The Concept of Equilibrium

Transcription

1 The Concept of Equilibrium

2 Reversible reactions As the concentrations of the reactants decrease the rate of reaction in the forward direction decreases. As the concentrations of the products increase the rate of reaction in the reverse direction increases.

3 Chemical Equilibrium reached when rates of the forward and reverse reactions are equal and the concentrations of the reactants and products no longer change with time Chemical equilibrium is dynamic: chemical reactions take place,but concentrations of reactants and products remain unchanged

4 A Consider a simple case: k k ʹ B Rate of forward reaction = k [A] Rate of backward reaction = k ʹ [B] At equilibrium: k [A] = k ʹ [B] (assume single unimolecular elementary step for forward and reverse processes) [B] [A] = k = K k ʹ equilibrium constant

5 The Law of Mass Action a A + b B c C + d D equilibrium constant k = [C] c [D] d [A] a [B] b k = products reactants

6 The NO 2 - N 2 O 4 system at 25 C Init: M M N 2 O 4 (g) 2NO 2 (g) equil: M M [NO 2 ] 2 [N 2 O 4 ] = (0.0547) = 4.65 x 10-3

7 The NO 2 - N 2 O 4 system at 25 C Init: M M N 2 O 4 (g) 2NO 2 (g) equil: M M [NO 2 ] 2 [N 2 O 4 ] = ( M) M = 4.60 x 10-3

8 The NO 2 - N 2 O 4 system at 25 C Init: M M N 2 O 4 (g) 2NO 2 (g) equil: M M [NO 2 ] 2 [N 2 O 4 ] = 4.63 x 10-3

9 Change in concentrations Initially only N 2 O 4 is present 2N 2 O 4 (g) reactant N 2 O 4 (g) 2NO 2 (g) Conc NO 2 (g) Time product Equilibrium concentrations

10 Change in concentrations Initially only NO 2 is present NO 2 (g) product N 2 O 4 (g) 2NO 2 (g) Conc Equilibrium concentrations 2N 2 O 4 (g) reactant Time

11 Equilibrium Can Be Reached from Either Direction It doesn t matter whether we start with NO 2 or whether we start with N 2 O 4, we will have the same proportions of the two substances at equilibrium.

12 The magnitude of the equilibrium constant depends on temperature depends on reaction 2300 C 2O 3 (g) 3O 2 (g) K = 2.5 x C Cl 2 (g) 2Cl (g) K = 1.4 x CO(g) + H 2 O (g) 830 C H 2 (g) + CO 2 (g) K = 5.1

13 Ways of Expressing Equilibrium Constants

14 Homogeneous equilibria Heterogeneous equilibria

15 Homogeneous Equilibria all reacting species are in the same phase gas phase equilibrium constant can be expressed in terms of pressure or concentration solution phase

16 Example: homogeneous gas-phase N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] Where K c is the equilibrium constant when concentrations are expressed in moles/liter

17 Alternatively: N 2 O 4 (g) 2NO 2 (g) K p = P 2 NO 2 P N 2 O 4 Where K p is the equilibrium constant when concentrations are expressed in pressure units

18 Practice Exercise The equilibrium concentrations for the reaction CO(g) + Cl 2 (g) COCl 2 (g) Are [CO] = 1.2 x 10-2 M, [Cl 2 ] = M and [COCl 2 ] = 0.14 M. Calculate the equilibrium constant (K c ). K c = [COCl 2 ] [CO] [Cl 2 ] 0.14 = (1.2 x 10-2 ) (0.054) = 2.2 x 102

19 Practice Exercise 2NO 2 (g) 2NO (g) + O 2 (g) atm atm? The equilibrium constant K p for the reaction shown as 158 at 1000 K. calculate the partial pressure of O = ( ) 2 P ( O2) ( ) 2 ( ) 2 (158) ( ) 2 = P ( O2) P (O2) = 347 atm

20 Relationship between K p and K c Gas-phase concentrations can also be expressed in terms of partial pressures PV = nrt Therefore, = P i.e.: n V R T P = (molar concentration) R T K p = K c ( RT ) Δ n

21 Δn = the sum of the coefficients of the gaseous products - the sum of the coefficients of the gaseous reactants

22 Homogeneous Equilibria all reacting species are in the same phase gas phase equilibrium constant can be expressed in terms of pressure or concentration solution phase concentration term for the solvent does not appear in the expression for the equilibrium constant

23 Example: homogeneous aqueous-phase reaction O CH 3 COH (aq) + H 2 O( l ) O CH 3 CO - (aq) + H 3 O + ( aq)

24 Example: homogeneous aqueous-phase reaction HOAc (aq) H 2 O( l ) + Kʹ = [ - OAc ] [H 3 O + ] [HOAc] [H 2 O] - OAc (aq) H 3 O + ( aq) +

25 Example: homogeneous aqueous-phase reaction HOAc (aq) H 2 O( l ) + K ʹ [H 2 O] = [ - OAc ] [H 3 O + ] [HOAc] - OAc (aq) H 3 O + ( aq) +

26 Example: homogeneous aqueous-phase reaction HOAc (aq) H 2 O( l ) + K c = [ - OAc ] [H 3 O + ] [HOAc] - OAc (aq) H 3 O + ( aq) +

27 Heterogeneous Equilibria all reacting species are not in the same phase concentration term for solid or liquid does not appear in the expression for the equilibrium constant

28 Example CaCO CaO 3 (s) (s) + CO 2 (g) K c = [CO 2 ]

29 CaCO 3 (s) CO 2 (g) + CaO(s) Increasing the surface area of the solid increases the rate forward and reverse reactions equally. The amount of CO 2 above the solid remains the same. 29 Equilibrium

30 Practice Exercise Ni (s) + 4CO (g) Ni(CO) 4 (g) What is the expression for K c and K p for the reaction shown [Ni(CO) 4 ] K c = K p = [CO] 4 P (Ni(CO)4) P 4 (CO)

31 Practice Exercise What is the value for K c and K p for the equilibrium shown at 295 K? K p NH 4 HS (s) = P NH3 K p = (0.265) 2 = K p P H2S NH 3 (g) + H 2 S (g) 0.265atm 0.265atm K p = K c ( RT ) Δ n K c = K p ( RT ) 2

32 Practice Exercise What is the value for K c and K p for the equilibrium shown at 295 K? NH 4 HS (s) NH 3 (g) + H 2 S (g) 0.265atm 0.265atm K p = P NH3 K p = (0.265) 2 = K p P H2S K p = K c ( RT ) Δ n K c = (.0821 Latm/mol K(295K) ) 2

33 Practice Exercise What is the value for K c and K p for the equilibrium shown at 295 K? K p NH 4 HS (s) = P NH3 K p = (0.265) 2 = K p P H2S NH 3 (g) + H 2 S (g) 0.265atm 0.265atm K p = K c ( RT ) Δ n K c =

34 Multiple Equilibria If a reaction can be expressed as the sum of two or more individual reactions, the equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions.

35 Multiple Equilibria If a reaction can be expressed as the sum of two or more individual reactions, the equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions.

36 Practice Exercise Given: H 2 S (aq) H + (aq) + HS - (aq) Kʹ c = 9.5 x 10-8 HS - (aq) H + (aq) + S 2- (aq) K ʹ ʹ c = 1.0 x Calculate the equilibrium constant for: H 2 S (aq) 2H + (aq) + S 2- (aq)

37 Practice Exercise H 2 S(aq) 2H + (aq) + S 2- (aq) K = [H+ ] 2 [S 2- ] [H 2 S] [H + ] [HS - ] [H + ] [S 2- ] K = = [H [HS - Kʹ c K ʹ ʹ c 2 S] ] HS - (aq) H + (aq) + S 2- (aq) H H + 2 S (aq) (aq) + HS - (aq)

38 Practice Exercise H 2 S (aq) H + (aq) + HS - (aq) HS - (aq) H + Kʹ c = 9.5 x 10-8 (aq) + S 2- (aq) K ʹ ʹ c = 1.0 x H 2 S (aq) 2H + (aq) + S 2- (aq) K ʹ ʹ ckʹ c = (9.5 x 10-8 )(1.0 x ) = 9.5 x 10-27

39 The Form of K and the Equilibrium Equation When the equation for a reversible reaction is written in the opposite direction,the equilibrium constant becomes the reciprocal of the original equilibrium constant. The value of K also depends on how the equilibrium constant is balanced.

40 Practice Exercise N 2 (g) + 3H 2 (g) K c = 1.2 at 375 C 2NH 3 (g) K c = [N 2 ] [NH 3 ] 2 [H 2 ] 3 = 1.2 What is the expression for K c for: 2NH 3 (g) N 2 (g) + 3H 2 (g) [N 2 ] [NH 3 ] 2 [H 2 ] 3 = [NH 3 ] 2 ( ) [N 2 ] [H 2 ] 3-1 = 0.83

41 Practice Exercise N 2 (g) + 3H 2 (g) K c = 1.2 at 375 C What is the expression for K c for: 1 2 [NH 3 ] = [N [H 2 ] 3/2 2 ] 1/2 [NH 3 ] 2 K c = [N 2 ] ( ) [N 2 ] 2NH 3 (g) [H 2 ] 3 1/2 [NH 3 ] 2 = [H 2 ] 3 N 2 (g) H 2 (g) NH 3 (g) = 1.2

42 Practice Exercise N 2 (g) + 3H 2 (g) K c = 1.2 at 375 C 2NH 3 (g) 4 moles of gases on the left; 2moles of gases on the right K p = K c ( RT ) 2 What is K p for the reaction K p = 1.2[ ( Latm/molK )(648 K)] 2 K p =

43 Summary of Guidelines for Writing Equilibrium Constant Expressions The concentrations of the reacting species in solution are expressed in mol/l. In the gas phase, concentrations are expressed in mol/l or in atm. K p = K c ( RT ) Δ n

44 Summary of Guidelines for Writing Equilibrium Constant Expressions The concentrations of pure solids, pure liquids, and solvents do not appear in the equilibrium constants expressions. Equilibrium constants do not have units. The balanced equation must be shown when the value for an equilibrium constant is given.

45 Summary of Guidelines for Writing Equilibrium Constant Expressions If a reaction can be expressed as the sum of two or more individual reactions, the equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions.

46 Summary of Guidelines for Writing Equilibrium Constant Expressions When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant become the reciprocal of the original equilibrium constant.

47 47

48 3.6 x 10^8 at 25 C N 2 (g) + 3H 2 (g) 2NH 3 (g) [NH 3 ] 2 [N 2 ] [H 2 ] 3