1	Equilibrium
2	Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Eventually the rates are equal
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4	What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. or if the reaction is verrrry slooooow.
5	Law of Mass Action For any reaction jA + kB lC + mD K = [C]^l[D]^m PRODUCTS^power [A]^j[B]^k REACTANTS^power K is called the equilibrium constant. is how we indicate a reversible reaction
6	Playing with K If we write the reaction in reverse. lC + mD jA + kB Then the new equilibrium constant is K’ = [A]^j[B]^k = 1/K[C]^l[D]^m
7	Playing with K If we multiply the equation by a constant njA + nkB nlC + nmD Then the equilibrium constant is _{K’ =}[C]^nl[D]^nm([C]^l[D]^m)ⁿ ₌ Kⁿ[A]^nj[B]^nk ₌ ([A]^j[B]^k)ⁿ
8	The units for K Are determined by the various powers and units of concentrations. They depend on the reaction.
9	K is CONSTANT At any temperature. Temperature affects rate. The equilibrium concentrations don’t have to be the same, only K. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number.
10	Equilibrium Constant One for each Temperature
11	Calculate K N₂ + 3H₂ 2NH₃ Initial At Equilibrium [N₂]₀ =1.000 M [N₂] = 0.921M [H₂]₀ =1.000 M [H₂] = 0.763M [NH₃]₀ =0 M [NH₃] = 0.157M
12	Calculate K N₂ + 3H₂ 2NH₃ Initial At Equilibrium [N₂]₀ = 0 M [N₂] = 0.399 M [H₂]₀ = 0 M [H₂] = 1.197 M [NH₃]₀ = 1.000 M [NH₃] = 0.203M K is the same no matter what the amount of starting materials
13	Equilibrium and Pressure Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT
14	Equilibrium and Pressure 2SO₂(g) + O₂(g) 2SO₃(g) _{Kp =} (P_SO3)²(P_SO2)² (P_O2) _{K =} [SO₃]² [SO₂]² [O₂]
15	Equilibrium and Pressure _{K =} (P_SO3/RT)² (P_SO2/RT)²(P_O2/RT) _{K =} (P_SO3)²(1/RT)² (P_SO2)²(P_O2) (1/RT)³ _{K = Kp}(1/RT)² _{= Kp RT}(1/RT)³
16	General Equation jA + kB lC + mD K_p= (P_C)^l(P_D)^m₌(C_CxRT)^l(C_DxRT)^m(P_A)^j(P_B)^k(C_AxRT)^j(C_BxRT)^k K_p=(C_C)^l(C_D)^mx(RT)^l+m(C_A)^j(C_B)^kx(RT)^j+k K_p=K (RT)⁽^{l+m)-(j+k) =}K (RT)^Dⁿ Dn=(l+m)-(j+k)=Change in moles of gas
17	Homogeneous Equilibria So far every example dealt with reactants and products where all were in the same phase. We can use K in terms of either concentration or pressure. Units depend on reaction.
18	Heterogeneous Equilibria If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. As long as they are not used up we can leave them out of the equilibrium expression. For example
19	For Example H₂(g) + I₂(s) 2HI(g) K = [HI]² [H₂][I₂] But the concentration of I₂does not change. K[I₂]= [HI]^{2 =}K’ [H₂]
20	"Write the equilibrium constant for..." Write the equilibrium constant for the heterogeneous reaction
21	The Reaction Quotient Tells you the direction the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Q = [Products]^coefficient[Reactants] ^coefficient Compare value to equilibrium constant
22	What Q tells us If Q<K Not enough products Shift to right If Q>K Too many products Shift to left If Q=K system is at equilibrium
23	Example for the reaction 2NOCl(g) 2NO(g) + Cl₂(g) K = 1.55 x 10^-5M at 35ºC In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl₂ are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium?
24	Solving Equilibrium Problems Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Learn to create a table of initial and final conditions.
25	"Consider the following reaction at..." Consider the following reaction at 600ºC 2SO₂(g) + O₂(g) 2SO₃(g) In a certain experiment 2.00 mol of SO₂, 1.50 mol of O₂ and 3.00 mol of SO₃ were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO₃ were found to be present. Calculate The equilibrium concentrations of O₂ and SO₂, K and K_P
26	"Consider the same reaction at..." Consider the same reaction at 600ºC 2SO₂(g) + O₂(g) 2SO₃(g) In a different experiment .500 mol SO₂ and .350 mol SO₃ were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O₂ are present. Calculate the final concentrations of SO₂ and SO₃ and K
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28	Solving Equilibrium Problems Type 1
29	What if you’re not given equilibrium concentration? The size of K will determine what approach to take. First let’s look at the case of a LARGE value of K ( >100). Allows us to make simplifying assumptions.
30	Example H₂(g) + I₂(g) 2HI(g) K = 7.1 x 10² at 25ºC Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.8 g of H₂ 294 g I₂ . [H₂]₀ = (15.8g/2.02)/5.00 L = 1.56 M [I₂]₀ = (294g/253.8)/5.00L = 0.232 M [HI]₀ = 0
31	"Q" Q= 0<K so more product will be formed. Set up table of initial, final and change in concentrations. Assumption since K is large- reaction will almost go to completion. Stoichiometry tells us I₂ is LR, it will be smallest at equilibrium let it be x
32	"Choose X so it is..." Choose X so it is small. For I₂ the change in X must be X-.232 M Final must = initial + change
33	"Using to stoichiometry we can..." Using to stoichiometry we can find Change in H₂= X-0.232 M Change in HI = -twice change in H₂ Change in HI = 0.464-2X
34	"Now we can determine the..." Now we can determine the final concentrations by adding.
35	"Now plug these values into..." Now plug these values into the equilibrium expression K = (0.464-2X)² = 7.1 x 10² (1.328+X)(X)
36	Why we chose X K = (0.464-2X)² = 7.1 x 10² (1.328+X)(X) Since X is going to be small, we can ignore it in relation to 0.464 and 1.328 So we can rewrite the equation 7.1 x 10²= (0.464)² (1.328)(X) Makes the algebra easy
37	"When we solve for X..." When we solve for X we get 2.3 x 10^-4 So we can find the other concentrations I₂ = 2.3 x 10^-4 M H₂ = 1.328 M HI = 0.464 M
38	Checking the assumption The rule of thumb is that if the value of X is less than 5% of all the smallest concentrations, our assumption was valid. If not we would have had to use the quadratic equation More on this later. Our assumption was valid.
39	Practice For the reaction Cl₂ + O₂ 2ClO(g) K = 156 In an experiment 0.100 mol ClO, 1.00 mol O₂ and 0.0100 mol Cl₂ are mixed in a 4.00 L flask. If the reaction is not at equilibrium, which way will it shift? Calculate the equilibrium concentrations.
40	"At an elevated temperature" At an elevated temperature, the reaction: has a value of K_eq = 944. If 0.234 mol IBr is placed in a 1.00 L. flask and allowed to reach equilibrium, what is the equilibrium concentration in M. of I₂?
41	Type 2 Problems with small K K< .01
42	Process is the same Set up table of initial, change, and final concentrations. Choose X to be small. For this case it will be a product. For a small K the product concentration is small.
43	For example For the reaction 2NOCl 2NO +Cl₂ K= 1.6 x 10^-5 If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl₂ are mixed in a 1 L container What are the equilibrium concentrations Q = [NO]²[Cl₂] = (0.45)²(0.87) = 0.15 M [NOCl]² (1.20)²
44	"Choose X to be small" Choose X to be small NO will be LR Choose NO to be X
45	"Figure out change in NO" Figure out change in NO Change = final - initial change = X-0.45
46	"Now figure out the other..." Now figure out the other changes Use stoichiometry Change in Cl₂ is 1/2 change in NO Change in NOCl is - change in NO
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51	Practice Problem For the reaction 2ClO(g) Cl₂ (g) + O₂ (g) K = 6.4 x 10^-3 In an experiment 0.100 mol ClO(g), 1.00 mol O₂and 1.00 x 10^-2 mol Cl₂ are mixed in a 4.00 L container. What are the equilibrium concentrations?
52	Type 3 Mid-range K’s .01<K<10
53	No Simplification Choose X to be small. Can’t simplify so we will have to solve the quadratic (we hope) H₂(g) + I₂ (g) 2HI(g) K=38.6 What is the equilibrium concentrations if 1.800 mol H₂, 1.600 mol I₂ and 2.600 mol HI are mixed in a 2.000 L container?
54	Problems Involving Pressure Solved exactly the same, with same rules for choosing X depending on K_P For the reaction N₂O₄(g) 2NO₂(g) K_P = .131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N₂O₄?
55	Le Châtelier’s Principle If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. 3 Types of stress Concentration Pressure Temperature
56	Change amounts of reactants and/or products Adding product makes Q>K Removing reactant makes Q>K Adding reactant makes Q<K Removing product makes Q<K Determine the effect on Q, will tell you the direction of shift
57	Change Pressure By changing volume System will move in the direction that has the least moles of gas. Because partial pressures (and concentrations) change, a new equilibrium must be reached. System tries to minimize the moles of gas if volume is reduced And visa versa
58	Change in Pressure By adding an inert gas Partial pressures of reactants and product are not changed No effect on equilibrium position
59	Change in Temperature Affects the rates of both the forward and reverse reactions. Doesn’t just change the equilibrium position, changes the equilibrium constant. The direction of the shift depends on whether it is exo- or endothermic
60	Exothermic DH<0 Releases heat Think of heat as a product Raising temperature push toward reactants. Shifts to left.
61	Endothermic DH>0 Produces heat Think of heat as a reactant Raising temperature push toward products. Shifts to right.