Equilibrium. Forward and Backward Reactions. Hydrogen reacts with iodine to make hydrogen iodide: H 2 (g) + I 2 (g) 2HI(g)

Transcription

1 Equilibrium Forward and Backward Reactions Hydrogen reacts with iodine to make hydrogen iodide: H 2 (g) + I 2 (g) 2HI(g) forward rate = k f [H 2 ][I 2 ] 2HI(g) H 2 (g) + I 2 (g) backward rate = k b [HI] 2 1

2 Equilibrium Constant k f [H 2 ][I 2 ] = k b [HI] 2 All the constants together make the equilibrium constant, K: k f [H 2 ][I 2 ] = k b [HI] 2 k f k b = K = [HI] 2 [H 2 ][I 2 ] [HI] 2 [H 2 ][I 2 ] K is a property of the reaction, and is something you could look up in a table. K will also (typically) change with temperature. Equilibrium Constant Expressions We extend the above idea to reactions with two products: aa + bb cc + dd for which the equilibrium constant is given by so, for example: K = [C]c [D] d [A] a [B] b 6CO 2 + 6H 2 O C 6 H 12 O 6 + 6O 2 2CH 3 Cl + 3O 2 3H 2 O + CO 2 + COCl 2 2

3 gives K = [C 6H 12 O 6 ][O 2 ] 6 [CO 2 ] 6 [H 2 O] 6 K = [H 2O] 3 [CO 2 ][COCl 2 ] [CH 3 Cl] 2 [O 2 ] 3 2H 2 + O 2 2H 2 O K = [H 2O] 2 [H 2 ] 2 = [O 2 ] 3

4 Gas Equilibria When a reaction just involves gases with aa + bb cc + dd K P = P c C P d D P a A P b B For H 2 (g) + I 2 (g) 2HI(g) P 2 HI K P = = P H2 P I2 If, at equilibrium, we have partial pressures of H 2 and I 2 of 0.1 atm, we get P 2 HI P H2 P I2 = PHI 2 = 0.29P H2 P I2 = 0.29(0.1 atm)(0.1 atm) = atm 2 P HI = atm 4

5 Homogeneous and Heterogeneous Equilibria We define Homogeneous equilibria are those where all reagents and all products are in the same phase: solid, liquid, or gas. Heterogeneous equilibria are everything else: at least one reagent or product is in a different state from the others. For we can write CO 2 (g) + C(s) 2CO(g) K = [CO] 2 [CO 2 ][C] K [C] = [CO]2 [CO 2 ] K = [CO]2 [CO 2 ] or K P = P 2 CO P CO2 Generally, we assume that pure solid and liquid components have been eliminated from an equilibrium expression. 5

6 Sequential Reactions Combustion of nitrogen occurs stepwise: N 2 (g) + O 2 (g) 2NO(g) K 1 = NO(g) + O 2 (g) 2NO 2 (g) K 2 = K 1 = K 2 = K 3 = P 2 NO P N2 P O2 P 2 NO 2 P 2 NO P O 2 P 2 NO 2 P N2 P 2 O 2 and so and K 1 K 2 = P 2 NO P N2 P O2 P 2 NO 2 P 2 NO P O 2 = P 2 NO 2 P N2 P 2 O 2 = K 3 K 3 = K 1 K 2 = ( )( ) = And, in general: Adding two reactions (reagents and products) to make a new reaction gives an equilibrium constant that is the product of the two equilibrium constants. Here are the tricks we could use to get from equilibrium constants we know to ones we don t: 1. If the forward reaction is K, the backward is 1/K. 2. If we multiply the moles of all species by n, then K goes to K n. (e.g., K 2, K) 3. If we add two reactions, we multiply their equilibrium constants. 6

7 Concentrations Equilibrium Constant 2SO 2 (g) + O 2 (g) 2SO 3 (g) at 800 K, we measure an equilibrium mixture of M SO 3, M O 2, and M SO 2. K = = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] ( ) 2 ( ) 2 ( ) = Let s say we start with 0.05 M SO 3 alone at 800 K, and we allow it to reach equilibrium. Once it does, we measure a concentration of M O 2. Can we calculate K now? 7

8 and [SO 3 ] [SO 2 ] [O 2 ] Initial: Change: Final: K = = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] ( ) 2 ( ) 2 ( ) = Equilibrium Constant Concentrations H 2 (g) + I 2 (g) 2HI(g) [HI] 2 [H 2 ][I 2 ] = 54 [H 2 ] [I 2 ] [HI] Initial: Change: x x +2x Final: 0.05 x 0.05 x 2x 8

9 [HI] 2 [H 2 ][I 2 ] (2x) 2 (0.05 x)(0.05 x) = 54 = 54 (2x) 2 (0.05 x) 2 = 54 This gives us [HI] 2 [H 2 ][I 2 ] (2x) 2 (0.05 x)(0.05 x) = 54 = 54 4x 2 = 54(0.05 x)(0.05 x) 2x 2 = 27( x + x 2 ) 0 = 25x 2 2.7x x = b ± b 2 4ac 2a = 2.7 ± x = or So x = [H 2 ] = 0.05 x = [I 2 ] = 0.05 x = [HI] = 2x =

10 Same problem, with [H 2 ] = 0.5 and [I 2 ] = and so [H 2 ] [I 2 ] [HI] Initial: Change: x x +2x Final: 0.5 x 0.05 x 2x [HI] 2 [H 2 ][I 2 ] (2x) 2 (0.5 x)(0.05 x) = 54 = 54 4x 2 = 54( x + x 2 ) 0 = 25x x x = ± = 1 (14.85 ± 12.37) 50 = or [H 2 ] = 0.5 x = 0.45 [I 2 ] = 0.05 x = [HI] = 2x =

11 The Reaction Quotient aa + bb cc + dd { [C] c [D] d } K = [A] a [B] b eq We define the reaction quotient: Q = [C]c [D] d [A] a [B] b which is exactly the same thing as K, but the system does not have to be at equilibrium: At equilibrium, Q = K. Away from equilibrium, Q can take any (non-negative) value. If Q < K, this means that [C] c [D] d [A] a [B] b < { [C] c [D] d } [A] a [B] b eq and the reaction proceeds, as written, to the right. If Q > K, the reaction proceeds, as written, to the left. [C] c [D] d { [C] c [A] a [B] b > [D] d } [A] a [B] b eq 11

12 and N 2 (g) + 3H 2 (g) 2NH 3 (g) K P = K If we introduce a equal mixture of all three gases atm each of H 2, N 2, and NH 3 what happens? Q P = = P 2 NH 3 P N2 PH 3 2 ( 1 2 3) ( 1 ) ( ) = 9 Q P = 9 > = K P 12

13 LeChâtelier s Principle If a system is at equilibrium, the concentrations (by definition) of reactants and products will no longer change. Changes in conditions stresses can alter the position of equilibrium, however: 1. Adding or removing reactants or products changes the composition of the system, taking it away from equilibrium. 2. Changing the pressure or volume changes the concentration of gas-phase components, but not liquid or solid components; this can change relative concentrations for some reactions. 3. Changing temperature will usually change K. LeChâtelier s principle tells us If there are more moles of gas in the reactants than the products, the reaction decreases pressure, and increasing pressure will favor reaction. If there are more moles of gas in the products than the reactants, the reaction increases pressure, and increasing pressure 13

14 will inhibit reaction. If there are the same number of moles of gas in reagents and products, changing pressure or volume is unlikely to affect the reaction. 14