Ch16_Equilib_1

Ch16_Equilib_1 - 1 2 Attention Online Users CHEMICAL...

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Unformatted text preview: 1 2 Attention Online Users CHEMICAL EQUILIBRIUM Chapter 16 Properties of an Equilibrium 3 Equilibrium systems are Equilibrium systems are •• DYNAMIC (in constant DYNAMIC (in constant motion) motion) •• REVERSIBLE REVERSIBLE •• can be approached from can be approached from either direction either direction • The font used to create the double arrows used in chemical equations to designate an equilibrium cannot be imbedded in an Acrobat document. Acrobat substitutes a comma or similar funny character wherever there should be double arrows. • Keep tuned -- we’ll try to solve the problem. Pink to blue Co(H2O)6Cl2 ---> Co(H 2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O ---> Co(H 2O)6Cl2 4 Chemical Equilibrium 3+ Fe3+ + SCN -- ¸ FeSCN 2+ FeSCN 2+ Chemical Equilibrium 3+ Fe3+ + SCN -- ¸ FeSCN 2+ FeSCN 2+ 5 6 Examples of Chemical Equilibria Phase changes such as H2O(s) ¸ H2O(liq) O(s) + Fe(H2O)63+ + SCN- ¸ Fe(SCN)(H 2O)53+ + H2O • After a period of time, the concentrations of reactants and products are constant. • The forward and reverse reactions continue after equilibrium is attained. Page 1 7 Examples of Chemical Equilibria 8 Chemical Equilibria For any type of chemical equilibrium of the type aA + bB ¸ cC + dD the following is a CONSTANT (at a given T) conc. of products CaCO 3(s) + H2O(liq) + CO2(g) ¸ Ca2+(aq) + 2 HCO 3-(aq) Formation of stalactites and stalagmites K= [Ca2+] At a given T and P of CO 2, and [HCO3-] can be found from the CaCO 3(s) + H2O(liq) + CO2(g) ¸ Ca2+(aq) + 2 HCO 3-(aq) Ca [C]c [D]d [A]a [B]b equilibrium constant conc. of reactants If K is known, then we can predict concs . of products or reactants. EQUILIBRIUM CONSTANT. 10 Determining K 9 THE EQUILIBRIUM CONSTANT THE EQUILIBRIUM CONSTANT 11 12 Determining K 2 NOCl(g) ¸ 2 NO(g) + Cl 2(g) (g) 2 NOCl(g) ¸ 2 NO(g) + Cl 2(g) (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] Before 2.00 0 0 Change Equilibrium 0.66 Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] Before 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 Determining K Page 2 2 NOCl(g) ¸ 2 NO(g) + Cl 2(g) (g) [NOCl] 2.00 -0.66 1.34 Before Change Equilibrium K K [NO] 2[Cl2 ] [NOCl]2 [NO] 0 +0.66 0.66 [Cl2] 0 +0.33 0.33 [NO]2 [Cl2 ] [NOCl]2 = (0.66)2 (0.33) (1.34)2 = 0.080 13 Writing and Manipulating K Expressions 14 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. Solids and liquids NEVER appear in equilibrium expressions. NH 3(aq) + H 2O(liq) ¸ NH NH4+(aq) + OH-(aq) S(s) + O 2(g) S(s) (g) ¸ SO2(g) K [SO2 ] [O2 ] 16 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3(aq) + H 2O(liq) ¸ NH NH4+(aq) + OH-(aq) K [NH4+ ][OH- ] [NH3 ] Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2(g) S(s) (g) ¸ SO2(g) 15 17 The Meaning of K 1. Can tell if a reaction is productfavored or reactant-favored. For N2(g) + 3 H2(g) ¸ 2 NH3(g) (g) Kc = [NH3 ]2 [N 2 ][H2 ]3 = 3 .5 x 108 Conc. of products is much greater of much than that of reactants at equilibrium. The reaction is strongly The productproduct- favored. Page 3 18 The Meaning of K The Meaning of K For AgCl(s) ¸ (s) Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 Conc. of products is much of much less than that of reactants at equilibrium. The reaction is strongly Ag++(aq) + Cl--(aq) Ag (aq) + Cl(aq) ¸¸ AgCl(s) AgCl(s) reactant-favored. is product-favored. is product-favored. The Meaning of K The Meaning of K 19 The Meaning of K The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. n-butane HHHH H—C—C—C—C—H HHHH K= [iso] [n] n-butane HHHH H—C—C—C—C—H HHHH iso-butane HHH H—C—C—C—H H H HCH H K= In general, all reacting chemical systems are characterized by their REACTION are REACTION QUOTIENT, Q . product concentrations reactant concentrations If Q = K, then system is at equilibrium. conc. of iso 0.35 Q= = = 2.3 conc. of n 0.15 Q (2.33) < K (2.5). Reaction is NOT at equilibrium, so [ Iso] Reaction NOT must become ________ and [n] must ____________. [n] Q= = 2.5 22 The Meaning of K product concentrations reactant concentrations If Q = K, then system is at equilibrium. If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? conc. of iso 0.35 Q= = = 2.3 0.15 conc. of n Q (2.3) < K (2.5) 23 Typical Calculations PROBLEM: Place 1.00 mol each of H 2 and I2 in a 1.00 L flask. Calc. equilibrium concentrations. Q= H2(g) + I 2(g) ¸ 2 HI(g) (g) Kc = [HI]2 = 5 5.3 [H2 ][I2 ] Page 4 24 H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] [H Initial Initial Change Equilib ¸ 21 In general, all reacting chemical systems are characterized by their REACTION are REACTION QUOTIENT, Q . iso-butane HHH H—C—C—C—H H H HCH H If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction “shift” to approach equilibrium? See Screen 16.9 = 2.5 The Meaning of K The Meaning of K [iso] 20 [I2] [HI] 1.00 1.00 1.00 0 25 H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c 26 H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c Step 1. Set up table to define EQUILIBRIUM concentrations. Step 2. Put equilibrium concentrations into K c expression. [H2] [H [I2] [HI] Initial Initial 1.00 1.00 1.00 0 Change -x -x +2x Equilib 1.00-x 1.00-x Step 3. Solve K c expression - take square root of both sides. 2x Kc = [2x]2 = 55.3 [1.00-x][1.00-x] 7.44 = [H22] = [I22] = 1.00 -- x = 0.21 M [H ] = [I ] = 1.00 x = 0.21 M [HI] = 2x = 1.58 M [HI] = 2x = 1.58 M 28 N2O4(g) N2O4(g) ¸ ¸¸ 2 NO 2(g) 2 NO 2(g) 2x 1.00-x x = 0.79 Therefore, at equilibrium where x is defined as am’t of H2 and I2 where consumed on approaching equilibrium. Nitrogen Dioxide Equilibrium 27 H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c 29 Nitrogen Dioxide Equilibrium N2O4(g) ¸¸ 2 NO2(g) 2 4(g) 2 Kc = [NO2 ]2 = 0.0059 at 298 K [N2O4 ] If initial concentration of N 2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] [N Initial 0.50 0 Change Equilib Page 5 30 Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO2(g) (g) Kc = [NO2 ]2 = 0.0059 at 298 K [N2O4 ] If initial concentration of N 2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] [N Initial 0.50 0 Change -x +2x Equilib 0.50 - x 2x 31 32 Nitrogen Dioxide Equilibrium N2O4(g) ¸¸ 2 NO2(g) 2 4(g) 2 Step 2. Substitute into Kc expression and solve. [NO2 ]2 (2x) 2 K c = 0 .0059 = = [N2O 4 ] (0.50 - x) Rearrange: Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0059 0.0029 - 0.0059x = 4x 2 0.0029 4x2 + 0.0059x - 0.0029 = 0 4x This is a QUADRATIC EQUATION This QUADRATIC ax2 + bx + c = 0 a=4 b = 0.0059 c = -0.0029 0.0059 -0.0029 Nitrogen Dioxide Equilibrium N2O4(g) ¸¸ 2 NO2(g) 2 4(g) 2 Solve the quadratic equation for x. ax2 + bx + c = 0 a=4 b = 0.0059 c = -0.0029 0.0059 -0.0029 2 - 4 ac -b b x= x= -0.0059 2a (0.0059) 2 - 4 (4)(-0.0029) 2(4) x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 Page 6 33 Nitrogen Dioxide Equilibrium N2O4(g) ¸¸ 2 NO2(g) 2 4(g) 2 x= -0.0059 (0.0059) 2 - 4 (4)(-0.0029) 2(4) x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion Conclusion [N2O4] = 0.050 - x = 0.47 M [N [NO2] = 2x = 0.052 M [NO ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Mcquade during the Fall '08 term at FSU.

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