)Toward formation of reactantNone Pressure (volume) Increase PToward formation of lower amount (mol) of gas1None Decrease PToward formation of higher amount (mol) of gasNone Temperature Increase TToward absorption of heat Increases  H°rxn>0 Decreases if  H°rxn<0 Decrease TToward release of heatIncreases  H°rxn<0 Decreases  H°rxn>0 Catalyst addedNone; rates of forward and reverse reactions increase equallyNone

VAN’T HOFF EQUATION Changes in K due to T In K 2 =  H° rxn ( 1 - 1) K 1 R T 1 T 2 R = J/mol K T = Kelvin The formation of methanol is an important industrial reaction in the processing of new fuels. At 298K, K p = 2.25 x 10 4 for the reaction CO (g) + 2 H 2(g)  CH 3 OH (l) If  H° rxn = -128 kJ/mol CH 3 OH, calculate K p at 0°C.

The van’t Hoff Equation The Effect of T on K ln K2K2 K1K1 = -  H 0 rxn R 1 T2T2 1 T1T1 - Temperature Dependence R = universal gas constant = J/mol*K K 1 is the equilibrium constant at T 1 ln k2k2 k1k1 = - EaEa R 1 T2T2 1 T1T1 - ln P2P2 P1P1 = -  H vap R 1 T2T2 1 T1T1 - ln K2K2 K1K1 = -  H 0 rxn R 1 T2T2 1 T1T1 -

Lecture Question CH 4(g) + CO 2(g)  2CO (g) + 2 H 2(g) A. What is the theoretical yield of H 2 when equimolar mixture of CH 4 and CO 2 with a total pressure of 20.0 atm reaches equilibrium at 1200K at which K p = x 10 6 ? B.What is the [H 2 ] eq for this system at 1300K, at which K p = x 10 7 ? C. Use the Van’t Hoff equation to find  H° rxn.