Chemical Equilibrium and Le Châtelier’s Principle

Chemical Equilibrium CHEM-102 1 Chemical Equilibrium Le Châtelier’s Principle Mansour Almatarneh University of Jordan CHEM-102 2 The concept of Equilibrium Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: • the rates of the forward and reverse reactions are equal and • the concentrations of the reactants and products remain constant Physical equilibrium NO2 H2O (l) H2O (g) Chemical equilibrium N2O4 (g) CHEM-102 2NO2 (g) 3 N2O4 (g) 2NO2 (g) equilibrium equilibrium Start with N2O4 Start with NO2 & N2O4 equilibrium Start with NO2 CHEM-102 4 constant CHEM-102 5 N2O4 (g) K= [NO2]2 [N2O4] aA + bB K= CHEM-102 [C]c[D]d 2NO2 (g) = 4.63 x 10-3 cC + dD Law of Mass Action [A]a[B]b 6 K= [C]c[D]d aA + bB cC + dD [A]a[B]b Equilibrium Will CHEM-102 K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants 7 ways of expressing equilibrium constants Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) Kc = 2NO2 (g) [NO2]2 Kp = [N2O4] 2 PNO 2 PN2O4 In most cases Kc  Kp aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b) CHEM-102 8 Homogeneous Equilibrium CH3COOH (aq) + H2O (l) [CH3COO-][H3O+] Kc′ = [CH3COOH][H2O] CH3COO- (aq) + H3O+ (aq) [H2O] = constant [CH3COO-][H3O+] = Kc′ [H2O] Kc = [CH3COOH] General practice not to include units for the equilibrium constant. CHEM-102 9 Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium: (a) HF(aq) + H2O(l) (b) 2NO(g) + O2(g) H3O+(aq) + F−(aq) 2NO2(g) (c) CH3COOH(aq) + C2H5OH(aq) CHEM-102 CH3COOC2H5(aq) + H2O(l) 10 (a) (b) + [H O ][F ] ' 3 Kc = [HF][H 2 O] [H 3O + ][F - ] Kc = [HF] 2 [NO2 ] Kc = [NO]2 [O 2 ] (c) The equilibrium constant K c' Kp = 2 PNO 2 2 PNO PO2 is given by [CH3COOC2 H5 ][H 2 O] = [CH3COOH][C 2 H5OH] [CH 3COOC2 H 5 ] Kc = [CH 3COOH][C2 H 5OH] CHEM-102 11 Example The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g) PCl3(g) + Cl2(g) is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C? Kp = PPCl3 PCl2 PPCl5 1.05 = PCl2 CHEM-102 (0.463)(PCl2 ) (0.875) (1.05)(0.875) = = 1.98 atm (0.463) 12 Example: Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) CH3OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of KP at this temperature? Δn = moles of gaseous products - moles of gaseous reactants KP = Kc(0.0821T )Δn Because T = 273 + 220 = 493 K and Δn = 1 - 3 = -2, we have KP = (10.5) (0.0821 x 493)-2 = 6.41 x 10-3 CHEM-102 13 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 (s) [CaO][CO2] Kc′ = [CaCO3] Kc = [CO2] = Kc′ x [CaCO3] [CaO] CaO (s) + CO2 (g) [CaCO3] = constant [CaO] = constant Kp = PCO2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. CHEM-102 14 CaCO3 (s) CaO (s) + CO2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of CaCO3 or CaO The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. CHEM-102 15 Example: Consider the following heterogeneous equilibrium: CaCO3(s) CaO(s) + CO2(g) At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc for the reaction at this temperature. (a) KP = PCO2 = 0.236 (b) KP = Kc(0.0821T)Δn T = 800 + 273 = 1073 K and Δn = 1, so 0.236 = Kc(0.0821 x 1073) Kc = 2.68 x 10-3 CHEM-102 16 A+B C+D Kc′ C+D E+F Kc′′ A+B E+F [C][D] Kc′ = [A][B] Kc [E][F] Kc′′= [C][D] [E][F] Kc = [A][B] Kc = Kc′ x Kc′′ If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. CHEM-102 17 N2O4 (g) K= [NO2]2 [N2O4] 2NO2 (g) = 4.63 x 10-3 2NO2 (g) N2O4 (g) [N2O4] 1 = 216 K′ = = 2 K [NO2] When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. CHEM-102 18 Writing Equilibrium Constant Expressions 1. The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. 2. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. 3. The equilibrium constant is a dimensionless quantity. 4. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. 5. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. CHEM-102 19 What does the equilibrium constant tell us? Chemical Kinetics and Chemical Equilibrium A + 2B kf kr ratef = kf [A][B]2 AB2 rater = kr [AB2] Equilibrium ratef = rater kf [A][B]2 = kr [AB2] kf [AB2] = Kc = kr [A][B]2 CHEM-102 20 The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF • Qc < Kc system proceeds from left to right to reach equilibrium • Qc = Kc the system is at equilibrium • Qc > Kc system proceeds from right to left to reach equilibrium CHEM-102 Figures from Chang’s General Chemistry book 21 Example: At the start of a reaction, there are 0.249 mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50-L reaction vessel at 375°C. If the equilibrium constant (Kc) for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed. CHEM-102 22 Solution 0.249 mol [N 2 ]o = = 0.0711 M 3.50 L 3.21  10 −2 mol [H 2 ]o = = 9.17  10−3 M 3.50 L 6.42  10−4 mol [NH3 ]o = = 1.83  10 −4 M 3.50 L [NH3 ]o2 (1.83 × 10-4 ) 2 Qc = = = 0.611 3 -3 3 [N 2 ]o [H 2 ]o (0.0711)(9.17 × 10 ) Because Qc is smaller than Kc (1.2), the system is not at equilibrium. The net reaction will proceed from left to right until equilibrium is reached. CHEM-102 23 Calculating Equilibrium Concentrations 1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3. Having solved for x, calculate the equilibrium concentrations of all species. CHEM-102 24 Calculate the [H+] of 0.01 M HCl HCl(aq) 0.01 M H+(aq) + Cl-(aq) 0.01 M 0.01 M Calculate the [OH-] of 0.02 M Ba(OH)2 solution. Ba(OH)2(aq) 0.02 M Ba2+(aq) + 2OH-(aq) 0.02 M 0.04 M Example A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask at 430°C. The equilibrium constant Kc for the reaction [ H2(g) + I2(g) 2HI(g) ] is 54.3 at this temperature. Calculate the concentrations of H2, I2, and HI at equilibrium. Initial (M): Change (M): Equilibrium (M): CHEM-102 H2 + I2 2HI 0.500 -x (0.500 - x) 0.500 -x (0.500 - x) 0.000 + 2x 2x 26 Step 2: The equilibrium constant is given by [HI]2 Kc = [H 2 ][I 2 ] Substituting, we get (2x) 2 54.3 = (0.500 - x)(0.500 - x) Taking the square root of both sides, we get 2x 7.37 = 0.500 - x x = 0.393 M CHEM-102 27 Step 3: At equilibrium, the concentrations are [H2] = (0.500 - 0.393) M = 0.107 M [I2] = (0.500 - 0.393) M = 0.107 M [HI] = 2 x 0.393 M = 0.786 M Check – check your answers by calculating Kc using the equilibrium concentrations. Remember that Kc is a constant for a particular reaction at a given temperature. CHEM-102 28 Example: For the same reaction and temperature as in Example 15.6, H2(g) + I2(g) 2HI(g), suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium. [HI]02 (0.0224)2 Qc = = = 19.5 [H 2 ]0 [I 2 ]0 (0.00623)(0.00414) Because Qc (19.5) < Kc (54.3), the net reaction will proceed from left to right until equilibrium is reached; there will be a depletion of H2 and I2 and a gain in HI. CHEM-102 29 Initial (M): Change (M): H2 + I2 2HI 0.00623 0.00414 -x -x 0.0224 + 2x Equilibrium (M): (0.00623 - x) (0.00414 - x) [HI]2 Kc = [H 2 ][I 2 ] (0.0224 + 2x) (0.0224 + 2x)2 54.3 = (0.00623 - x)(0.00414 - x) 54.3(2.58 x 10-5 - 0.0104x + x2) = 5.02 x 10-4 + 0.0896x + 4x2 50.3x2 - 0.654x + 8.98 x 10-4 = 0 This is a quadratic equation CHEM-102 -b ± b 2 - 4ac x= 2a 30 x= (-0.654) 2 - 4(50.3)(8.98 × 10-4 ) 2 × 50.3 x = 0.0114 M or x = 0.00156 M 0.654 ± At equilibrium, the concentrations are [H2] = (0.00623 - 0.00156) M = 0.00467 M [I2] = (0.00414 - 0.00156) M - 0.00258 M [HI] = (0.0224 + 2 x 0.00156) M = 0.0255 M CHEM-102 31 Factors that affect chemical equilibrium Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. • Changes in Concentration N2 (g) + 3H2 (g) 2NH3 (g) Equilibrium shifts left to offset stress Add NH3 CHEM-102 32 Le Châtelier’s Principle • Changes in Concentration continued Remove Add Remove Add aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) left right right left CHEM-102 33 Le Châtelier’s Principle • Changes in Volume and Pressure A (g) + B (g) CHEM-102 C (g) Change Shifts the Equilibrium Increase pressure Decrease pressure Increase volume Decrease volume Side with fewest moles of gas Side with most moles of gas Side with most moles of gas Side with fewest moles of gas 34 Example: Predict the direction of the net reaction in each case as a result of increasing the pressure (decreasing the volume) on the system at constant temperature. (a) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) (b) PCl5(g) PCl3(g) + Cl2(g) (c) H2(g) + CO2(g) H2O(g) + CO(g) Solution (a) The rxn. will shift toward the products (to the right). (b) The rxn. will shift to the left, toward the reactant. (c) The number of moles is equal on both sides, so a change in pressure has no effect on the equilibrium. CHEM-102 35 Le Châtelier’s Principle Changes in Temperature Change Increase temperature Decrease temperature N2O4 (g) Exothermic Rx K decreases Shift to Left K increases Shift to Right K increases K decreases Shift to Right Shift to Left 2NO2 (g) K= CHEM-102 Endothermic Rx [Prod.] [React.] 36 Le Châtelier’s Principle •Adding a Catalyst • does not change K • does not shift the position of an equilibrium system • system will reach equilibrium sooner Adding a Catalyst Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. CHEM-102 37 Le Châtelier’s Principle - Summary Change Equilibrium Constant no Change Shift Equilibrium Concentration yes Pressure yes* no Volume yes* no Temperature yes yes Catalyst no no *Dependent on relative moles of gaseous reactants and products CHEM-102 38 Example: Consider the following equilibrium process between dinitrogen tetrafluoride (N2F4) and nitrogen difluoride (NF2): N2F4(g) 2NF2(g) ΔH° = 38.5 kJ/mol Predict the changes in the equilibrium if Shift to right (a) the reacting mixture is heated at constant volume; (b) some N2F4 gas is removed from the reacting mixture at Shift to left constant temperature and volume; (c) the pressure on the reacting mixture is decreased at Shift to right constant temperature; and (d) a catalyst is added to the reacting mixture. No effect [NF2 ]2 Kc = [N 2 F4 ] CHEM-102 39 Solution (a) The rxn. is an endothermic process (ΔH° > 0), the system will adjust to remove some of the added heat by undergoing a decomposition reaction (from left to right). The equilibrium constant will therefore increase with increasing temperature. (b) It will shift from right to left until equilibrium is reestablished. As a result, some NF2 combines to form N2F4. (c) The system will shift from left to right to reestablish equilibrium. The equilibrium constant will remain unchanged because temperature is held constant. (d) The function of a catalyst is to increase the rate of a reaction. If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of NF2 and N2F4 or the equilibrium constant. CHEM-102 40 Example: At 720°C, the equilibrium constant Kc for the reaction N2(g) + 3H2(g) 2NH3(g) is 2.37 x 10-3. In a certain experiment, the equilibrium concentrations are [N2] = 0.683 M, [H2] = 8.80 M, and [NH3] = 1.05 M. Suppose some NH3 is added to the mixture so that its concentration is increased to 3.65 M. (a) Use Le Châtelier’s principle to predict the shift in direction of the net reaction to reach a new equilibrium. (b) Confirm your prediction by calculating the reaction quotient Qc and comparing its value with Kc. Solution (a) To offset this stress, some NH3 reacts to produce N2 and H2 until a new equilibrium is established. The net reaction therefore shifts from right to left; that is, N2(g) + 3H2(g) 2NH3(g) CHEM-102 41 (b) At the instant when some of the NH3 is added, the system is no longer at equilibrium. The reaction quotient is given by [NH 3 ]02 Qc = [N 2 ]0 [H 2 ]30 (3.65) 2 = (0.683)(8.80)3 = 2.86  10−2 Because this value is greater than 2.37 x 10-3, the net reaction shifts from right to left until Qc equals Kc. CHEM-102 42