Chemical Equilibrium. Chapter 8

Transcription

1 Chemical Equilibrium Chapter 8

2 Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H 2 O (l) H 2 O (g) Chemical equilibrium N 2 O 4 (g) 2NO 2 (g) 14.1

3 N 2 O 4 (g) 2NO 2 (g) equilibrium equilibrium equilibrium Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O

4 constant 14.1

5 N 2 O 4 (g) K = [NO 2] 2 [N 2 O 4 ] 2NO 2 (g) = 4.63 x 10-3 aa + bb K = [C]c [D] d [A] a [B] b cc + dd Law of Mass Action 14.1

6 K = [C]c [D] d [A] a [B] b aa + bb cc + dd Equilibrium Will K >> 1 K << 1 Lie to the right Lie to the left Favor products Favor reactants 14.1

7 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2] 2 [N 2 O 4 ] In most cases K p = P 2 NO 2 P N2 O 4 K c K p aa (g) + bb (g) cc (g) + dd (g) K p = K c (RT) Dn n = moles of gaseous products moles of gaseous reactants = (c + d) (a + b) 14.2

8 Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3COO - ][H 3 O + ] [CH 3 COOH] = K c [H 2 O] General practice not to include units for the equilibrium constant. 14.2

9 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) K c = [COCl 2] [CO][Cl 2 ] = x = 220 K p = K c (RT) Dn Dn = 1 2 = -1 R = T = = 347 K K p = 220 x ( x 347) -1 =

10 The equilibrium constant K p for the reaction 2NO 2 (g) 2NO (g) + O 2 (g) is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = atm and P NO = atm? 2 K p = 2 P NO P O2 2 P NO 2 2 P P NO O2 = K p 2 2 P NO P O2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm 14.2

11 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x [CaCO 3 ] [CaO] K p = P CO2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2

12 CaCO 3 (s) CaO (s) + CO 2 (g) P CO2 = K p P CO2 does not depend on the amount of CaCO 3 or CaO 14.2

13 Consider the following equilibrium at 295 K: NH 4 HS (s) NH 3 (g) + H 2 S (g) The partial pressure of each gas is atm. Calculate K p and K c for the reaction? K p = P NH3 P H 2 S = x = K p = K c (RT) Dn K c = K p (RT) - Dn Dn = 2 0 = 2 T = 295 K K c = x ( x 295) -2 = 1.20 x

14 A + B C + D A + B C + D E + F E + F K c K c K c K c = [C][D] [A][B] K c = [E][F] [A][B] K c = [E][F] [C][D] K c = K c x K c If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

15 N 2 O 4 (g) 2NO 2 (g) 2NO 2 (g) N 2 O 4 (g) K = [NO 2] 2 [N 2 O 4 ] = 4.63 x 10-3 K = [N 2O 4 ] = 1 [NO 2 ] 2 K = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 14.2

16 Writing Equilibrium Constant Expressions 1. The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. 2. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. 3. The equilibrium constant is a dimensionless quantity. 4. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. 5. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

17 Chemical Kinetics and Chemical Equilibrium k f A + 2B AB 2 k r rate f = k f [A][B] 2 rate r = k r [AB 2 ] Equilibrium rate f = rate r k f [A][B] 2 = k r [AB 2 ] k f [AB 2 ] = K k c = r [A][B]

18 The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

19 Calculating Equilibrium Concentrations 1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3. Having solved for x, calculate the equilibrium concentrations of all species. 14.4

20 At C the equilibrium constant (K c ) for the reaction Br 2 (g) 2Br (g) Is 1.1 x If the initial concentrations are [Br 2 ] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) Br 2 (g) 2Br (g) x +2x x x [Br] K 2 c = K [Br 2 ] c = ( x) x = 1.1 x 10-3 Solve for x 14.4

21 ( x)2 K c = = 1.1 x x -3 4x x = x 4x x = 0 Initial (M) Change (M) Equilibrium (M) ax 2 + bx + c =0 Br 2 (g) x = 2Br (g) x +2x x x -b ± b 2 4ac 2a x = x = At equilibrium, [Br] = x = M or M At equilibrium, [Br 2 ] = x = M 14.4

22 Le Châtelier s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N 2 (g) + 3H 2 (g) Equilibrium shifts left to offset stress 2NH 3 (g) Add NH

23 Le Châtelier s Principle Changes in Concentration continued Remove Add Remove Add aa + bb cc + dd Change Shifts the Equilibrium Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) left right right left 14.5

24 Le Châtelier s Principle Changes in Volume and Pressure A (g) + B (g) C (g) Change Increase pressure Decrease pressure Increase volume Decrease volume Shifts the Equilibrium Side with fewest moles of gas Side with most moles of gas Side with most moles of gas Side with fewest moles of gas 14.5

25 Changes in Temperature Le Châtelier s Principle Change Increase temperature Decrease temperature Exothermic Rx K decreases K increases Endothermic Rx K increases K decreases colder hotter 14.5

26 Le Châtelier s Principle Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner uncatalyzed catalyzed Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. 14.5

27 Chemistry In Action Life at High Altitudes and Hemoglobin Production Hb (aq) + O 2 (aq) HbO 2 (aq) K c = [HbO 2] [Hb][O 2 ]

28 Chemistry In Action: The Haber Process N 2 (g) + 3H 2 (g) 2NH 3 (g) H 0 = kj/mol

29 Le Châtelier s Principle Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no 14.5

30 Thermodynamic Description of The Equilibrium State Thermodynamics views a chemical reaction as a process in which atoms flow from reactants to products. If the reaction is spontaneous and is carried out at constant T and P, thermodynamics requires that ΔG<0 for the process. Consequently, G always decreases during a spontaneous chemical reaction. When a chemical reaction has reached equilibrium, ΔG=0; that is, there is no further tendency for the reaction to occur in either the forward or the reverse direction. We will use the condition ΔG=0 in the following three subsections to develop the mass action law and the thermodynamic equilibrium constant for gaseous, solution, and heterogeneous reactions.

31 Reactions Among Ideal Gases Dependence of Gibbs free energy of an ideal Gas on pressure