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 as 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.

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. HC 2 H 3 O 2  H + + C 2 H 3 O 2 - b. H 2 O + H 2 O  H 3 O+ + OH - c. 4NH 3(g) (g)  2N 2(g) + 6H 2 O (g) d. N 2(g) + 3H 2(g)  2NH 3(g)

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.

K as either K c or K p K c = the equilibrium constant using concentrations. K p = the equilibrium constant using pressure P = n  P x n RT v v K p = K c (RT)  ngas Write K p the 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 ?

N 2 + O 2  2NO a. Since Q is very small Q<<<1, very little, NO will 25°C. The equilibrium lies to the left favoring reactants. N 2 + O 2 = 2NO b. Kp = (P NO ) 2  n = = 0 P N2 P O2 kp = Kc Kc = [NO] 2 [N 2 ][O 2 ] c. Kp = P N2 P O2 (P NO ) 2 d.  n = = 0 kp = kc kc = [N 2 ][O 2 ] [NO] 2

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, Kc C. Write the equilibrium expression for 2NO (g)  N 2(g) + O 2(g) D. Determine the Kc for “C”

HOMOGENEOUS EQUILIBRIA H 2(g) + I 2(g)  2HI (g) Kp = (P HI ) 2 (P I2 )(P H2 ) Kc = ? 2O 3(g)  3O 2(g) Kp = (P O2 ) 3 (P O3 ) 2 Kc = ? HETEROGENEOUS EQUILIBRIA 2H 2 O (l)  H 3 O + (aq)_ + OH -(aq) Kc = [H 3 O + ][OH - ] C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (g) Kc = ?

HETEROGENEOUS EQUILIBRIA Substance in more then 1 phase 1.CaCO 3(s)  CaO (s) + CO 2(g) Kc = [CaO][CO 2 ] [CaCO 3 ] How do the [ ] of solid express? A. D = g/cm 3 = mol MW g/mol cm 3 Pure solids & liquid have constant [ ] Kc = constant [CO 2 ] constant Kc = Kc con = [CO 2 ] con

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 equil, expressed by 1 a)pure CaCO 3 b)CaO & P CO2 > Kp c)solid CaCO 3 & P CO2 > Kp d)CaCO 3 & CaO

CALCULATING THE K eq 1. In one experiment, Haber introduced a mixture of H 2 & 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 started with partial pressures of atm for H 2 and atm for N 2, what is the partial pressure of NH 3 ? 4. A 1.00 L flask is filled with 1.00 mol of H 2 and 2.00 mol I 2 at 448°C is What are the equilibrium concentrations of H 2, I 2 & HI?

CALCULATING K eq 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 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.

APPLICATION OF K eq 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

1. At 448°C the equilibrium constant Kc for the reaction is H 2(g) + I 2(g)  2HI (g) Predict how the Rx will proceed to reach equil at 448°C if the initial amount of HI is 2.0 x mol, 1.0 x mol, H 2 and 3.0 x mol I 2 in at 2.0 L container. [HI] ° = 2 x mol/2.0L = 1.0 x M [H 2 ] ° = 1.0 x mol/2.0L = 5.0 x M [I 2 ] ° = 3.0 x mol/2.0L = 1.5 x M Q = Prod = [HI] 2 = (1.0 x ) 2 = 1.3 React [H 2 ][I 2 ] (5 x )(1.5 x ) since K = 50.5, Q = 1.3, Q < K [HI] will need to increase and [H 2 ][I 2 ] will decrease to reach equilibrium.

2. At 1000K the value of Kc for the reaction 2S0 3(g)  2SO 2(g) + O 2(g) is 4.07 x Calculate the value for Q and predict the direction in which the reaction will proceed towards equil if the initial concentration of reactants are: [SO 3 ] = 2 x M [SO 2 ] = 5 x M [O 2 ] = 3 x M Q = 0.2 reaction will proceed from Rt to left forming SO 3.

CALCULATING K eq 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 Kc. 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

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.

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)

Example: 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: 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

N 2 O 4  2NO 2 A) The system will adjust so to decrease [N 2 O 4 ] shifts to  products B) Shifts to  reactants (NO 2 removed) C) N 2 will increase total pressure but since it is not involved in Rx the partial pressures of N 2 O 4 and NO 2 are unchanged, no shift. D)  volume shifts  to more moles of gas E)  Temp - Rx is endo heat + N 2 O 4  2 NO 2 Temp  shifts so more heat produced  K is affected

1.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) 2. 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 DISTURBANCENET DIRECTION OF REACTIONEFFECT 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, Kp = 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 kp at 0°C.

CH 4 ( g) + CO 2(g)  2CO (g) + 2 H 2(g) A. What is the percent 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 Kp = x 10 6 ? B.What is the percent yield of H 2 for this system at 1300K, at which Kp = x 10 7 ? C. Use the Van’t Hoff equation to find  H° rxn.