Chemical reactions with large K c (also K p) effectively go 100% to products.

Transcription

1 th 7 Homework: Reading, M&F, ch. 13, pp (applications of equilibrium constants, Le Chatelier s Principle). Problems: Nakon, ch. 17, #9, 10, 12-15, 17, 24, 34; M&F, ch. 13, #41, 42, 44, 49, 51, 53, 57. IV. Chemical Equilibria (cont.) C. Le Chatelier s Principle. Chemical reactions with large K c (also K p) effectively go 100% to products. 81 Ex. Zn(s) + S(s) 6 ZnS(s) + heat, light, smoke... K c = 10 at 298 K 230 Ex. Mg(s) + 1/2O 2 (g) 6 MgO(s) + heat, light, smoke... K c = 10 at 298 K 3 As a rule-of-thumb, when K c is greater than 10, reactions largely go to products. Likewise,!3 when K c is less than 10, reactions barely happen.!3 3 In the intermediate case (10 < K c < 10 ), significant concentrations of reactants and products exist at equilibrium. It is often desirable to force a reaction toward the product side. To do that, we use: Le Chatelier s Principle: if a stress is applied to a reaction mixture at equilibrium, reaction occurs in the direction that relieves the stress. Stresses include changes of concentration of reactant or product, addition of heat, or change of total pressure for a gas-phase reaction. The classic example: the Haber process for synthesizing ammonia: N (g) + 3H (g) º 2NH (g) + heat 3 (exothermic reaction) At 700 K, K c = 0.3. How can we generate more NH 3? (a) Raise the conc. of either N 2 or H 2 or both. The stress is excess reactant; the stress is relieved by shifting the equilibrium towards the product (i.e., reacting N 2 and H 2 to generate more NH 3). (b) Lower the conc. of NH 3 (can be done by liquefying the NH 3). The stress is insufficient product; the stress is relieved as in (a). (c) Lower the temperature. The stress is insufficient heat in the reaction; the stress is relieved as in (a). This implies that K c increases as the temperature decreases. -20-

2 General rule: For an exothermic reaction, K increases as the temperature decreases. For an c endothermic reaction, K increases as the temperature increases. c Note that the opposite stresses (lowering the conc. of N or H, raising the conc. of NH, or 3 raising the temperature) all cause the equilibrium to shift towards reactants (less NH is 3 produced). 2+!!! Ex. Exp. 3: FeCl + SCN º FeNCS + Cl yellow red 3+! Adding more Fe or adding more SCN both cause the color of the solution to become deeper! red. Adding Cl causes the color to lighten towards yellow. What about heterogeneous equilibria? Adding or removing a pure solid or liquid has no effect on the equilibrium concentrations because the solid or liquid has a constant concentration. Ex. CaCO 3 (s) º CaO(s) + CO 2(g) Stress: raise the conc. of CO 2 6 amount of CaO decreases, amount of CaCO 3 increases. Adding or removing either CaCO 3 or CaO has no effect on CO 2 conc. as long as the solids are both present. For gas-phase reactions, the equilibrium can be stressed by changing the pressure of all gases through a volume change. Rule: an increase in pressure shifts the equilibrium towards the side with the fewer moles of gas. Ex. N (g) + 3H (g) º 2NH (g) (4 moles gas on left; 2 moles gas on right) 3 Halve the volume, double all partial pressures 6 equilibrium shifts towards products. o Ex. Given: H 2 (g) + I 2(g) º 2HI(g) )H =!9.4 kj Predict the change in conc. of HI when: (a) More H 2 is added. (b) More HI is added. (c) The temperature is raised. (d) The pressure is decreased by doubling the volume. Answers given in lecture. -21-

3 What about adding a catalyst? Adding a catalyst accelerates both the forward and the reverse rates of reaction, but the presence of a catalyst has no effect on the equilibrium concentrations of a reaction. D. Applications of the equilibrium constant. 1. Predicting the direction of reaction. Given a reaction, K c or K p, and any arbitrary set of concentrations. Is the reaction at equilibrium? If not, which direction will the reaction go? Define reaction quotient Q = same expression as K c or K p ([products]/[reactants]). Evaluate Q and compare it to K c or K p. (a) If Q = K c or K p, then the reaction is at equilibrium (no shift). (b) (c) If Q < K c or K p, then the reaction will shift towards more products, less reactants. If Q > K c or K p, then the reaction will shift towards more reactants, less products. 5 Ex. Given: 2NO(g) + O 2 (g) º 2NO 2(g) K c = at 500 K A 5.00 L vessel is filled with mol NO, 1.0 mol O 2, and 0.80 mol NO 2. Is the reaction at equilibrium? If not, which direction will the reaction go? Q = [NO 2] /{[NO] [O 2]} [NO] = mol/5.00 L = M; [O 2] = 1.0 mol/5.00 L = 0.20 M; [NO 2] = 0.80 mol/5.00 L = 0.16 M 2 5 Q = (0.16) /{(0.012) (0.20)} = < (not at equilibrium) The reaction will shift towards more NO 2, less NO and O Calculating equilibrium concentrations. A problem-solving bonanza. Variations of these types of problems will be important for the next 3 weeks of lecture. PAY ATTENTION! The problems are organized by types. Type 1: A reaction is given, along with the equilibrium conc. of all components. Calculate K or c K. Easy. Write the correct equilibrium expression, plug in the equilibrium conc., and solve. p Ex. M&F, Given: PCl 5 (g) º PCl 3 (g) + Cl 2(g)!3!2!2 An equilibrium mixture contains M PCl 5, M PCl 3, and M Cl 2. What is K c?!2!2!3!2 K = {[PCl ][Cl ]}/[PCl ] = {( )( )}/( ) = c

4 Variation: K c or K p and all but one equilibrium conc. are given; calculate the unknown conc. No example given here. Type 2: Given a reaction, an initial set of concentrations, and one equilibrium concentration, calculate the other equilibrium concentrations. More difficult. Set up a table of initial, change and final conc. for each component. Use the stoichiometry of the reaction to calculate the changes in all concentrations. Then calculate the final concentrations and K. Ex. Nakon, 17.13: Given H 2 (g) + I 2(g) º 2HI(g) A mixture of 2.00 mol of H 2 and 2.00 M I 2 were sealed in a 1.00 L flask. At equilibrium, the flask contained 0.58 mol HI. What is K c? Table H 2 + I 2 º 2HI Init. conc (units M) (note: 2.00 mol/1.00 L) )conc. (units M) Final conc (units M) )conc. for HI is clearly M (0.58-0). )conc. for H =!(0.58 M HI)(1 mol H /2 mol HI) =!0.29 mol (the! sign is required because if [HI] increases, then [H ] must decrease). 2 Similarly, )conc. for I =!(0.58 mol HI)(1 mol I /2 mol HI) =!0.29 mol. The table becomes: Table H + I º 2HI Init. conc (units M) )conc.!0.29! (units M) Final conc (units M) Then: K c = [HI] /{[H 2][I 2]} = (0.58) /{(1.71)(1.71)) = 0.12 Type 2': The problem states that a certain percentage of a reactant or product disappears. Calculate equilibrium conc. and K. Ex. Nakon, Given: 2NO 2 (g) º 2NO(g) + O 2(g) 1.00 mol of NO 2 placed in a 1.00 L container is 15% decomposed at equilibrium. Calculate K c. Table 2NO 2 º 2NO + O2 Init. conc (units M) (note: 1.00 mol/1.00 L) )conc.! (units M) Final conc (units M) -23-

5 15% decomposed means that )conc. for NO 2 is!0.15(1.0 M) = 0.15 M. The final conc. for NO 2 = 1.00! 0.15 = 0.85 M. )conc. for NO = +(0.15 M NO 2 )(2 mol NO/2 mol NO 2) = M NO (note + because [NO] must increase). Similarly, )conc. for O 2 = +(0.15 M NO 2)(1 mol O 2 /2 mol NO 2 ) = M O 2. The final conc. of NO and O 2 are 0.15 M and M.!3 K c = {[NO] [O 2]}/[NO 2] = {(0.15) (0.075)}/(0.85) = Type 3: Given the reaction, the value of K c or K p, and a set of initial concentrations, calculate the equilibrium concentrations. Most difficult. Set up a table of initial, change and final conc. for each component. Let x be the )conc. of the component with the smallest coefficient. Use the stoichiometry of the reaction to calculate the changes in all concentrations in terms of x. Then calculate the final concentrations of all components in terms of x. Write the equilibrium expression, put in the variables with x, and solve for x. Note: If the result is a quadratic equation, choose the solution that makes chemical sense. Finally, calculate the required equilibrium conc. Ex. Nakon, 17.14: Given H 2 (g) + I 2 (g) º 2HI(g) K c = moles of HI were placed in a 1.00 L vessel. What are all the conc. at equilibrium? Table H 2 + I 2 º 2HI Init. conc (units M) (note: 2.00 mol/1.00 L) )conc. +x +x -2x (units M) Final conc. +x +x 2.00! 2x (units M) Let +x = )conc. of H 2 ; the +sign is chosen because [H 2] must increase. Then, from the stoichiometric coefficients, the )conc. are +x for I 2 and!2x for HI. The final conc. are shown above. K c = 0.12 = [HI] /{[H 2][I 2]} = (2.00! 2x) /{(x)(x)} The right side is a perfect square. Take the square root of both sides; choose the positive root for x because the conc. of H 2 and I 2 must be positive. ½ (0.12) = 0.35 = (2.00! 2x)/x; 0.35x = 2.00! 2x; 2.35x = 2.00; x = The final conc. are: [H 2 ] = [I 2] = x = M; [HI] = 2.00! 2x = 2.00! 2(0.851) = M. Ex. M&F, Given: PCl (g) º PCl (g) + Cl (g) K = c!2 Calculate the final conc. of all gases if only PCl is present initially at M. 5 Let +x = )conc. of PCl (+ sign because [PCl ] must increase)

6 Table PCl 5 º PCl 3 + Cl2 Init. conc (units M) )conc.!x +x +x (units M) Final conc ! x +x +x (units M) K = = {[PCl ][Cl ]}/[PCl ] = {(x)(x)}/(0.160! x) c Oh, bother! This leads to a quadratic equation: x = (0.058)(0.160! x) 2 x x! = 0 To solve, use the quadratic formula. Choose the positive root because [PCl 3 ] and [Cl 2] must be positive. 2 ½ 2 ½ x = {!b ± (b! 4ac) }/(2a) = (! ((.058)! 4(1)(!0.0093)) )/2 = [PCl ] = 0.160! = M; [PCl ] = [Cl ] = M Hooboy! It is unlikely that you would see very many problems this complicated on quizzes or hour exams. Later, we will explore shortcuts which avoid the need for a full quadratic formula. -25-