Unformatted text preview: Chapter 15 Chemical Equilibrium Chemical equilibrium is similar to the equilibrium states reached in evaporation of a liquid in a closed container or the dissolution of a solid to form a saturated solution. Use a double arrow ( ) to indicate the process is dynamic and reversible.
2NO2 N2O4 red + colorless 1 Analogy to Chemical Equilibrium 2 15.1 The Concept of Equilibrium At equilibrium, both reactants and products are present simultaneously. N2O4 2NO2 3 1
1 Butane Isomerization butane isobutane
CH3CH2CH2CH3 CH3CHCH3 CH3 Initially Midway Equilibrium 4 Sample Problem At which point does this system reach a state of equilibrium? 5 Rates at Equilibrium At equilibrium, Ratef = Rater 6 2
2 Rate Constant, k AB Rate laws can be used to describe the rate of the forward reaction and the rate of the reverse reaction: Rate forward= kf[A] f d Ratereverse = kr[B] At Equilibrium: Ratef = Rater kf[A] = kr[B] kf/kr = [B]/[A] = constant = Kc
7 Rates at Equilibrium No matter how complex the rate laws for the forward and reverse reactions, they always combine to give the simple equilibrium constant expression, with Kc = kf/kr 8 Equilibrium and Energy 9 3
3 15.2 The Equilibrium Constant, K
aA + bB cC + dD Kc = [C]c[D]d/[A]a[B]b What is the equilibrium constant expression for: 2NO2(g) N2O4(g) Kc = ?
10 15.2 The Equilibrium Constant, K
Kc = [C]c[D]d/[A]a[B]b To calculate Kc: measure equilibrium concentrations and substitute into the Kc expression. 11 What is the correct equilibrium constant expression for this reaction? CO(g) + 3H2(g) CH4(g) + H2O(g)
[CH 4 ][H 2 O] 1. Kc = [CO][H 2 ] [CH 4 ][H 2 O] 2. Kc = [CO][H ]3 2
25% 25% 25% 25% 3. Kc = [CH ][H O] 4 2 4. Kc = [CH ][H O] 4 2
[CO][H 2 ]3
1 2 3 4 [CO][H 2 ] 12 4
4 The Value of Kc 2NO2(g) N2O4(g) At 100C what is the value of Kc? 100C, [NO2]eq = 0.0172 M [N2O4]eq = 0.00140 M 13 Is this reaction at equilibrium?
Kc = 25
1. Yes 2. No + 50% +
50% 1 2 14 Equilibrium Constants in Terms of Pressure In a gasphase gasreaction, we can also describe the equilibrium in terms of ilib i i f the partial pressures of the reactants and products (instead of concentrations):
aA + bB cC + dD Kp ( PC )c ( PD ) d ( PA ) a ( PB )b 15 5
5 Equilibrium Constants in Terms of Pressure In most cases, Kp Kc K p K c ( RT ) n n = (mole gas products moles gas reactants) R = 0.0821 L atm/K mol (derived from ideal gas law: P=nRT/V) 16 Equilibrium Constants in Terms of Pressure 2NO2(g) N2O4(g) Kc = 4.74 at 100 C . What is Kp? K p K c ( RT ) n n = (mole gas products moles gas reactants) R = 0.0821 L atm/K mol atm/K (derived from ideal gas law: P=nRT/V) P=nRT/V)
17 15.3 Interpreting and Working with Equilibrium Constants What does the equilibrium constant tell us? 18 6
6 Position of Equilibrium Since Kc is a ratio of products/reactants: Kc >> 1 Lots of Products Equilibrium lies to the right Kc <<1 Lots of Reactants Equilibrium lies to the left 19 Position of Equilibrium At what temperature are there mostly products at equilibrium?
T = 25 C N2(g) + 3H2(g) T = 470 C N2(g) + 3H2(g) 2NH3(g) 2NH3(g) Kc = 4.0x101 Kc = 1.0x108
20 Manipulation of K Equation Manipulation Reverse Equation i Add 2 Equations Double or Triple Equations Effect on K K' = 1/K ' 1/ K' = K1 x K2 K' = K2; K' = K3 21 7
7 Manipulation of K 2H2O 2H2 + O2
K1 = [H2]2[O2]/[H2O]2 2H2 + O2 2H2O
K2 = [H2O]2/ [H2]2[O2] 22 What is K3 for the reaction below?
CO + 3H2 CH4 + H2O K1 = 4 CH4 + H2O CH3OH + H2 K2 = 0.005 CO + 2H2 CH3OH K3 = ? 1. 2. 3. 4. 800 4.005 0.02 There's no way anyone could know that! 25% 25% 25% 25% 1 2. 3. 4 23 Manipulation of K
CO + 3H2 CH4 + H2O K=4 What is K for double this reaction?
2CO + 6H2 2CH4 + 2H2O K' = ? 24 8
8 15.4 Heterogeneous Equilibria The concentrations of pure solids and liquids are constant, so they do not show up in the equilibrium expression. p q p CaCO3(s) CaO(s) + CO2(g) CaO(s)
Kc CaO CO constant CO 2 2 CaCO3 25 K c K c constant CO 2 Heterogeneous Equilibria
Write the equilibrium constant expression (Kc) for each of the following (K processes. CO(g) + H2O(l) CO2(g) + H2(g) 26 What is the Kc expression for this reaction?
Mg(OH)2(s) Mg2+(aq) + 2OH(aq) aq) aq)
1. 2. 2 3. 4. Kc = [Mg 2+ ][OH  ]2 Kc = [Mg(OH)2 ] Kc = [Mg ][OH ] [Mg(OH) ] Kc =
2+ 2 25% 25% 25% 25% [Mg 2+ ][OH  ] [Mg 2+ ][OH  ]2 1 2 3 4 27 9
9 15.5 Calculating Equilibrium Constants If K value is unknown, but not all equilibrium concentrations are known, must use stoichiometry to calculate equilibrium concentrations first 28 Approach for Solving Equilibrium Problems If K unknown, but know starting and equilibrium concentrations of some substances: Write a balanced chemical equation. g Select one of the concentration changes and call it x. Use stoichiometry to determine all the concentration changes in terms of x. Use ICE table to calculate equilibrium concentrations Plug into eq. constant expression to solve for K 29 ICE Table What is K for this reaction: H2 (g) + I2 (g) 2HI (g)
1.00x103 2.00x103 0 At eq, [HI] = 1.87x103M eq, 30 10
10 What is the equilibrium constant of this reaction, given the data below?
2SO3 2SO2 + O2 1. 1 2. 3. 4. 0.225 0 225 0.338 2.96 4.44
Initially, PSO3=0.500 atm, atm, PSO2=0, PO2=0
25% 25% 25% 25% At Equilibrium, A E ilib i PSO3=0.200 atm 1. 2. 3. 4. 31 Predicting Direction of a Reaction Suppose you start out with nonzero nonconcentrations of reactants and products in a reaction container. How do we know if the reaction will proceed in the forward direction or the reverse direction? 32 Predicting Direction of a Reaction If the concentrations are not equilibrium concentrations, then plugging their values into the K expression will not g p give the true value of K. We call this value Q, the reaction quotient. quotient. 33 11
11 Predicting Direction of a Reaction (Example) CO + H2O CO2 + H2
1.00 2.00 Q=0 0 .00 Kc = 4.00 0.00 M Reaction proceeds to the right 34 Predicting Direction of a Reaction (Example) CO + H2O CO2 + H2
1.00 1.00 1.00 1.00 M Kc = 4.00 Q = 1.00 Q < K Rxn proceeds to the right 35 Predicting Direction of a Reaction (Example) CO + H2O CO2 + H2 Kc = 4.00
1.00 1.00 Q=4
Q=K 2 .00 2.00 M
At Equilibrium 36 12
12 Predicting Direction of a Reaction
CO + H2O CO2 + H2 Kc = 4.00 0.500 0.500 2.00 2.00 M (initial conc.) Q=? 37 Predicting Direction of Reaction K tells us how far toward products the position of equilibrium lies. Q tells us in which direction the reaction will proceed. Q = K equilibrium Q < K spontaneous forward Q > K spontaneous reverse
38 Direction of Reaction 39 13
13 Which direction will this reaction go if Kc=25? 33% 33% 33% 1. Forward 2. Reverse 3. Neither, it is at equilibrium
1 2 3 40 15.6 Applications of Equilibrium Constants Solving for one equilibrium concentration: If we know K, and all but one equilibrium concentration, plug values into equilibrium expression and solve 41 15.6 Application of Equilibrium Constants Given this reaction:
2SO3 2SO2 + O2 (Kp =0.345) If the equilibrium pressures of SO2 and O2 are 0.135atm and 0.455atm, what is the pressure of SO3? 42 14
14 Applications of Equilibrium Constants What if we don't know any equilibrium concentrations? If we know initial concentrations and K, concentrations, K we can find equilibrium concentrations using ICE tables, stoichiometry, and a stoichiometry, healthy dose of algebra 43 ICE Approach Write a balanced chemical equation. Select one of the concentration changes and call it x. Use stoichiometry to determine all the concentration changes in terms of x. Write expressions for equilibrium concentrations in terms of x 44 ICE Approach continued Write the equilibrium constant expression. Insert the equilibrium concentration expressions from the table into the equilibrium constant expression. expression Solve the equation for x, applying any possible simplifications (perfect squares or dropping terms when small). Check simplifications for validity (5% error is tolerable). If not valid, return to step 7 and do not use the simplification.
45 15
15 ICE Approach continued
Check your answer: Substitute the value of x into the expressions for the equilibrium concentrations and determine their values. Use the equilibrium concentrations to calculate the reaction quotient, and compare it to the equilibrium constant to verify the accuracy of the answers.
46 Typical Problem: Known Initial Concentrations
What are equilibrium concentrations? N2 + 3H2 2NH3 Kc = 2.5 105
1.00 1.00 0M initial concentrations Make table of concentrations: Substance: N2 3H2 2NH3 Initial Conc.: 1.00 1.00 0.00 Change: x 3x +2x Equil. Conc.: 1.00  x 1.00  3x 2x quil.
47 Typical Problem Solution
N2 + 3H2 2NH3 Kc = 2.5 x 105 Assume 3x << 1.00 and x << 1.00 to simplify [NH3]2 (2x)2 Kc =  =  = 2.5 105 [N2][H2]3 (1.00  x)(1.00  3x)3  = 2.5 105 = 4x2
(1.00)(1.00)3 (2x)2 2x = (2.5 x 105)1/2 = 5.0 103 x = 2.5 103 (This is <<1, good sign!) 48 16
16 Solution [NH3]eq = 2x = 5.0 x 103 M [N2]eq = 1.00  x = 0.9975 M [H2]eq = 1.00  3x = 0.9925 M q Check calculations:
Kc'=(5.0 103)2/(0.9975)(0.9925)3 = 2.56 x 105 Kc = 2.5 105 These are close to equal, so the calculations are valid.
49 Group Work N2 + 3H2 2NH3 Kc = 2.5 x 105 0.100 0.100 0 M initial concentrations What are the equilibrium concentrations? 50 Given the data below, what is the equilibrium concentration of NH3?
N2 + 3H2 2NH3 Kc = 2.5 x 105 0.100 0.100 0 M initial concentrations
25% 25% 25% 25% 1. 1 2. 3. 4. [NH3] = 1 00 x 1.00 [NH3] = 5.00 x 109M [NH3] = 1.25 x 104M [NH3] = 5.00 x 105M
1 2 3 4 104M 51 17
17 Avoid using the quadratic A(g) + B(g) 2AB(g) 2AB(g) How does this simplify if initial concentrations of A and B are equal? 52 What if assumptions fail? The most common assumption is that x << Co If this assumption is not correct, the equation in x becomes more complex. The most difficult situation that we meet is generally one in which the equation can be rearranged into a quadratic equation in x: ti i ax2 + bx + c = 0
x = b The solution to this equation is: b2  4 a c 2a Of the two solutions, one will generally not make chemical sense. For example, x might be larger than the initial concentration. (See Sample Ex. 15.12) 53 Quadratic Example CO + H2O CO2 + H2 Kc = 4.00
1.00 2.00 0 .00 0.00 M What are the equilibrium concentrations? Substance: CO H2O CO2 H2 Initial Conc: 1.00 Conc: 2.00 0.00 0.00 Change: x x +x +x Equil Conc: Conc: 1.00  x 2.00  x +x +x Kc = [CO2][H2]/[CO][H2O] = 4.00
54 18
18 Quadratic Example Solution (x)(x)/(1.00 x)(2.00  x) = 4.00 Assume x << 1.00, x << 2.00 x2/(1.00)(2.00) = 4.00 ( )( ) 2 = 8.00 x x = 2.83, so the assumption was bad. (x)(x)/(1.00  x)(2.00  x) = 4.00 (x)(x) = 4.00 (1.00  x)(2.00  x) x2 = 8.00  12.00x + 4.00 x2 3.00x2  12.00 x + 8.00 = 0 55 Quadratic Example Solution x = b 3.00x2  12.00 x + 8.00 = 0 b2  4 a c 2a 2 x = 12.00 (12.00) 4(3.00)(8.00) 2(3.00) x = 0.845 0 845 [CO2] = [H2] = x = 0.845 M K' = (0.8450.845)/(0.1551.155) = 3.99, which is a [CO] = 1.00  x = 0.155 M [H2O] = 2.00  x = 1.155 M good match to K = 4.00, so the calculations are valid.
56 Product Present Initially The procedure remains the same if any product is in the mixture initially. The only difference will be in the value of the y initial concentration. It is also possible that reaction will occur in the reverse direction, in which case x direction, will have a negative value. value. 57 19
19 What is the equilibrium concentration of PCl3?
PCl5
1.00 PCl3 + Cl2
0.0100 0.500 M initial concentrations Kc = 0.00900 1. 2. 3. 4. 1.76 x 102M 2.40 x 103M 0.516M .992M 25% 25% 25% 25% 1 2 3 4 58 15.7 LeChatelier's Principle If we place a stress on a system at equilibrium, the system will respond by shifting to a new equilibrium position in such a way as to remove that stress. 59 15.7 LeChatelier's Principle Stresses: change reactant concentration change product concentration change pressure change volume change temperature (changes K value) add a catalyst (no effect) 60 20
20 Change in Reactant Concentration
N2 + 3H2 2NH3 Add H2 to system at equilibrium 61 Change in Reactant Concentration
N2 + 3H2 2NH3 Add H2 to system at equilibrium [H2] so Q < Kc Causes [NH3] ? [N2] [ NH 3 ]2 Kc [ N 2 ][ H 2 ]3 62 Effect of Concentration Changes
CO(g) + 3H2(g) CH4(g) + H2O(g) (g) Kc =[CH4][H2O]/[CO][H2]3 Add CO(g), Q < K, net forward reaction occurs until Q = K, (g), K, , thereby removing some CO(g) Add CH4(g), Q > K, net reverse reaction occurs until Q = K, K, thereby removing some CH4(g) K, Add CaO(s) to remove H2O(g), Q < K, forward reaction CaO(s) occurs to form some H2O(g)
63 21
21 Which of these changes will cause the equilibrium to shift to the left?
O3 + NO O2 + NO2 1. 2. 3. 4. Adding O3 Adding NO Adding O2 Removing NO2
25% 25% 25% 25% 1 2 3 4 64 Effect of Pressure 2CO2(g) 2CO(g) + O2(g) Which direction will the equilibrium shift if pressure is increased? 65 Effect of Pressure Increase pressure by decreasing volume; same as increasing all concentrations (of gases) by a constant factor. An increase in pressure shifts the equilibrium in the direction that has the fewer number of gas molecules. If ngas = 0, there is no change in the position of equilibrium. No effect if there are no gases in the system.
66 22
22 Effect of Pressure 2CO2(g) 2CO(g) 2CO(g) + O2(g) Initial Composition Equilibrium Composition
67 Effect of Pressure Increase Which direction does the equilibrium shift 2NO2(g) N2O4(g) if pressure increases? (g) (g) 68 2NO2 N2O4 Experiment
Change the volume to increase the pressure 69 23
23 Effect of Pressure CO(g) + 3H2(g) CH4(g) + H2O(g) (g) Kc =[CH4][H2O]/[CO][H2]3 Change the volume of the mixture to 1/2 the original value. All the concentrations then g increase by a factor of 2. Before the change, Q = K After the change, Q = K x (2)(2)/(2)(2)3 = K/4 Q < K because the reactant change was greater than the product change and the equilibrium will shift towards products.
70 Effect of Pressure Changing pressure by adding an inert gas has no effect because that gas is not involved in the reaction. 71 Effect of Temperature The effect of temperature changes depends on whether the reaction is exothermic or endothermic. endothermic. The effect results from a change i th value of K with temperature. h in the l f ith t t Sign of Ho determines whether K increases or decreases as temperature increases.
72 24
24 Effect of Temperature N2O4(g) 2NO2(g) (g)
colorless dark endothermic Low T High T
Does K increase or decrease?
73 Effect of Temperature CO + 3H2 CH4 + H2O Ho = 206.2 kJ Which direction does the equilibrium shift q if the temperature is increased? 74 Effect of Temperature We can imagine that heat is a reactant or product. If temperature is increased, the amount of heat is increased. The reaction will shift in such a way as to remove some of the added heat. Effect of increase in T? N2 + 3H2 2NH3 + heat Ho =  92 kJ 2H2O + heat 2H2 + O2 Ho = 484 kJ
75 25
25 LeChatelier's Principle PCl3(g) + Cl2(g) PCl5(g) (g) H = 87.9 kJ Predict the shift in equilibrium for each stress (Right, Left, or No Change): Add Cl2 Increase pressure Increase volume Increase temperature
76 Which of the following changes will increase the amount of O2 at equilibrium?
2Cl2(g) + 2H2O(g) 4HCl(g) + O2(g) Ho = +113 kJ
20% 20% 20% 1. Lower the temperature 2. Add HCl 3. Add Cl2 4. Remove H2O 5. Decrease the volume 20% 20% 1 2 3 4 5 77 26
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This note was uploaded on 04/06/2011 for the course CHM 116 taught by Professor Unknown during the Spring '08 term at ASU.
 Spring '08
 Unknown
 Equilibrium

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