Chapter 13: Chemical Equilibrium

Transcription

1 Chapter 13: Chemical Equilibrium 13.1 The Equilibrium Condition Equilibrium: a state in which no observable changes occur H 2 O (l) H 2 O (g) Physical equilibrium: no chemical change. N 2(g) + 3H 2(g) 2NH3(g) the reaction rate declines to a value of zero, at which time there is still N 2 and H 2 remaining The Equilibrium Constant N 2 O 4(l) 2NO 2(g) Initial Equilibrium Ratio of s at equilibrium [NO2] [N2O4] [NO 2 ] [N 2 O 4 ] [NO 2 ]/[N 2 O 4 ] [NO 2 ] 2 /[N 2 O 4 ] equilibrium constant K = [NO 2] 2 [N 2 O 4 ] = NO 2(g) N 2 O 4(l) Initial Equilibrium Ratio of s at equilibrium [NO2] [N2O4] [NO 2 ] [N 2 O 4 ] [N 2 O 4 ]/[NO 2 ] [N 2 O 4 ]/[NO 2 ]

2 equilibrium constant K = [N 2O 4 ] [NO 2 ] 2 = 45.9 In general, for aa + bb cc + dd, the equilibrium constant is given by K = [C]c [D] d [A] a [B] b This is called the law of mass action If K is large, then there will be much more C and D, and little A or B 2O 3(g) 3O2(g) K = [O 2] 3 2 = [O 3 ] If K is very small, then there will be little product Cl 2(g) 2Cl (g) K = [Cl]2 [Cl 2 ] = If K is near 1, there will be significant quantities of both reactants and products CO (g) + H 2 O (g) H 2(g) + CO 2(g) K = [H 2][CO 2 ] [CO][H 2 O] = 5.10 Eample: For the reaction 2H 2(g) + S 2(g) 2H 2 S (g) at equilibrium, there are 2.50 mol H 2, mol S 2, and 8.70 mol H 2 S in a 12.0 L flask. Calculate K. [H 2 ] = 2.50 mol/12.0 L = M [S 2 ] = mol/12.0 L = M [H 2 S] = 8.70 mol/12.0 L = M [H 2 S] 2 K = [H 2 ] 2 [S 2 ] (0.7250) 2 = (0.2083) 2 ( ) = Equilibrium Epressions Involving Pressures K c = [NO 2] 2 [N 2 O 4 ] where subscript c indicates molar. Since P (P=(n/V)RT), we can also write K p = P2 NO2 P N 2O4 where subscript p indicates pressure. ([NO 2 ]RT) 2 2 K p = ([NO]RT) 2 ([O 2 ]RT) = [NO 2 ] [NO] 2 [O 2 ] (RT) 2 (RT) 2 (RT) = K (RT) moles prod c (RT) moles react = K c(rt) n 13-2

3 K p = K c (RT) n, where n = moles gaseous product - moles gaseous reactant When n = 0, K p = K c eamples: N 2(g) + 3H 2(g) 2NH 3(g) 2H2(g) + O2(g) 2H 2 O (g) NH 3 (aq) + H 2 O (l) 2NH 4 + (aq) + OH - (aq) K c for the reaction 2NO (g) + O 2(g) 2NO 2(g) is Calculate K p Relation of K eq to the Form of the Chemical Equation 2H 2(g) + O 2(g) 2H 2 O (g) [H 2 O] 2 K 1 = [H 2 ] 2 [O 2 ] H 2(g) + 1/2O 2(g) H 2 O (g) [H 2 O] K 2 = [H 2 ][O 2 ] 1/2 = K 1 2H 2 O (g) 2H 2(g) + O 2(g) K 3 = [H 2] 2 [O 2 ] [H 2 O] 2 = 1 K 1 The form of the equilibrium epression depends on the balanced equation Note that for any reaction, K rev = 1/K for, and that the value of K depends on how the equation is balanced: N 2 O 4(l) 2NO 2(g) K for = [NO 2] 2 [N 2 O 4 ] = at equilibrium, [NO 2 ] = , [N 2 O 4 ] = NO 2(g) N 2 O 4(l) K rev = [N 2O 4 ] [NO 2 ] 2 = = 216 = (0.0204) 2 NO 2(g) 1/2N 2 O 4(l) K rev = [N 2O 4 ] 1/2 [NO 2 ] = (0.0898)1/ = 14.7 = K rev Multiple Equilibria If a reaction can be epressed as the sum of two or more reactions, the equilibrium constrant for the overall reaction is equal to the product of the equilibrium constants of the individual steps. A + B C + D K 1 = [C][D] [A][B] C + D E + F K 2 = [E][F] [C][D] A + B E + F K C = [E][F] [A][B] = [C][D] [A][B] [E][F] [C][D] 13-3

4 2N 2 + O [N 2 O] 2 2 2N 2 O K 1 = [N 2 ] 2 [O 2 ] 2N 2 O + 3O [NO 2 ] 4 2 4NO 2 K 2 = [N 2 O] 2 [O 2 ] 3 2N 2 + 4O [NO 2 ] NO 2 K c = [N 2 ] 2 [O 2 ] 4 = [N 2 O] [N 2 ] 2 [O 2 ] [NO 2 ] [N 2 O] 2 [O 2 ] Heterogeneous Equilibria When one or more of the reactants or products of a reaction is a solid, [solid] is a constant, and is not included in the equilibrium epression CaCO 3(s) CaO (s) + CO 2(g) K = [CaO][CO 2] [CaCO 3 ] K c = K[CaCO 3] [CaO] * [CO 2 ] or, K p = P CO2 = K c RT 2NaHCO 3(s) Na 2 CO 3(s) + H 2 O (g) + CO 2(g) 2Hg (l) + Cl 2(g) Hg 2 Cl 2(s) NH 3(g) + HCl (g) NH 4 Cl (s) 13.5 Applications of the Equilibrium Constant Reaction Quotient Q c : value obtained by substituting initial s into the equilibrium epression Q c > K c : ration of products to reactants is too large, reaction will proceed in reverse direction to reach equilibrium Q c = K c : the system is at equilibrium Q c < K c : ration of products to reactants is too small, reaction will proceed in forward direction to reach equilibrium Eample: CH 4(g) + H 2 O (g) CO (g) + 3H 2(g) K c = 5.67 Initial conc Q c = (0.100)(0.200)3 (0.500)(0.800) = < K c so reaction goes forward Initial conc Q c = (0.500)(0.800)3 (0.100)(0.200) = 12.8 > K c so reaction goes backward 13-4

5 Calculating Equilibrium Pressures and Concentrations 13.6 Solving Equilibrium Problems If the initial s of CO and H 2 O are each 0.100, what are the equilibrium s? CO + H 2 O CO 2 + H 2 K c = 4.06 initial change eq. con Kc = 4.06 = [CO 2][H 2 ] [CO][H 2 O] = ()() (0.100-)(0.100-) = = ; = ; = [CO 2 ] = [H 2 ] = ; [CO] = [H 2 O] = = PCl 5 PCl 3 + Cl 2 i C -y +y +y e y y y K c = 0.6 = [PCl 3][Cl 2 ] [PCl 5 ] 0.6 = y y y y = 0 = (y)(y) ( y) y = -0.6 ± (1)(-0.06) 2 = M [PCl 3 ] = [Cl 2 ] = M [PCl 5 ] = = M PCl 3 + Cl 2 PCl5 i C e = ( ) = 0 = ± (8.18)( ) 2(8.18) [PCl 5 ] = = M [PCl 3 ] = [Cl 2 ] = = M =

6 2SO 3 2SO2 + O 2 i C -2y +2y +y e y 2y y K p = 1150 = [SO 2] 2 [O 2 ] [SO 3 ] 2 K c = K (1/RT)2 (1/RT) 1150 p (1/RT) 2 = (0.0821)(1500) = 9.34 if this reaction goes to completion, then 2y 0.80 and y = (0.80)2 (0.40) [SO 3 ] 2 [SO 3 ] 2 = [SO 3 ] = 0.17 M [SO 2 ] = = 0.63 M [O 2 ] = = 0.32 M y is not small compared to SO 2 and O 2, so reiterate: 9.34 = ( )2 ( ) [SO 3 ] 2 [SO 3 ] 2 = [SO 3 ] = 0.12 M [SO 2 ] = = 0.68 M [O 2 ] = = 0.34 M PCl 3 + Cl 2 PCl5 i C e = [PCl 5 ] [PCl 3 ][Cl 2 ] = = 0 ( )( ) = 9.18 ± (1.81)(8.18) 2(8.18) [PCl 5 ] = = 0.26 M [PCl 3 ] = =.07 M [Cl 2 ] = = 0.41 M = 0.26 M 13-6

7 K c = 0.50 = [CH 2O] [CO][H 2 ] initially, n(co) = (2.0 L)(0.10 M) = 0.20 mol CO [CO] = 0.20 mol/7.0 L = M n(h 2 ) = (5.0 L)(0.20 M) = 1.0 mol H 2 [H 2 ] = 1.0 mol/7.0 L = 0.14 M CO + H 2 CH 2 O i C e = ( )( ) = 0 = ± (0.50)( ) 2(0.50) = [CH 2 O] = = M [CO] = = M [H 2 ] = = 0.14 M 13.7 Le Chatelier's Principle LeChatelier s Principle: If an eternal stress is applied to a system at equilibrium, the system adjusts itself in such a way to reduce the stress and reestablish equilibrium. The Effect of a Change in Concentration CO 2(g) + 2H 2 O (l) H 3 O + (aq) + HCO 3 - (aq) an increase in ph will cause increase in CO 2 The Effect of a Change in Pressure 2SO 3(g) 2SO 2(g) + O 2(g) increase in pressure favors SO 3, increase in volume favors SO 2 The Effect of a Change in Temperature CO (g) + 2H 2(g) CH 3 OH (g) H = -18 kj reaction is eothermic (heat is a product): increase in T favors reactants Effects of a catalyst: changes E a and E arev equally, changes rate but no change in equilibrium 13-7