# Equilibrium Notes Ch 14:

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1 Equilibrium Notes Ch 14: Homework: E q u i l i b r i u m P a g e 1 Read Chapter 14 Work out sample/practice exercises in the sections, Bonus Chapter 14: 23, 27, 29, 31, 39, 41, 45, 51, 57, 63, 77, 83, 97 Check for the MasteringChemistry.com assignment and complete before due date For fun: How is chemical equilibrium similar to a successful relationship? Consider pressures, temperature and the environment. Equilibrium State: 1. Chemical Equilibrium occurs for reversible reactions when the forward and reverse reactions rates are equal and there is no net conversion of reactants to products. 2. These equilibrium reactions do not go to completion, all species from reactants and products exist at equilibrium. 3. A double arrow is used as reactions occur in both directions at the same rate,. 4. It does not matter if you start with only reactants, or only products, or some mixture of all. Most reactions end in a dynamic equilibrium condition. 5. Irreversible reactions are those that proceed nearly to completion so the end result contains almost all products and almost none of the limiting reactants.

2 E q u i l i b r i u m P a g e 2 Chemical Equilibrium: Chemical Equilibrium answers the question To what extent will a reaction go? Most chemical reactions do not go to completion. Closed systems after a time generally reach a point of dynamic equilibrium in which the forward rate and the reverse rate are equal. For the generic reaction aa + bb cc + dd The forward rate at equilibrium is equal to the reverse rate, making the following rate law expressions equal to each other Rate = k forward [A] a [B] b = k reverse [C] c [D] d Rearranging the equilibrium rate law expressions creates a new constant with a capitalized K that we call the equilibrium constant, or the Law of Mass Action. K = k forward /k reverse = [C] c [D] d /[A] a [B] b Heterogeneous Equilibria: When reactions have solid, pure liquids, or a solvent as either a reactant or product, these species have an activity = 1 and are not part of the equilibrium constant, providing some of each species must be present to establish equilibrium. The Equilibrium Constant: Equilibrium constant K = product concentrations to the power of their coefficients divided by reactant concentrations to the power of their coefficients K = k forward /k reverse = [C] c [D] d /[A] a [B] b Unlike the rate law expression from kinetics in which the slowest step affects the orders, a reaction at equilibrium will have all steps ( forward, reverse, and elementary steps ) going at the same rate while remaining in a dynamic equilibrium. For this reason the coefficients in the overall reaction will be in the equilibrium constant, capital K. All the K constants use activities and are therefore unitless. The activity value for pure solids, pure liquids and solvents are equal to 1. The activities for concentrations in M divide by 1M, for pressure it is divided by 1 atmosphere. Therefore, no units are involved in a capital K, equilibrium constant.

3 E q u i l i b r i u m P a g e 3 The important values to determine K come from gases and solutions. K c : concentrations bases on Molarity K p : concentrations based on partial pressures, atmospheres. All species in the equilibrium constant must be gas (No aqueous Molarities, but solid whose activity is 1 is OK) K eq : Molarity for aqueous solutions and partial pressure for gases. K is similar to the rate constant k, and will vary with temperature, but it is independent of time and initial concentrations. An important difference is that it is unaffected by a catalyst. Equilibrium can be reached from either direction. K will be reached whether you start with all reactants, all products, or a little of each substance in the reaction. Understanding K will help in determining how a reaction may be manipulated in order to increase a desired substance. PV = nrt from the gas laws so P = MRT We can calculate and find K p = K c (RT) n or K c = K p (RT) - n Where R is the gas constant (L atm)/(mol K) and n is the change in number of moles of gases only. Magnitude of K: The equilibrium constant is always a positive number between 0 to infinite. Large constant, K >> 1, indicates mostly products will exist at equilibrium. Small equilibrium constant, K << 1, indicates mostly reactants If the reaction doubles, K is squared K 2 If the reaction reverses K is inversed K -1 In a multistep reaction, multiply each K to get the overall K 1 x K 2 x K 3

4 E q u i l i b r i u m P a g e 4 K can be determined experimentally if all equilibrium concentrations are known, or calculated from thermochemical data (K eq = e - G /RT ) discussed later when we study thermochemistry. RICE represents Reaction: write the balanced chemical reaction Initial: initial concentrations/amounts Change: change taking place to reach equilibrium Equilibrium: equilibrium concentrations/amounts Applying Equilibrium: Example 1: For the reaction H 2 (g) + I 2 (g) 2 HI (g) at 229 C in a 1.00 L vessel a) Initially, mole of H 2 gas and mole of I 2 gas is injected into the 1.00L vessel. Once equilibrium has been established, mol of HI is found in the vessel. Apply RICE to determine the equilibrium Molarities of all substances in this reaction and the equilibrium constant, K c. b) This same reaction H 2 (g) + I 2 (g) 2 HI (g) at 229 C is now initiated by injecting 1.00 mol of HI in a 1.00 L evacuated vessel. Using the calculated value of K c, apply RICE to determine the equilibrium Molarities of all substances.

5 E q u i l i b r i u m P a g e 5 For the same reaction H 2 (g) + I 2 (g) 2 HI (g) at 229 C in a 1.00 L vessel c) Graph concentration of all substances verses time for both part a and b. d) Calculate the value of K p for this reaction at 229 C using the equation K p = K c (RT) n (this R is L atm/mol K) e) Using the K c that was calculated in part (a), determine the value of K c for the reaction written below HI (g) ½ H 2 (g) + ½ I 2 (g) at 229 C

6 E q u i l i b r i u m P a g e 6 Example 2: Write all possible (K c, K p, K eq ) equilibrium constant expressions for the following. (Not all reactions have a K p ) a) PCl 5 (g) PCl 3 (g) + Cl 2 (g) b) 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) c) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) d) CaCO 3 (s) CaO (s) + CO 2 (g) e) Mg (s) + 2 HCl (aq) MgCl 2 (aq) + H 2 (g) f) Cu(NO 3 ) 2 (aq) + 2KOH (aq) Cu(OH) 2 (s) + 2KNO 3 (aq)

7 E q u i l i b r i u m P a g e 7 Example 3: Prove that K overall = K 1 x K 2 for the two step reaction below. (assume all are K c ) Step 1: NO 2 (g) + O 3 (g) NO 3 (g) + O 2 (g) K 1 Step 2: NO 3 (g) + NO 2 (g) N 2 O 5 (g) K 2 Overall: K overall Example 4: For the reaction 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) ; K p = 2.5 x 10 9 a) Do you expect mostly reactants or products at equilibrium? Explain. b) What is the numerical value of K p for the reverse reaction? c) What is the numerical value of K p if the original reaction is doubled? d) What is the numerical value of K c for the reaction at 25 C?

8 E q u i l i b r i u m P a g e 8 Example 5: At 27 C, 0.80 mol N 2 and 0.90 mol H 2 were injected in an empty 1.00 L vessel. Once equilibrium was established, 0.20 mol NH 3 (g) was present. Calculate K c and K p for the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) The Reaction Quotient, Q, and Predicting Direction of the Reaction: The Reaction Quotient, Q, is the same formula as the equilibrium constant, K, except Q is obtained using any starting concentrations that have generally not reached the equilibrium point. Q < K, indicates that reactant concentrations are higher than those needed for equilibrium. Which direction will the reaction shift to reach equilibrium? Q = K, indicates that all concentrations have reached equilibrium. Q > K, indicates that product concentrations are higher than those needed for equilibrium. Which direction will the reaction shift to reach equilibrium?

9 E q u i l i b r i u m P a g e 9 Example 6: For the reaction H 2 (g) + I 2 (g) 2 HI (g) at 450 C, K c = mole of I 2, 0.22 mole H 2 and 0.66 mole HI are injected in an evacuated 4.00 L vessel. a) Solve for the reaction Quotient, Q c b) What must happen to establish equilibrium? c) Solve for the equilibrium concentrations.

10 E q u i l i b r i u m P a g e 10 Le Chatelier s Principle and How to Manipulate the Reaction: If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position as to counteract the effect of the disturbance. Changes to Consider: Concentration of any substance: Adding a reactant or removing a product will shift the reaction (which direction?) Removing a reactant or adding a product will shift the reaction (which direction?) Volume / Partial Pressures of any gas substance: Increasing the volume decreases all the partial pressures and the reaction will shift to create more pressure if possible will shift in the direction that has (Choose one: more/less) moles of gas. No effect occurs when an inert gas is added that does not participate in the reaction or when the number of moles of gas are equal on either side of the chemical reaction. Temperature: Increasing the temperature will favor the (Choose one: endothermic/exothermic) reaction to reduce the extra energy added by increasing the temperature. Decreasing the temperature will favor the (Choose one: endothermic/exothermic) reaction. Catalyst/Inhibitor: How will this affect a system at equilibrium? Explain.

11 E q u i l i b r i u m P a g e 11 For the Reaction below, N 2 O 4 (g) 2 NO 2 (g) H = When N 2 O 4 gas is added to a reaction vessel at equilibrium, more NO 2 gas will form.

12 E q u i l i b r i u m P a g e 12 Heat also will effect this reaction: Cause and Effect for a System at Equilibrium that undergoes a Change Example 7: Consider the synthesis of ammonia from the elements nitrogen and hydrogen, the Haber Process N 2 (g) + 3 H 2 (g) 2 NH 3 (g); H = -92 kj/mol What conditions will increase the production of ammonia on an industrial scale? Explain. a) Add or Remove NH 3 b) Add or Remove N 2 and H 2 c) Changing the pressure of the system by (i) Addition of an inert gas (ii) Altering the volume that contain the gases (Increase or decrease the overall pressure?) (d) and (e) are tricky: Consider kinetics as well as equilibrium. Which may be more important in the industrial production of ammonia? d) Increase or decrease the temperature? e) Addition of a catalyst.

13 E q u i l i b r i u m P a g e 13 Mol% NH 3 at The Haber Process: N 2 (g) + 3 H 2 (g) 2 NH 3 (g); H = -92 kj/mol Effect of Temperature and Pressure on the Yield of NH 3 C K c Mol% NH 3 at Mol% NH 3 at 10 atm P total 100 atm P total 1000 atm P total % 82% 98% % 25% 80% % 5% 13% The production of ammonia is generally run at conditions around 450 C and between atm pressure. The reaction is much too slow at lower temperatures, and if the temperature is raised too high the mole % of ammonia decreases dramatically. The Quadratic Equation may be necessary for some calculations: Example 8: For the reaction A (g) B (g) + C (g) a) An equilibrium mixture from the reaction above is found to contain 0.20 M A, 0.30 M B, and 0.30M C. Determine the equilibrium constant, K c. R: A (g) B (g) + C (g) E: 0.20M K c =

14 For the same reaction A (g) B (g) + C (g) E q u i l i b r i u m P a g e 14 b) If the above equilibrium reaction at the same temperature is expanded into a vessel double the original size Determine the reaction quotient, Q. Will the reaction shift toward products, reactants, or stay the same? Solve for the new equilibrium concentrations. R: A (g) B (g) + C (g) E: 0.20M C: double the vessel volume has affect on M I: 0.10M C: E: c) If the first equilibrium reaction at the same temperature is compressed into a vessel half the original size Determine the reaction quotient, Q. Will the reaction shift toward products, reactants, or stay the same? Solve for the new equilibrium concentrations.

15 E q u i l i b r i u m P a g e 15 Example 9: For the reaction CO (g) + Cl 2 (g) COCl 2 (g) A 2.00 L vessel in equilibrium contains 1.20 mol COCl 2, 0.60 mol CO, and 0.20 mol Cl 2 a) Determine the equilibrium constant, K c. b) If 0.80 mol Cl 2 is added to the above equilibrium reaction at the same temperature, will the reaction shift toward products, reactants, or stay the same? Determine the new equilibrium concentrations. Example 10: NOBr is 34.0% dissociated at 25 C in a vessel whose total pressure is 0.25 atm. What is K p? 2 NOBr (g) 2 NO (g) + Br 2 (g)

16 E q u i l i b r i u m P a g e 16 Example 11: BrCl may replace Cl 2 as water disinfectant someday mol Br 2 and 0.75 mol Cl 2 are added to a 5.00 L container and allowed to reach equilibrium. Cl 2 (g) + Br 2 (g) 2 BrCl (g) K c = 4.7 x 10-2 a) Solve for the equilibrium concentrations. b) What % of the original Cl 2 has reacted? Example 12: For the reaction 2 SO 3 (g) O 2 (g) + 2 SO 2 (g) ; H = kj If this reaction is already at equilibrium, what will the following do to it? a) Increase Temperature b) Increase pressure c) Remove O 2 d) Add a catalyst e) Increase the volume

17 E q u i l i b r i u m P a g e 17 Example Answers: 1a,b) M, 0.110M, 0.780M, c) graph, 1d) 50.3, 1e) 0.141, 4a) products, 4b) 4.0 x 10-10, 4c) 6.3 x 10 18, 4d) 6.1 x 10 10, 5) K c = 0.26, K p = 4.4 x 10-4, 6a) Q c = 9, 6b)shift to product, 6c) M H 2, 0.031M I 2, 0.214M HI, 7a) remove, 7b) add, 7ci) no affect, 7cii) increase, 7d,e) needs thoughtful explanation 8a) b) forward to more products, Q = 0.225, A = 0.07M, B = 0.18M, C = 0.18M 8c) reverse toward more reactants, Q = 0.90, A = 0.52M, B = 0.48M, C = 0.48M 9a) 20 9b) CO = 0.12 M, Cl 2 = 0.32 M, COCl 2 = 0.78 M, 10) 0.25 atm =[ P NOBr + P NO + P Br2 ] = x[ ], x = 0.21 atm, K p = 9.3 x a) 0.13 M Cl 2, 0.18 M Br 2, M BrCl 11b) 11%, 12 a,c,e) toward product, 12b) toward reactant, 12d) no affect

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