Chemical Equilibrium. Foundation of equilibrium Expressing equilibrium: Equilibrium constants Upsetting equilibrium Le Chatelier

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1 Chemical Equilibrium Foundation of equilibrium Expressing equilibrium: Equilibrium constants Upsetting equilibrium Le Chatelier

2 Learning objectives Write equilibrium constant expressions for both solutions and gas phase reactions Use K eq expressions to calculate concentrations Deduce relationship between K c and K p Use reaction quotients to predict direction of reaction Apply LeChatelier s principle to predict consequences of changing equilibrium conditions

3 Chemical equilibrium is: The state reached when the concentrations of reactants and products remain constant over time Equilibrium is also evident in physical processes: Vapour pressure over a liquid Solid and liquid coexisting at the freezing point Undissolved solute in a saturated solution Equilibrium

4 Equilibrium in a chemical change Not all reactions proceed to completion reactants completely converted into products As reactants convert into products, the products themselves convert back into reactants REACTANTS PRODUCTS PRODUCTS REACTANTS As reactant concentration declines, the forward rate decreases As product concentration increases, the backward rate increases

5 Different start: same finish In the reaction N O4( g) NO ( g) 2 2 Same final concentrations of NO 2 and N 2 O 4 are obtained regardless of the initial conditions pure NO 2 or pure N 2 O 4 The reactions don t stop, rates become equal in both directions

6 Reaction rate Rate perspective As concentration of reactant (product) decreases (increases), forward (backward) rate decreases (increases) Equilibrium is reached when rates are equal Time

7 Reactants and products no more In equilibrium mixture, there is constant, dynamic cycling of materials and original identities of reactant and product are lost By convention, the substance(s) used initially are called reactants and appear on left side REACTANTS PRODUCTS

8 Equilibrium constant Reactions take place in solution or in the gas phase In solution equilibrium constant is K c concentrations are in mol/l For the equilibrium aabb K c [ C] [ A] cc c a dd [ D] [ B] d b Concentration M Products Coefficients from equation Reactants

9 Notes about K c K c is a constant at a given temperature, regardless of concentrations of reactants used Substances may be gases or solutions K c has no units: each concentration is considered as a ratio to the standard state (1 M)

10 Equilibrium constant K p For reactions in gas phase, partial pressure can be used instead of molarity P and M are related by PV = nrt K p P P c C a A P P d D b B K p is unitless because the ratio with respect to the standard state (1 atm) is used

11 Relationship between K c and K p n c p b a d c b a d c p RT K K RT B A D C K ) ( ) ( ] [ ] [ ] [ ] [ ) ( ) ( RT n V P A A A RT RT V n P A A ] [ b B a A d D c C p P P P P K a a a A RT A P ) ( ] [ RT = 24.4 at 25 o C

12 Heterogeneous equilibria Reactions often involve solid or liquid phases in addition to gas and solution Concentrations of liquids and solids are constants independent of the amount Since they are constant they become folded into K c and are excluded from expression

13 Solids and pure liquids are ignored CaCO ( s) CaO( s) CO ( g) 3 2 [ CaO( s)][ CO ( g)] K 2 c [ CaCO ( s)] K K c c [ CaCO ( s)] x 3 [ CaO( s)] [ CO 2 ( g)] ( g)] N.B. Include only gas or solution entities in K c 3 [ CO 2 Solid concentrations are constant

14 Using equilibrium constants The value of K c or K p indicates the extent of a reaction Generally: K c > 10 3, reaction goes to completion K c < 10-3, reaction does not proceed 10 3 > K c > 10-3 reactants and products all present

15 Reaction Quotient Reaction Quotient is instanteous value of expression for K c for combination of reactants and products not at equilibrium It can be used to predict direction reaction takes from current conditions: Right, left or stay the same

16 K c and Q c : Onward or back? In the reaction H ( g) I ( g) 2HI ( g), K 2 2 c A mixture made up with [H 2 ] = 0.1 M, [I 2 ] = 0.2 M and [HI] = 0.4 M Q c [ HI [ H 2 ] ][ I K c > Q c so reaction will go towards HI 2 2 ]

17 Predicting reactions with Q c Q c < K c, reaction goes towards products Q c > K c, reaction goes towards reactants Q c = K c, reaction is at equilibrium

18 Calculations with K c Simple calculations: K c and all equilibrium concentrations but one are given. Use K c to calculate concentration of unknown Complex calculations: K c and initial concentrations are given, calculate equilibrium concentrations

19 The I.C.E. Man cometh ICE is cool: key to solving equilibrium problems Given K c = 57 and initial concentrations of [H 2 ] = [I 2 ] = 0.1 M, find equilibrium concentrations of H 2, I 2 and HI H 2 I 2 HI Initial conc Change -x -x 2x Equilibrium conc x x 2x

20 K c 57 x [ H Solve for x [ HI 2 4x (0.1 ] 2 ][ I 2 x) 2 2 ] ( (2x) x) M, or0.136m 2 (0.1 X = > 0.1 so is physically unreasonable X =0.0791: [H 2 ] = [I 2 ] = M; [HI] = M x)

21 Le Chat sat on the mat Conditions of reactions must be manipulated to optimize yield Concentrations of reactants or products Temperature Pressure/volume Le Chatelier s Principle (of the pointed stick): If stress is applied to reaction at equilibrium, system adjusts to relieve stress

22 The Haber process Important industrial processe for synthesis of NH 3 from elements: N 2 (g) + 3H 2 (g) At 700 K, K c = 0.29 Altering reactant concentrations: Increasing [reactant]: reactants products Increasing [product]: products reactants 2NH 3 (g)

23 Effect of addition of N 2 on equilibrium K remains unchanged Concentrations must change if overall composition changes [NH 3 ] increases Reactants products [H 2 ] decreases Reactants products [N 2 ] increases

24 Effects of pressure Pressure is only important if number of gas moles changes: N 2 (g) + 3H 2 (g) 2NH 3 (g) In Haber process there are 4 moles of reactants vs 2 moles of products Increasing pressure: converts reactants to products (fewer moles) Decreasing pressure: converts products to reactants (more moles)

25 At constant T, increasing P improves yield

26 On a molecular scale Reaction at equilibrium At new pressure reaction out of equilibrium Reaction adjusts to equilibrium

27 Effects of temperature K c depends on T. Increasing or decreasing T causes equilibrium to adjust to new value of K c. Treat ΔH of a reaction as a reactant/product Raising T causes heat input Lowering T causes heat withdrawal

28 The Haber process as Function of T Process is exothermic: heat is a product N 2 (g) + 3H 2 (g) 2NH 3 (g) kj Raising T causes heat input: Equilibrium adjusts to reduce heat output moves towards reactants Lowering T removes heat: Equilibrium adjusts to increase heat output moves towards products Endothermic reactions will show the opposite T- dependence

29 Increasing T in Haber process reduces yield

30 Catalysts and equilibrium Equilibrium depends on initial and final states Catalyst lowers the transition state Position of equilibrium is unaffected by addition of a catalyst

31 From the rate perspective The catalyst increases the forward reaction rate by lowering the energy barrier The rate of the backward reaction is lowered by the same amount

32 Linking rate equations with K c For the general reaction: A + B = C + D Assume single bimolecular steps Rate of forward reaction = k f [A][B] Rate of backward reaction = k r [C][D] At equilibrium k f [ A][ B] k [ C][ D] r k k f r [ C][ D] [ A][ B] K c