Chapter 15 – Equilibrium

Equilibrium in Chemical Reactions Lets’ look back at our hypothetical reaction from the kinetics chapter. Chapter 15 – Equilibrium Amount of reactant/product A + 2B → C [A] Why doesn’t the concentration of A ever go to zero? What if we waited longer? [C] What is happening here? time In reality, a better was to represent our reaction would be… A + 2B ⇌ C Reversible Reactions & Equilibrium Equilibrium – Some Reactions are Reversible ⇌ + Tells us there is more than one reaction taking place. 1 N2O4 ⇌ 2NO2 Reaction 1 N2O4 → 2NO2 “Forward” reaction Fwd worker breaks reactant in two, sends pieces down the line…. Reactants Reaction 2 Products 2NO2 → N2O4 and “Reverse” reaction ⇌ 2 Will there be more reactants or more products at equilibrium? Rev worker takes the two pieces, puts back together. Sends back…. However, it takes REV worker twice as long to put the pieces together…. How will the amounts of products and reactants change over time? Reversible Reactions & Equilibrium Reversible Reactions & Equilibrium As the workday progresses….. 3 Because the fwd and rev workers are working at different rates, over time, there will be a build-up of products Reactants 4 Eventually… Products Since the REV worker is being overwhelmed with more products, the foremen sends in help!!! 5 Because there are now two REV workers, they together they work at the same rate as the FWD worker Reactants Products From this point on, the amounts of reactants and products will stay the same…they are in dynamic equilibrium Once dynamic equilibrium is reached, the amounts of reactants and products DO NOT CHANGE. The reaction appears to have stopped, but the forward and reverse reaction are taking place at the same rate. 1 Equilibrium Constant – K Experimental Evidence of Equilibrium INITIAL FINAL Notice that although each trial started with a different ratios of reactant/products, the “ratio” of concentrations of reactants/products at equilibrium will be the same Trial 1 INITIAL FINAL Trial 2 Do not confuse the two kay’s k K • lower-case / small k • Related to “how fast” a reaction proceeds • Determined based on the rate law • UNITS!! • Upper-case / large K • Denotes “how far” are reaction proceeds • Determine by equilibrium amounts of reactants / products • NO UNITS!! Equilibrium Constant Equilibrium Constant – K What does the numerical value of Kc actually mean? The quantitative measure of “how “far a reaction proceeds is represented by a constant, K. Kc  K can be thought of as the result of two reactions: the forward reaction and the reverse reaction. Whether we end up with more reactant and more product in the end is determined by the rate of each reaction. [product]coefficient atequilibriu m [reactant]coefficient kfwd [N2O4 ]  Kc krev [NO2 ]2 kfwd [NO2 ]2   Kc krev [N2O4 ] Similar rxn rates Kc  Fwd faster than rev [product]coefficient atequilibriu m [reactant]coefficient Substantial amount of both reactants and products Mostly products, very small amount of reactants Rev faster than fwd Product Favored Reactant ~ Product Mostly reactants,, very small amount of products Product(s) Rate = krev[NO2]2 Reactant(s) “Reverse” reaction At equilibrium Rate = kfwd[N2O4] Product(s) N2O4⇌ 2NO2 “Forward” reaction Reactant(s) Let’s assume the following reaction is a one-step process. We can determine the rate laws since it is an elementary step: Reactant Favored Equilibrium Constant and concentration – Kc K < < < < 1 Negligible K < < 1 “Reactant Favored” K ≈1 K >>1 “Product Favored” K >>>>1 Complete K values and equilibrium concentration practice For aqueous species and gases we can use concentration in our constant Oxygen gas can be converted into ozone based on the following reaction: [C]c [D]d K c  a b atequilibriu m [A] [B] aA + bB ⇌ cC + dD We can determine Kc experimentally by measuring concentrations of reactants and products at equilibrium N2O4 ⇌ 2NO2 Experiment 1 2 3 Initial (M) [N2O4] [NO2] 0.000 0.0200 0.000 0.0300 0.000 0.0400 [NO2 ]2 [N2O4 ] • If the equilibrium concentration of O 2 is 0.50M and O3 is 2.2x10-7M, determine the value of the equilibrium constant, Kc Equilibrium (M) [N2O4] [NO2] Kc 0.00140 0.0172 0.211 0.00280 0.0243 0.211 0.00452 0.0310 0.213 Since Kc  3O2 (g) ⇌ 2O3 (g) Kc  (0.0172M)2  0.211 (0.00140M) • If the equilibrium concentration of O 2 is 0.10M, what would be the equilibrium concentration of O3? -8 M 2 Heterogeneous Equilibria Equilibrium Constant and Pressure – Kp Just as we can have an equilibrium constant based on pressure of products and reactants (Kc), we can also use partial pressures when dealing with gases How do we deal with reactions that have different states of matter within the same reaction? For example: Kc  Fe(OH)3(s) → Fe3+(aq) + 3OH-(aq) 3 A solid or ,liquid does not change concentration - 3 [Fe ][OH ] [Fe(OH)3 ] Kp  (PartialPressureof Products) atequilibriu m (PartialPressureof Reactants) Although a gas can be represented by Kc or Kp, how do we represent solids? What is the concentration of a solid? Well, there is no need to calculate it since it will not change even if you change the amount do we. So, we take it out of the K expression. Kc ' [Fe(OH)3 ]  [Fe3 ][OH- ]3 Pure liquids and solids are not represented in the equilibrium equation Kc  [Fe3 ][OH- ]3 So for the following equation… Kp  H+(aq) +OH-(aq) → H2O(l) [CO]2 [CO2 ] Kc  …solid carbon will not be a part of the equilibrium expression 1 [OH- ][H ] …liquid H2O will not be a part of the equilibrium expression Kp  K c (RT)Δ n Based on partial pressures 2O3⇌ 3O2 [O ]3 Kc  2 2 [O3 ] (PO2 )3 Kp  (PO3 )2 Relationships between related K’s Remember that for the reaction…. K c1  aA + bB ⇌ cC + dD [C]c [D]d [A]a [B]b But what about the reverse reaction? Mol gas products – mol gas reactants from balanced equation For example, the following reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g) has a Kc of 4.09x10-2 at 1000.K. To determine Kp: Latm molK Pgas = the partial pressure the gas exerts in the reaction mixture So, K’s for the reaction: cC + dD ⇌ aA + bB atm 0.08206 Lmol K Kp  (4.09x10 )[(0.08206 (PC ) (PD ) atequilibriu m (PA )a (PB )b Based on concentrations The Relationship between Kc and Kp The relationship between Kc and Kp is determined by the following equation 2 d And for the following equation… CO2(g) +C(s) → 2CO(g) Kc  aA + bB ⇌ cC + dD c (23) )(1000.K)]  When we deal with reversible reactions, we can write it any direction we choose. Since we have written the equation in reverse, products and reactants are switched. Thus, K is different. Kc2  [A]a [B]b [C]c [D]d 4.98x10-4 The reaction: H2(g) + Cl2(g) ⇌ 2HCl(g) has a Kp of 4.0x1031 at 300.K. What is Kc? K2  (2-2) atm (4.0x1031)  (Kc ) [(0.08206 Lmol  4.0x1031 K )(300.K)] 1  K1 1 K1 When the number of moles of gas are the same on both sides of the equation, Kc = Kp Relationships between related K’s Let’s say we have two reactions that have a reactant/product in common: aA + bB ⇌ xX xX ⇌ cC We can add the two reaction together aA + bB ⇌ cC K c1  [X]x [A]a [B]b Kc2  Reaction Quotient and Equilibrium Constant What do we do when we know the concentration of reactants before equilibrium (or before we mix them to perform a reaction)? Do we have an expression that can quantify this? Qc  [C]c [X]x [C]c [D]d atanytimeNOT atequilibriu m [A]a[B]b Forward More Products Formed How do we combine K expressions? K Q K c1 K c2    [X]x  [C]c   x  a b   [A] [B]  [X]  Q< K K3  K1 K 2  [C]c K3  [A]a [B]b c d [C] [D] Qc  a b [A] [B] Reverse More Reactants Formed Equilibrium Ratio of reactants higher than at equilibrium Q Q>K [C]c [D]d Qc  a b [A] [B] Ratio of products higher than at equilibrium 3 Le Chatelier’s Principle Le Chatelier’s Principle What happens when a equilibrium is “disturbed”? Once it is disturbed, it will shift toward reactants or products to return to equilibrium N2O4 + 2NO2 Dark orange colorless When an equilibrium is disturbed, the reaction will proceed in the direction that reestablishes an equilibrium . Equilibrium mixture of NO2 and N2O4 gives a yellowish color. Wait 10 minutes N2O4 ? 2NO2 The addition of more NO2 turns the mixture orange (NO2 is orange in color). N2O4 ⇌ 2NO2 After 10 minutes, the color returns as before, suggesting that the equilibrium mixture has returned. Which way did the reaction shift to reach equilibrium? Le Chatelier’s Principle Only reactants and/or products that are part of the equilibrium equation, will affect equilibrium. Kc  [Fe3 ][OH- ]3 OH- OH- Add more Fe(OH)3 OH- Fe3+ Fe3+ OH- OHOH- OH- OH- OHConcentration of Fe(OH)3 ions does not change Adding more Fe(OH)3 will not cause more ions to dissolve. No shift in equilibrium OHFe3+ OH- Fe(OH)3 Increase the amount of a product A+B⇌C+D A+B⇌C+D A+B⇌C+D A+B⇌C+D Fe(OH)3 OH- Fe(OH)3 is not part of the equilibrium expression OH- CaCO3(s) → CO2(g) + CaO(s) CO2 CO2 CO2 Q>K Reaction shifts toward reactants A+B→C+D A+B→C+D A+B←C+D K is reestablished K is reestablished K is reestablished A+B⇌C+D A+B⇌C+D Reaction shifts toward reactants A+B←C+D K is reestablished A+B⇌C+D A+B⇌C+D To make a reaction shift toward reactants •Decrease the amount of reactant •Increase the amount of product Le Chatelier’s Principle – Change in Temp Exothermic reactions (-ΔH) A + B ⇌ C + HEAT Reaction shifts toward reactants Endothermic reactions (+ΔH) A + B ⇌ C + HEAT HEAT + A + B ⇌ C Reaction shifts toward products Reaction shifts toward products HEAT + A + B ⇌ C Reaction shifts toward reactants Kc  [CO2 ] CO2 CO2 CO2 CO2 CO2 CaCO3 Q>K Reaction shifts toward products A + B ← C + HEAT Add more CaCO3 CO2 CO2 Q< K Reaction shifts toward products In order to determine the effect of temperature on equilibria, it is convenient to view heat either as a product or reactant. Fe(OH)3(s) → Fe3+(aq) + 3OH-(aq) OH- Decrease the amount of a reactant To make a reaction shift toward products •Decrease the amount of product •Increase the amount of reactant N2O4← 2NO2 Fe3+ Decrease the amount of a product Q< K More NO2 added N2O4 ⇌ 2NO2 Increase the amount of a reactant Adding more CaCO3 will not cause more CO2 or CaO to be created. No shift in equilibrium CO2 CO2 Concentration of CO2 does not change Equilibrium is established w/ smaller K A + B ⇌ C + HEAT CaCO Ca(CO 3 3)2 A + B → C + HEAT Equilibrium is established w/ larger K A + B ⇌ C + HEAT HEAT + A + B → C HEAT + A + B ← C Equilibrium is established w/ larger K Equilibrium is established w/ smaller K HEAT + A + B ⇌ C HEAT + A + B ⇌ C CaO Le Chatelier’s Principle Decrease in volume 1 atm The “I.C.E” Table An increase in total pressure will shift a reaction toward the side with the fewest moles of gas. 2 atm 2 atm A certain reaction has a balanced equation of A ⇌ 2B + 3C. After placing 6.00M A in reaction vessel @ 1000.K, the reaction begins. During the reaction, the reactant decreased by1.33M . What are the equilibrium concentrations and what is Kc? A •Write the balanced equation N2O4 ⇌ 2NO2 N2O4 ← 2NO2 Equilibrium shifts to the side with the fewest moles of gas. N2O4 ⇌ 2NO2 More of that gas (product or reactant depending on the balanced equation) is produced 2 atm 2 atm CO2 CO2 CO2 CO2 CO2 CO2 CaCO3 CaCO3(s) ⇌ CO2(g) + CaO(s) CO2 CO2 CO2 CO2 CO CO2 2 CaCO3 CaCO3(s) ← CO2(g) + CaO(s) CO2 CO2 CO2 CaCO Ca(CO 3 3)2 CaCO3(s) ⇌ CO2(g) + CaO(s) •We plug everything we know into the table. •Since A changed -1.33M, how should the concentration of B and C change in comparison? Look at the balanced equation I C E nitial hange ⇌ 6.00M 2B 0M -1.33M +(2)1.33 + 3C 0M +(3)1.33 quilibirum (6.00M – 1.33M) (0M+2.66M) (0M+3.99M) •B should change 2 times as much and C should change 3 times as much a A. Kc  4.67M [B]2 [C]3 [A] 2.66M Kc  3.99M (2.66)2 (3.99)3  96.2 4.67 4 Calculating Equilibrium Concentration using K Determining Change from Initial and Equilibrium Values Gas “A” is put in a reaction flask and heated to 40.0°C until it changes to gas “B” in a 3-to-2 ratio. Before the reaction, gas “A” had a partial pressure of 1.00atm. At the end of the reaction, after it has cooled gas “A” had a partial pressure of 0.40atm. Determine the equilibrium pressure of gas “B” and determine Kp for the reaction. •Write the balanced equation •Plug in what we know I C E nitial •Since we don’t know the change, we, can substitute with a variable “x” (temporarily). •We can actually work backward from equilibrium. hange quilibirum 1.00atm 0atm -3x -0.60atm +2x +0.40atm 0.40atm 0.40atm nitial •Since we don’t know the change, we, can substitute with a variable x. •Equilibrium is just the initial plus the change Therefore, we know that -3x = -0.60 x = 0.20 atm and Change = -3x So 2x = 0.40atm (0.40)2 Kp   2.5 (0.40)3 [B]  0.111 [A] x  0.111 3.33 - x H2 •Write the balanced equation I C E •Plug in what we know nitial •Since we don’t know the change, we, can substitute with a variable x. •Equilibrium is just the initial plus the change •Substitute equilibrium concentrations Since… Kc  hange + ⇌ 2HI I2 0.100M 0.100M 0M -x -x +2x 0.100M - x 0.100M - x [HI]  50.5 [H2 ][I1 ] or 2  2x     50.5  0.100 - x  2 H2 I C E nitial hange quilibirum + ⇌ 2HI I2 0.100M 0.100M 0M -x -x +2x 0.100M - x 0.100M - x 2  2x     50.5  0.100 - x  2x  7.11 (0.100 - x) 0M + 2x 2x  0.711 - 7.11x [H2]eq and [I2]eq = 0.022M 9.11x  0.711 9.11 [HI]eq = 2x [HI]eq = 0.156M Double-check by working backward Kc  Perfect Square Calculating Equilibrium Concentration using K – Quadratic Formula We can calculate equilibrium concentrations if we know initial concentrations and the equilibrium constant, Kc. If we start with an initial concentration of N2O4 of 5.0M before decomposition , how much NO2 will there be at equilibrium? Kc for the reaction is 4.6x10-3. •Write the balanced equation •Plug in what we know N2O4 I C E nitial •Since we don’t know the change, we, can substitute with a variable x. •Equilibrium is just the initial plus the change •Substitute equilibrium concentrations hange quilibirum ⇌ 2NO2 5.0M 0M -x +2x 5.0M - x 2x [NO2 ]2  4.6x10 3 [N2O4 ] Kc  (2x)2  4.6x10 3 (5.0 - x) Solve for x… (0.156)2  50.3 (0.022)(0.022) Calculating Equilibrium Concentration using K – Quadratic Formula I C E N2O4 ⇌ 2NO2 5.0M 0M hange -x +2x quilibirum 5.0M - x nitial Kc  (2x)2  4.6x10 3 (5.0 - x) (2x)2 = (5.0-x)(0.0046) 4x2 = -0.0046x + 0.023 2x 4x2 + 0.0046x - 0.023 = 0 General form of a Quadratic ax2 + bx + c = 0 4x2 + 0.0046x - 0.023 = 0 Since… K c  0.333  0.111 3.00 Calculating Equilibrium Concentration using K – Perfect Squares Since [H2]eq and [I2]eq = 0.100-x  2x     50.5  0.100 - x  Kc  [B]eq = 0.333M 0M + 2x (2x)2  50.5 (0.100 - x)(0.100 - x) 2 0M + x 1.111x  0.370 Then… [A]eq = 3.00M x = 0.0780 quilibirum +x 3.33M - x x  0.370  0.111x Since… [A]eq = 3.33M - x [B]eq = x The reaction between H2 and I2 produces HI. Assume 0.100mols of H2 and 0.100 moles of I2 are placed into a 1.00L container. The gas react and are allowed to reach equilibrium. Calculate the equilibrium concentration of all the gases. Kc for the reaction at 443°C is 50.5. 0M -x Double-check by working backward x  0.333 Calculating Equilibrium Concentration using K – Perfect Squares B 3.33M hange quilibirum •Substitute equilibrium concentrations Kc  Change = Initial – Equilibrium = 1.00atm-0.40atm = 0.60atm I C E •Plug in what we know (P )2 Kp  B 3 (PA ) ⇌ A •Write the balanced equation Don’t forget your coefficients ⇌ 2B 3A A certain reactant “A” rearranges into a product “B” at room temperature in a 1-to-1 ration. If you started with 3.33M “A”, what will be the concentration of both “A” and “B” at equilibrium if Kc = 1.11x10-1 This is in the form of a quadratic Quadratic Formula  b  b2  4ac 2a - 0.0046  (0.0046)2  4(4)(-0.023) 2(4) 5 Calculating Equilibrium Concentration using K – Quadratic Formula Calculating Equilibrium Concentration using K – Simplifying Assumptions Consider the decomposition of H2S, to give H2 and S2. If Kc = 1.67x10-7 at 800°C, determine the equilibrium concentrations for all gases if the initial amount of H 2S is 0.0250M. At this point, you had better know which steps to take!!!  0.0046  (0.0046)2  4(4)(-0.023) 2(4) This value for x does not make “chemical sense” because you cannot have negative concentrations x = 0.075 and x = -0.076 [N2O4]eq = 5.0M-x [NO2]eq = 2x [N2O4]eq = 4.9M Double-check by working backward [NO2]eq = 0.15M Kc  (0.15)2  0.004591 4.9 Kc  [H2 ]2 [S2 ]  1.67x10 7 [H2S]2 (2x)2 (x)  1.67x10 7 (0.0250 - 2x)2 2H2S I C E nitial hange ⇌ 0.0250M -2x quilibirum 0.0250M - 2x 2H2 + S2 0M 0M +2x +x 2x +x This looks like it is going to be a cubic function; 3rd degree polynomial!!! That is: ax3 + bx2 + cx +d = 0. I don’t feel like solving this!!! Although you can do this via a graphing calculator (which you can’t use on an exam, of course…..) or using successive approximations. Let’s make some simplifying assumptions. For example, since K is so small, I’m going to assume that the change in x will be small also. So….. (2x)2 (x)  1.67x10 7 (0.0250 - 2x)2 …Let’s get rid of this..since it should small (2x)2 (x)  1.67x10 7 (0.0250)2 HHMMM….Better…… Calculating Equilibrium Concentration using K – Simplifying Assumptions 2H2S I C E nitial hange ⇌ 0.0250M -2x quilibirum 0.0250M - 2x 2H2 + S2 0M 0M +2x +x 2x +x 4x 3  1.67x10 7 6.25x10 -4 x3  Is our approximation appropriate? Calculate what percentage x is of your original []’s. If it is 5% or smaller, it is OK. 2.97x10 -4 x100  1.19% 0.0250 (2x)2 (x)  1.67x10 7 (0.0250)2 Our approximation is appropriate [H2S]eq = 0.0250M – 2x [H2S]eq = 0.0244M [H2]eq = 2x [H2]eq = 5.94x10-4M [S2]eq = x [S2]eq = 2.97x10-4M 3 (1.67x107 )(6.25x104 ) 4 x 3  3 2.61x10 11 x  2.97x10 4 Slightly off because of approximation. But O.K. Double-check by working backward (5.94x10-4 )2 (2.97x10-4 ) Kc   1.76x10 -7 (0.0244)2 6