Chap 15A - Chem 210 / Wang / Chapter 15A Equilibrium...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chem 210 / Wang / Chapter 15A Equilibrium Constants Chapter 15A Chemical Equilibrium 15.2 Equilibrium Constant 15.3 Interpreting & Working with Equilibrium Constants 15.4 Heterogeneous Equilibria A dynamic equilibrium on the molecular level dynamic is achieved when concentration of reactants concentration and products remain constant over time. remain N2O4(g) (reactant) 2NO2(g) (product) 15.5 Equilibrium Calculations 15.6 Applications of Equilibrium Constants - the equilibrium process is indicated with the twotwo-way equilibrium arrows. Equilibrium Kinetic Region 15.7 Le Chatelier’s Principle Chemical Equilibrium = dynamic dynamic equilibrium I’m in static static equilibrium. A state where the forward and reverse conditions occur at the same rate. Dynamic Equilibrium Law Law of Mass Action “… the combined efforts of the theorist Guldberg and the theorist empiricist empiricist Waage were needed to produce the first general, first exact, mathematical formulation of the role of the amounts of reactants in chemical equilibrium systems. … http://www.bookrags.com/biography/cato-guldberg-woc/ Guldberg and Waage were also related through two marriages. Waage married the sister of Guldberg's wife and had 5 children. After his first wife died, Waage married one of Guldberg's sisters, by whom he had 6 more children. Fall 2009 UM-SJTU JI Equilibrium Region Product concentration Remain the same. Concentration 15.1 The Concept of Equilibrium Chemical equilibrium Reactant concentration Remain the same. Time Reversible reactions and chemical equilibrium Dynamic equilibrium • Same as STATIC equilibrium, no net change STATIC is observed when a system is in CHEMICAL CHEMICAL equilibrium. • Different from STATIC equilibrium, reactions STATIC are still occurring in a DYNAMIC equilibrium equilibrium. – Forward and reverse reactions occur at the same rate in either directions. Rate forward = Rate reverse Rate reverse Law Law of Mass Action • Postulated in 1864 by Guldberg and Waage of the University of Oslo, Norway. • Result of extensive experimentation. • The law of mass action is the basis of the “equilibrium constant expression”, which shows that equilibrium is: equilibrium – Related to reaction stoichiometry (balanced stoichiometry (balanced chemical equation with definite ratios of reactants & products) – NOT related to reaction kinetics (the reaction process before reaching equilibrium). • Initial concentrations impact reaction rates, impact but but have no effect on the equilibrium state. equilibrium Law of Mass Action predicts an Equilibrium constant (K) All reactions end in an equilibrium constant (K) at constant temperature. K has this expression: aA + bB Keq = Equilibrium expressions Write equilibrium expressions for the following: H2(g) + Cl2(g) 2 HCl(g) cC + dD [C]c [D]d [A]a [B]b K= N2(g) + 3H2(g) 2 NH3 (g) K= [ ] represents concentration of the enclosed measured in moles per liter (M). K = [product] over [reactant], each raised to the power of its coefficient in the balanced equation. [HCl]2 [H2][Cl2] PbCl2(s) [NH3]2 [H2]3[N2] Pb+2(aq)+ 2Cl-1(aq) K = [Pb+2][Cl-1]2 page 1 Chem 210 / Wang / Chapter 15A Equilibrium Constants Equilibrium expressions Include all gaseous (g) and aqueous (aq) aq) reactants and products variable – These have variable moles per unit volume. Ignore all solids (s) and liquids (l) – These have fixed moles per unit volume. fixed CaCO3(s) 2H2O(l) CO2(g) + CaO(s) Question: This picture represents the equilibrium state for a gas phase reaction: 2AB A2 + B2 Calculate the equilibrium constant K. CH4(g) K = [CO2] 2HCl(g) → H2(g) + Cl2(g) K2 K3 [ HCl ] 2 ½H2(g)+½Cl2(g) [ HCl ] (b) How do K1 and K3 relate? K1 = 1/ K3 “reverse reaction” K = [CH42] [H2] Equilibrium constant (K or Keq) 10-5 Reaction hardly occurs large K 10-3 K value 1 103 reactants & products co-exist in equilibrium 105 goes to completion Both K and Q use the same “equilibriumequilibriumconstant expression” based on the law of law mass action for a chemical reaction. Equilibrium constant (K) = value calculated value by substituting equilibrium concentrations in equilibrium the equilibrium constant expression. Reaction quotient (Q) = value calculated by value substituting current concentrations in the current same expression. d [C]c Q or K = [C]a [D]b [A] [B] Fall 2009 UM-SJTU JI “square” K! “inverse” K! Questions involving Equilibrium Calculations K = 3.3×10-13 for AgBr(s) Ag+(aq) + Br-(aq) K = 1.5×10-16 for AgI(s) Ag+(aq) + I-(aq) (a) If [Ag+] = 5.7×10-7 M what is [Br-] at equilibrium? K = 3.3×10-13 = [Ag+] [Br-] = 5.7x10-7 [Br-] [Br-] = 5.8 x 10-7 M (b) Is AgI(s) more soluble than AgBr(s)? Explain. No, it is less soluble …. K is smaller! intermediate K Reaction Quotient, Q vs. Equilibrium Equilibrium Constant, K [ Cl2 ][ H2 ] [ HCl ] 2 [Cl2 ]½[ H2 ]½ (a)How do K1 and K2 relate? K1 = K22 “Double coeffcient” small K • → HCl(g) [ Cl2 ] [ H2 ] Actual value of K: • different for each reaction. [C]c [D]d K= • varies with temperature [A]a [B]b • usually given at 25° 25° C • Indicates “equilibrium position” or reaction reaction end point. • H2(g) + Cl2(g) → 2 HCl(g) K1 The range of equilibrium constants • Bal. EQ Kc 2H2(g)+ O2(g) K = [H ]2[O ] 2 2 C(s) + 2H2(g) Write equilibrium expressions for these reactions Q changes during the reaction. Q= Reaction Quotient, Q vs. Equilibrium Equilibrium Constant, K N 2 O4 ( g ) 2NO2(g) [ NO 2 ]2 [N 2 O4 ]1 Before equilibrium at equilibrium [NO2] > [N2O4] [NO2] = [N2O4] [NO2] =0 Q=0 When Q = K the system is at equilibrium. all concentrations remain constant. [ product ] When Q > K [product] is too high Q = [reactant ] reverse rate > forward rate the system is moving toward equilibrium by moving the the reverse reaction. [ product] When Q < K [reactant] is too high Q = [reactant ] forward rate > reverse rate the system is moving toward equilibrium by moving the the forward reaction. page 2 Chem 210 / Wang / Chapter 15A Equilibrium Constants Question: Given that K = 1 for this reaction N2(g) + 3H2(g) ⇔ 2 NH3(g) + 92 kJ. And [N2] = [H2] = [NH3] = 1.0 mol/L If the reaction vessel is suddenly decrease the to ½ of its existing size, what is Q at that “instant”? [ NH3 ] 2 [ 1.0 ] 2 Kc = =1 3= [ N2 ] [ H2 ] [ 1.0 ] [1.0 ] 3 Volume 1 L ½ L: 1.0 mol/L L: 1.0 [ 2.0 ] 2 Q= [ 2.0 ] [2.0] 3 2.0 2.0 mol/L = 2.0 M = 0.25 Q < K… system will move forward ( ) to make to more product and re-establish equilibrium. re- K= [ HI ] 2 [ 0.00234 ] 2 = [ I2 ] [ H2 ] [?] [? ] data set (1) H2(g)+I2(g)⇔ 2HI(g) H2(g) + I2(g) ⇔ 2 HI(g) Initial 0.00150 0.00150 0 Change -x -x + 2x Equil. quil. ? ? 0.00234 M 0.00234 0.00234 M HI comes from H2 and I2. 2x = 0.00234 M x = 0.00117 M 0.00117 At equilibrium: [H2]= [I2]= 0.00150M - 0.00117 M 0.00150M = 0.00033 M 2 K= K= [ 0.00234 ] 0.00234 [0.00033] [0.00033] [ HI ] 2 [ I2 ] [ H2 ] = Data set (2) [0.000165] [0.000165 ] H2(g) + I2(g) ⇔ 2 HI(g) INITIAL 0 0 0.00150 CHANGE +x +x - 2x Equil. Equil. 0.000165 0.000165 ?M 0.000165M H2 and I2 come from HI. x = 0.000165 M 0.000165 2x = 0.000330 M 0.000330 At equilibrium: [HI]= 0.00150M - 0.000330 M [HI]= 0.00150M = 0.00117 M 0.00117 K= [0.000165] [0.000165] Set (2) starting starting with products only Initially: Initially: [HI] = 0.00150 M At equilibrium:[H2] = [I2] = 0.000165 M equilibrium: Fall 2009 UM-SJTU JI Initially: [H2] = 0.0015 M and [I2] = 0.0015 M [0]2 Q= Q=0 [0.0015] [0.0015 ] At equilibrium: [HI] = 0.00234 M [ 0.00234 ] 2 K= [?] [? ] Set (3) (3) starting starting with both reactants & products Initially: Initially: [H2] = [I2] = [HI] = 0.00150 M At equilibrium:[H2] = 0.000496 M equilibrium: K= [ HI ] 2 [ I2 ] [ H2 ] [ 0.00234 ] 2 = [?] [? ] data set (1) H2(g)+I2(g)⇔ 2HI(g) H2(g) + I2(g) ⇔ 2 HI(g) Initial 0.00150 0.00150 0 Change -x -x + 2x Equil. quil. ? ? 0.00234 M 0.00234 0.00234 M HI comes from H2 and I2. ICE table 2x = 0.00234 M x = 0.00117 M 0.00117 At equilibrium: [H2]= [I2]= 0.00150M - 0.00117 M 0.00150M = 0.00033 M 2 K= [ 0.00234 ] 0.00234 [0.00033] [0.00033] [ I2 ] [ H2 ] Initially: [H2] = [I2] = [HI] = 0.00150 M [ 0.00150 ] 2 Q= = 1.00 [0.00150] [0.00150] At equilibrium:[H2] = 0.000496 M K= [?]2 Data set (2) H2(g)+I2(g)⇔2HI(g) [ HI ] 2 K= [ I2 ] [ H2 ] Initially: [HI] = 0.0150 M [ 0.00150 ] 2 Q= = divide by zero divide = infinity [0 ] [ 0 ] At equilibrium:[H2] = [I2] = 0.000165 M [?]2 K= [0.00165] [0.00165 ] = 50.3 50.3 Data set (3) H2(g)+I2(g)⇔2HI(g) [ HI ] 2 K= [0.000496] [? ] = 50.3 50.3 Same Same as data set (1)! Calculate K and Q, given experimental data data set (1) for: H2(g) + I2(g) ⇔ 2 HI(g) for: Set (1) starting starting with reactants only Initially: Initially: [H2] = 0.0015 M & [I2] = 0.0015 M. At equilibrium: [HI] = 0.00234 M equilibrium: = 50.3 50.3 [?]2 [ 0.00117 ] 2 0.00117 Verify this fact: “Initial concentrations impact “Initial impact but reaction rates, but have no effect on the final equilibrium state by 3 sets of experimental data state” by for this reaction: H2(g) + I2(g) ⇔ 2 HI(g) K= [ HI ] 2 [ I2 ] [ H2 ] = [?]2 [0.000496] [? ] Data set (3) H2(g) + I2(g) ⇔ 2 HI(g) INITIAL 0.00150 0.00150 0.00150 CHANGE -x -x +2x_______ Equil. Equil. 0.000496 0.00150-x 0.00150 + 2x 0.00150- [H2] = [I2] = 0.00150 – x = 0.000496 [H 0.00150 0.000496 x = 0.00150 - 0.000496 0.00150 = 0.001004 M 0.001004 [HI]= 0.00150 + 2x = 0.00150 + 2× 0.001004 0.00150 2x 0.00150 = 0.00351M 0.00351M 2 K= [ 0.00351 ] 0.00351 [0.000496] [0.000496] = 50.0 50.0 page 3 Chem 210 / Wang / Chapter 15A Equilibrium Constants Beginning equilibri um Set (1) Q=0 (reactants only) K = 50.3 Set (2) Q=∞ (products only) K = 50.3 Set (3) Q = 1.0 K = 50.0 (both reactants & products are present) Review summary • K values are about the same about (around 50) very • Q values are very different (0 to infinity). • This experiment proves the the law law of mass action: Whatever Whatever the starting concentrations concentrations may be, a reaction will reach the same “K” at the end. This This is true as long as temperature remains the same! Kc and Kp for N2(g) + 3H2(g) → 2 NH3(g) Kp = PNH3 2 PN2 PH23 Re-arrange the ideal gas equation: RePV = nRT (Ref. to chap.10) P = (n/V) RT P = (Molarity of gas) RT PNH32 = [NH3]2 (RT)2 [NH Kp = [NH3]2(RT)2 [N2]RT [H2]3(RT)3 = Kc (RT)-2 Ratio(Q) Initial Experiment [N2O4] [NO2] [NO2]2 [N2O4] Equilibrium [N2O4]eq [NO2]eq Ratio(K) [NO2]eq2 [N2O4]eq 1 0.1000 0.0000 0.0000 3.57x10-3 0.193 2 0.0000 0.1000 ∞ 9.24x10-4 9.83x10-3 10.4 3 0.0500 0.0500 0.0500 2.04x10-3 0.146 10.4 0.00833 2.75x10-3 0.170 Kc and Kp Normally we do not report any unit for K values, do for though the value changes depending on units used in the K expression. • Reactions in aqueous solutions often use Kc – Use “Molarity (M)” of each reactant or product in the K expression. • Reactions involving gases often use Kp: – Use “partial pressure” of each gas reactant “partial or product in the K expression. Example: for N2(g) + 3H2(g) → 2 NH3(g) 10.5 4 0.0750 0.0250 10.4 Kp = = PHI 2 P = [molarity] (RT) molarity] PI2 PH2 [ HI ]2 ([ HI ]RT)2 = [ I2 ]RT [ H2 ]RT [ I2 ] [ H2 ] = Kc PN2 PH23 [ NH3 ]2 NH K c= [ N2 ] [ H2 ]3 Kc and Kp Kc and Kp for H2(g) + I2(g) ⇔ 2 HI(g) Kp = PNH3 2 When both (aq) and (g) occurs in a reaction… (aq) HCl(aq)+NaHCO3(aq) aq) aq) Kc = NaCl(aq)+ NaCl(aq)+ H2O(l) + CO2(g) CO P= [Molarity] RT [Molarity] [NaCl][CO2 ] [HCl][NaHCO3] Kp = For For this particular gaseous reaction: For a gaseous reaction in general : a A(g) + b B(g) c C(g) + d D(g) Kp = Kc (RT)∆n (c+d) a+b) ∆n = (c+d) – (a+b) Ref. Table 15.1 Initial and Equilibrium Concentration Ratios for the N2O4-NO2 System at 2000C(473 K) R = 0.0821 atm·L/K·mole L/ T = absolute temp = oC + 273 ∆n = (1) – (0) = 1 H2(g) + I2(g) ⇔ 2 HI(g) Kp = Kc (RT)1 Kp = Kc (RT)∆n ∆n = (2) – (1+1) = 0 Kp = Kc Kp for reactions with both aqueous solution and gas: aqueous and gas Keep Keep aqueous solutions in “Molarity”. Express Express gas in terms of “Pressure” Calculate Kc and Kp for the Haber process: N2(g) + 3H2(g) → 2 NH3(g) Given: at 427°C and a total pressure of 10.0 atm, an equilibrium mixture contains 73.8% H2(g), 24.6% N2(g) and 1.66% NH3(g). H2 (g) at 73.8% 10.0 atm x 73.8% = 7.38 atm N2(g) at 24.6% = 2.46 atm; NH3(g) at 1.66% = 0.166 atm PNH3 2 0.166 2 = 2.79x10-5 = 2.46x 7.383 PN2 PH23 = Kc (RT)-2 T = 427 + 273 = 700 K 427 R = 0.0821 atm·L/K·mole atm L/ Kp = Kc = Kp (RT)2 = 2.79x10-5 x (0.0821x700)2 2.79x10 = 0.0921 Fall 2009 UM-SJTU JI [NaCl] PCO2 [HCl][NaHCO3] This picture represents two equilibrium mixtures of solid CaCO3, solid CaO, and gaseous CO2, as a result CaCO CaO CO of the endothermic decomposition of CaCO3: 1. Write an equilibrium constant expression for this reaction. 2. Do the relative amounts of CaCO3 and CaO affect the and value of the equilibrium constant? Why? -14 at 325oC, find K at 325oC. 3. Given Kc = 9.2 x 10 p page 4 ...
View Full Document

This note was uploaded on 07/30/2011 for the course CHEM 210 taught by Professor Zhang during the Spring '09 term at Shanghai Jiao Tong University.

Ask a homework question - tutors are online