29.03.2020 EQUILIBRIUM means - balance EQUILIBRIUM a state in which opposing forces or GENERAL CONCEPTS influences are balanced 1 2 1 2 CHEMICAL REACTTION REVERSIBLE CHEMICAL REACTIONS A substance changes into another can go in both directions substance the reactants can change into the products and the products can change back into the reactants reactants products 3 4 3 4 1 29.03.2020 CHEMICAL EQUILIBRIUM CHEMICAL EQUILIBRIUM the forward and reverse reactions are proceeding at the same rate and reaction never cease the concentration of reactants and products are rarely equal the concentrations of all species remain constant over time N2O4 N2O4 NO2 NO2 http://schoolbag.info/chemistry/central/136.html http://www.kchemistry.com/the_keq_family.htm 5 6 5 6 CONCENTRATION CHANGES MASS ACTION EXPRESION Relationship between the concentrations of the reactants and products for any chemical system at equilibrium is known as: REACTION QUOTIENT, Q. IT IS RELATIONSHP BETWEEN THE CONCENTRATION OF THE REACTANTS AND The decomposition of N O (g) into NO (g). 2 4 2 PRODUCTS FOR ANY CHEMICAL system AT Any The concentrations of N2O4 and NO2 change relatively quickly at time first, but eventually stop changing with time when equilibrium is reached. 7 8 7 8 2 29.03.2020 THE NUMERICAL VALUE Equilibrium among OF THE MASS ACTION EXPRESSION H2, I2, HI → is equal Q at equilibrium H2 (g) + I2 (g) 2HI(g) [F] f [G]g → Different amounts of the dD + eE fF + gG d e = Kc [D] [E] reactants and products are placed in a 10.0 L reaction The form is always “products over vessel at 440oC where the reactants” raised to the appropriate gases establish equilibrium. powers When equilibrium is The exponents in the mass action reached, different amounts of reactants and products expression are the same as remain. the stoichiometric coefficients. 9 10 9 10 EQUILIBRIUM AMONG H2, I2, HI EQUILIBRIUM AMONG H2, I2, HI ➢ the mass action expression = reaction quotient, Q at equilibrium The reaction quotient, Q, can be evaluated [HI]2 at any concentrations. 2 Expt. [H2 ] [I2 ] [HI] [HI] [H2 ][I2 ] Q = [H ][I ] I 0.0222 0.0222 0.156 49.4 2 2 II 0.0350 0.0450 0.280 49.8 At equilibrium (and 440oC) for the this III 0.0150 0.0135 0.100 49.4 reaction the reaction quotient has the value 49.5 (an unit less number). IV 0.0442 0.0442 0.311 49.5 Average = 49.5 11 12 11 12 3 29.03.2020 EQUILIBRIUM AMONG H2, I2, HI QUIZ The value 49.5 is called the equilibrium For the following reaction find the formula of constant, Kc, and characterizes the system KC: 2 [HI] o 3H2 + N2 ↔ 2NH3 Kc = = 49.5 (at 440 C) [H2 ][I2 ] a) b) This relationship is called the equilibrium law for the system. For chemical equilibrium to exist, the reaction c) quotient Q must be equal to the equilibrium constant Kc. 13 14 13 14 QUIZ SOLUTION Given the equilibrium concentrations for the reaction below, what is the value of the equilibrium constant in terms of concentrations? KC = 6 2A(aq) + 3 B(aq) ↔ C(aq) + 2 D(aq) At equlibrium: [A]= 1 M; [C]= 3M [B]= 2 M; [D] = 4 M 15 16 15 16 4 29.03.2020 QUIZ OPERATIONS PERFORMED ON EQUILIBRIUM EXPRESSIONS For the reaction below, the equilibrium Changing the direction of equilibrium constant is equal to one. Which of the these Multiplying coefficients sets of equilibrium concentrations is possible? Adding chemical equilibria 2A(aq) + B(aq) ↔ 4C(aq) A) [A]= 2 M, [B]= 1 M, [C]= 1 M B) [A]= 2 M, [B]= 2 M, [C]= 2 M C) [A]= 1 M, [B]= 1 M, [C]= 2 M D) [A]= 1 M, [B]= 1 M, [C]= 1 M 17 18 17 18 CHANGING THE DIRECTION OF Multiplying the coefficients by a factor EQUILIBRIUM ➢ When the coefficients in an equation are multiplied by a factor, the equilibrium constant ➢ When the direction of an equilibrium is reversed, is raised to a power equal to that factor; the new equilibrium constant is the reciprocal of the original: → [PCl5 ] → [PCl ] PCl + Cl PCl K = PCl + Cl PCl K = 5 3 2 5 c [PCl ][Cl ] 3 2 5 c [PCl ][Cl ] 3 2 3 2 2 2 → " [PCl3 ] [Cl2 ] 2 → ' [PCl3 ][Cl2 ] 1 2PCl 2PCl + 2Cl K = = K PCl PCl + Cl K = = 5 3 2 c 2 c 5 3 2 c [PCl5 ] [PCl5 ] Kc 19 20 19 20 5 29.03.2020 Adding chemical equilibriums EXAMPLE ➢When chemical equilibria are added, their equilibrium constants are multiplied Using the equilibrium constants from Table below calculate the value of K for the reaction [N O]2 + 2+ – → 2 CaCO3(s) + H (aq) → Ca (aq) + HCO3 (aq) 2N + O 2N O K = 2 2 2 c1 2 [N2 ] [O2 ] 4 → [NO2 ] 2+ 2– –7 2N O + 3O 4NO K = CaCO3(s) → Ca (aq) + CO3 (aq) K1 = 5.01 x 10 2 2 2 c2 [N O]2[O ]3 2 2 – + 2– –11 HCO3 (aq) → H (aq) + CO3 (aq) K2 = 5.01 x 10 Adding gives : 4 → [NO2 ] 2N2 + 4O2 4NO2 Kc3 = 2 4 = Kc1 Kc2 [N2 ] [O2 ] 21 22 21 22 SOLUTION: The magnitude of K and the position of equilibrium. The net reaction is the sum of reaction 1 and the reverse of reaction 2: K = 5.01 x 10–7 2+ 2– 1 CaCO3(s) → Ca (aq) + CO3 (aq) + 2– – H (aq) + CO3 (aq) → HCO3 (aq) + 2+ – A large amount of product and very little reactant at CaCO3(s) + H (aq) → Ca (aq) + HCO3 (aq) equilibrium gives K>>1 (large K). When K 1 , approximately equal amounts of reactant and product are present at equilibrium. When K<<1, mostly reactant and very little product are present at equilibrium. 23 24 23 24 6 29.03.2020 LARGE KC VALUES SMALL KC Large KC value means that reaction proceeds Small KC value means that reaction proceeds almost completely to the rights (products are almost completely to the lefts (reactants are favoured, products predominate over reactants) favoured, reactants predominate over products) e.g. e.g. 2H2 + O2↔ 2H2O N2 + O2 ↔2NO 25 26 25 26 INTERMEDIATE KC QUIZ Look at the following equilibrium constants. Do -2 2 When KC has a value between 10 and 10 it these equilibrium reactions lie to the left, to the tells that at equilibrium all components of right, or in the middle? reaction are present in significant amounts, A) H CO +H O↔HCO - + H O+ and that the equilibrium reaction is fairly evenly 2 3 2 3 3 K = 4.2 . 10-7 balanced (neither lying to the left nor to the C . 12 right). B) 2NO + O2 ↔ 2NO2 Kc = 2.25 10 o C) CH + H O↔ CO + 3H K = 9.4 .10-1 e.g. H2 + I2 ↔ 2HI at 400 C KC = 50.5 4 2 2 c - + (reactants and products are both present in D) CH3COOH + H2O↔ CH3COO + H3O . -5 approximately equal amounts). Kc = 1.8 10 27 28 27 28 7 29.03.2020 PREDICTING THE DIRECTION OF A REACTION 1. Q< KC Q (reaction quotient) Kc (equilibrium constant) aA+bB → cC + dD [C] and [D] increase [A] and [B] (must be 2 values: Kc (equilibrium constant) decrease formed) Q (reaction quotient) (must be Greater than at consumed) equilibrium Reaction proceeds from the left to the right 29 30 29 30 2. Q> KC 3. Q=KC Q (reaction quotient) The reaction quotient is the same as the K (equilibrium constant) c equilibrium constant. ➢ [C] and [D] The reaction is already at equilibrium, decrease; no changes in the balance between the forward must be [A] and [B] consumed and back reactions. increase must be Less than at formed equilibrium Reaction proceeds from the right to the left 31 32 31 32 8 29.03.2020 PROBLEM TO SOLVE SOLVING The concentration of the three substances in the following reaction: H2 + I2 ↔ 2HI -1 are as follows: [H2] = 1.5 mol L ’ [I ] = 0.5 mol L-1 2 Q = 1.3 [HI] = 1 mol L-1 KC = 50.5. 1.3 < 50.5 In which direction would the reaction need to proceed for equilibrium to be reached? The reaction must proceed from the left to the right (forward reaction) for equilibrium to be reached 33 34 33 34 EQUILIBRIUM CONSTANT IN TERMS TO SELF-CHECK OF PARTIAL PRESSURES The concentration of the three substances in the following reactions: The gas law can be used to write the equilibrium constant in terms of partial H2 + I2 ↔ 2HI -1 pressures are as follows: [H2] = 0.05 mol L ’ PV = nRT -1 [I2] = 1.05 mol L n [HI] = 1.9 mol L-1 P = RT = MRT molarity V KC = 50.5. In which direction would the reaction need to Equilibrium constants written in terms of proceed for equilibrium to be reached? partial pressures are given the symbol Kp 35 36 35 36 9 29.03.2020 EQUILIBRIUM CONSTANT KC AND KP EQUILIBRIUM CONSTANT KC AND KP The two different forms of the equilibrium → N2 (g) + 3H 2 (g) 2NH 3 (g) constants can be related [NH ]2 P2 PV = nRT K = 3 and K = NH3 c [N ][H ]3 P P P3 n 2 2 N2 H 2 P = RT = MRT so that V The size of the equilibrium constant gives ng a measure of how the reaction proceeds. KP = Kc (RT ) where General statements can be made about ng = (moles of gaseous products) the equilibrium constant (either K or K ).