CH 15 CHEM 2 NOTES YNGARD
CH 15 CHEM 2 NOTES YNGARD CHEM 1040
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Date Created: 09/14/16
Chapter 15 Chemical Equilibrium nd ChJulia Burdge & Jason Overby The McGraw-Hill Companies, Inc. 15.1 The Concept of Equilibrium 15.2 The Equilibrium Constant Calculating Equilibrium Constants Magnitude of the Equilibrium Constant 15.3 Equilibrium Expressions Heterogeneous Equilibria Manipulating Equilibrium Expressions Gaseous Equilibria 15.4 Chemical Equilibrium and Free Energy Using and K to Predict the Direction of Reaction Relationship Between ΔG and ΔG° Relationship Between ΔG° and K 15.5 Calculating Equilibrium Concentrations 15.6 Le Châtelier’s Principle: Factors that Affect Chemical Equilibrium Addition or Removal of a Substance Changes in Volume and Pressure Changes in Temperature Dynamic Equilibrium N 2 4 (g)⇌ 2 NO 2 (g) Colorless Brown ⇌ Starting Starting with with NO2 N O 2 4 Fig. 15.1 Fig. 15.2 3 The Reaction quotient and Equilibrium Constant products reactants 4 Q c Problem 1a: Worked example 15.1 Write the equilibrium expression for the following reactions. 2 a. N O2(g4 ⇌ 2NO (g) 2 Q c [NO ] 2[N O ]2 4 2+ - 2- b. Cd (aq) + 4Br (aq) ⇌ CdBr (aq) 4 2- 2+ - 4 Q c [CdBr ]/(4Cd ][Br ] ) c. H C O (aq) ⇌ 2H (aq) + C O (aq) 2- 2 2 4 2 4 Q = ([H ] [C O ])/([H C O ] c 2 4 2 2 4 1b. Write the equation for the reaction with the following equilibrium expression: + - Q c [HI]/([H ][I ]) + - Answer: H (aq) + I (aq) ⇌ HI(aq) 5 Q acd K c N 2 4g) ⇌ 2NO (g2 6 Q acd K c N 2 4g) ⇌ 2NO (g2 7 Magnitude of K c K large: “lie to the right” or “favor products” c Ag (aq) + 2NH (aq) ⇌ Ag(NH ) (aq) + K = 1.5 x 10 (at 25°C) 3 3 2 c K cmall: “lie to the left” or “favor reactants” –25 N2(g) + O 2g) ⇌ 2NO(g) K c 4.3 x 10 (at 25°C) K in between : c the equilibrium mixture will contain comparable amounts of both reactants and products. 8 Heterogeneous Equilibria Heterogeneous equilibria: species in a reversible reaction are in different phases The concentrations of solids and liquids do not appear in the equilibrium expression. • PbCl (s2 ⇌ Pb (aq) + 2Cl (aq) - 2+ - 2 K c [Pb ] [Cl ] - + • NH (a3) + H O(l2 ⇌ OH (aq) + NH (aq) 4 [OH ][NH ] + K = 4 c [NH ]3 9 Manipulating Equilibrium Expressions 1.If the equation is reversed, invert the equilibrium constant. A(g) + B(g)⇌ 2C(g) K c1 2C(g) ⇌ A(g) + B(g) K = 1/K c2 c1 2. If the coefficients in the reaction are multiplied by a factor, c1is raised to a power equal to that same factor. A(g) + B(g)⇌ 2C(g) K c1 2 2A(g) +2 B(g)⇌ 4C(g) K c2(K c1 10 Manipulating the Equilibrium Constant 3. The equilibrium constant for a net reaction made up of two or more reactions is the product of the equilibrium constants for the individual reactions. A (g) + B(g) ⇌ C(g) + D(g) Kc1 C(g) + F(g) ⇌ G(g) + A(g) K c2 B(g) + F(g) ⇌ D(g) + G(g) K = K K c3 c1 c2 11 Problem 2. The following reactions have the indicated equilibrium constants at 700 K, kc1 54.0 for H 2g) + I (2) ⇌ 2HI(g) -4 kc2 1.04 x 10 for N 2g) + 3H (g2 ⇌ 2NH (g) 3 What is the k fcr -2 a. 2HI(g) ⇌ H (g) +2I (g) 2 kc= 1.85x10 b. NH (g) ⇌ ½ N (g) + 3/2 H (g) k = 98.1 3 2 2 c c. 2NH (g) + 3I (g) ⇌ 6HI(g) + N (g) k = 1.51x10 9 3 2 2 c 12 Gaseous Equilibria aA + bB ⇌ cC + dD c d All gaseous: K = (P C (P D p (P ) (P ) b A B P = partial pressure Relationship Kcand K :p Kp= K (cT) n n = (moles of gaseous product) - (moles of gaseous reactant) R = 0.08206 (L•atm/K•mol) T = Temperature (Kelvin) K = K only when n = 0 p c 13 Problem 3: For the equilibrium 2SO (g) 3 ⇌ 2SO (2)+ O (g)2 -3 K cs 4.08 x 10 at 1000 K. R = 0.0821 L-atm/mol-K a. Does the equilibrium favor the formation of SO or SO 3, 2 and O a2 this temperature? b. Write the equilibrium expression (K ) c c. Calculate the value for K . p Answer: a. Favors SO sin3e K c ismall 2 2 b. K c ([SO ] 2O ])2[SO ] 3 c. 0.335 14 Chemical Equilibrium and Free Energy In general: aA + bB ⇌ cC + dD c d [C] [D] Kc= [A] [B] b A reaction with K very large (>> 1) has a negative ΔG° A reaction with K very small (<< 1) has a positive ΔG° 15 Chemical Equilibrium and Q Consider : H (2) + I (2) ⇌ 2HI(g) Kc= 54.3 at 430°C If we start an experiment with a mixture of 0.243 mole of H , 2 0.146 mole of I 2 and 1.98 mole of HI in a 1.00 L container at 430°C, would more HI be formed or consumed for the reaction to reach an equilibrium? Answer: HI consumed In general: aA + bB ⇌ cC + dD • If Q > K, the reaction proceeds from right to left until equilibrium is reached. • If Q < K, the reaction proceeds from left to right until equilibrium is reached. • If Q = K, the reaction is at equilibrium. 16 Relationship ∆G° and ∆G Reaction not at equilibrium: ΔG = ΔG° + RT lnQ R = gas constant (8.314 J/K•mol) T = absolute temperature (K) Q = reaction quotient Problem 4: Consider: H (2) + I 2s) ⇌ 2HI(g) Given, ΔG° at 25°C = 2.60 kJ/mol; P H = 4.0 atm; P HI1.0 atm, write Q , calculate ΔG, and determine if the reaction will p be spontaneously as written in the forward direction. Q = (P ) /(P ), ΔG = -0.83 kJ/mol. p HI H2 17 Relationship ∆G° and K Reaction at equilibrium: ΔG = 0; and Q =K 0 = ΔG° + RT lnK or ΔG° = -RT lnK (eq. 15.6) 18 Relationship ∆G° and K Problem 5: Practice problem 15.9B The K ffr Ag(NH ) (a3 2at 25°C is 1.5x10 . 7 Ag (aq) + 2NH (aq) ⇌ Ag(NH ) (aq) + 3 3 2 Using this and data from appendix 2, calculate the value of + ΔG° for Ag(NH ) (a3 2in kJ/mol. ΔG° fg = 77.1 kJ/mol ΔG° fH = -36.5 kJ/mol Answer = -16.8 kJ/mol 19 Calculating Equilibrium Concentrations Problem 6: Worked Example 15.12 K cor the reaction of hydrogen and iodine to produce hydrogen iodide, H (g2 + I (g2 ⇌ 2HI(g) is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with [I ] = 0.00414M, [HI] = 0.0424M, [H ] = 0.00623M? 2 2 Answer: [H ] = 0.00676M, [I ] = 0.00467M, [HI] = 0.0414M 2 2 20 Problem 7: Consider the equilibrium: -3 o 2 IBr(g) ⇌ I 2g) + Br (g)2 K = 8p5 x 10 at 150 C. If 0.025 atm of IBr is placed in a 1.0-L container, what is the partial pressure of all substances after equilibrium is reached? Answer: P = P =1.9 x 10 atm I2 Br2 PIBr= 2.1 x 10 -2 atm 21 Le Chatelier’s Principle Principle: when a stress is applied to a system at equilibrium, the system, will respond by shifting in the direction that minimizes the effect of the stress. Stress: addition/removal of a reactant or product pressure/volume change temperature change 22 Addition/Removal Effect N 2g) + 3H (2) ⇌ 2NH (3) 2 [NH 3 K = [N ][H ] 2 2 Removal of NH : 3 shifts right Addition NH : 3 shifts left Addition of N :2 shifts right Removal H : shifts left 2 23 Volume/Pressure Change Effect When the volume is decreased, the equilibrium is driven toward the side with the smallest number of moles of gas. N2(g) + 3H 2g) ⇌ 2NH (3) P with V : shifts right P with V : shifts left Addition inert gas that is neither reactant nor product H 2g) + I2(g) ⇌ 2HI(g) no effect 24 Temperature change 2- 2+ - CoCl 4aq) + 6H O2l) ⇌ Co(H O2 6 (aq) + 4Cl (aq) + heat Blue pink • Endothermic: reactants + heat ⇌ products • Exothermic: reactants ⇌ products + heat • T : shifts left • T : shifts right 25 Problem 8: Consider: 4 NH 3g) + 5 O (2) ⇌ 4NO(g) + 6H O(g), 2H = -904 kJ. How does each of the following changes affect the yield of NO at equilibruim? Answer with increase, decrease or no change. a. addition of NH 3 increase yield NO b. increase [H 2] decrease yield NO c. removal of O 2 decrease yield NO d. decrease volume container decrease yield NO e. add a catalyst no change f. increase temperature decrease yield NO 26
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