Ch16_Equilib_1 - 1 2 3 Properties of an Equilibrium...

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Unformatted text preview: 1 2 3 Properties of an Equilibrium Attention Online Users CHEMICAL EQUILIBRIUM Chapter 16 • 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. 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 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 3+ Fe3+ + SCN -- ¸ FeSCN 2+ SCN FeSCN 2+ Chemical Equilibrium 3+ Fe3+ + SCN -- ¸ FeSCN 2+ SCN FeSCN 2+ Chemical Equilibrium 5 6 Examples of Chemical Equilibria Phase changes such as H2O(s) ¸ H2O(liq) O(liq) + 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 8 9 Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO 3(s) + H2O(liq) + CO2(g) O(liq) ¸ Ca2+(aq) + 2 HCO 3-(aq) Ca aq) aq) Chemical Equilibria CaCO 3(s) + H2O(liq) + CO2(g) CaCO O(liq) ¸ Ca2+(aq) + 2 HCO 3-(aq) Ca aq) aq) At a given T and P of CO 2, and [HCO3-] can be found from the [Ca2+] THE EQUILIBRIUM CONSTANT THE EQUILIBRIUM CONSTANT 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 K= [C]c [D]d [A]a [B]b conc. of reactants equilibrium constant EQUILIBRIUM CONSTANT. If K is known, then we can predict concs. of concs. products or reactants. 10 11 12 Determining K 2 NOCl(g) ¸ 2 NO(g) + Cl 2(g) NOCl(g) NO(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. mol/L Calculate K. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] NOCl] Before 2.00 0 0 Change Equilibrium 0.66 Determining K 2 NOCl(g) ¸ 2 NO(g) + Cl 2(g) NOCl(g) NO(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. mol/L Calculate K. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] NOCl] Before 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 Determining K 2 NOCl(g) ¸ 2 NO(g) + Cl 2(g) NOCl(g) NO(g) Before Change Equilibrium [NOCl] NOCl] 2.00 -0.66 1.34 [NO] 0 +0.66 0.66 [Cl2] 0 +0.33 0.33 K [NO] 2[Cl2 ] [NOCl]2 [NO]2 [Cl2 ] [NOCl]2 = (0.66)2 (0.33) (1.34)2 = 0.080 K Page 2 13 14 15 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2(g) ¸ SO2(g) SO Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2(g) ¸ SO2(g) SO Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3(aq) + H 2O(liq) ¸ aq) O(liq) NH4+(aq) + OH-(aq) aq) aq) K [SO2 ] [O2 ] 16 17 18 Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3(aq) + H 2O(liq) ¸ aq) O(liq) NH4+(aq) + OH-(aq) aq) aq) The Meaning of K 1. Can tell if a reaction is productfavored or reactant-favored. For N2(g) + 3 H2(g) ¸ 2 NH3(g) NH The Meaning of K The Meaning of K For AgCl(s) ¸ AgCl(s) Ag+(aq) + Cl-(aq) aq) aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 [Cl Conc. of products is much Conc. much less than that of than reactants at equilibrium. The reaction is strongly Ag++(aq) + Cl--(aq) aq) aq) Ag (aq) + Cl(aq) ¸ AgCl(s) ¸ AgCl(s) AgCl(s) reactant-favored. Kc = [NH3 ]2 [N 2 ][H2 ]3 = 3 .5 x 108 Conc. of products is much greater Conc. much than that of reactants at equilibrium. The reaction is strongly K [NH4+ ][OH- ] [NH3 ] productproduct- is product-favored. is product-favored. favored. Page 3 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 [iso] [n] iso-butane HHH H—C—C—C—H H H HCH H = 2.5 19 The Meaning of K The Meaning of K n-butane HHHH H—C—C—C—C—H HHHH [iso] [n] iso-butane HHH H—C—C—C—H H H HCH H = 2.5 20 The Meaning of K In general, all reacting chemical systems are characterized by their REACTION REACTION QUOTIENT, Q . 21 Q= K= product concentrations reactant concentrations K= If [iso] = 0.35 M and [n] = 0.15 M, are you [iso] at equilibrium? If not, which way does the reaction “shift” to approach equilibrium? See Screen 16.9 If Q = K, then system is at equilibrium. If [iso] = 0.35 M and [n] = 0.15 M, are you [iso] at equilibrium? conc. of iso 0.35 Q= = = 2.3 0.15 conc. of n Q (2.3) < K (2.5) The Meaning of K The Meaning of K In general, all reacting chemical systems are characterized by their REACTION REACTION QUOTIENT, Q . 22 23 24 Typical Calculations PROBLEM: Place 1.00 mol each of H 2 and I2 in and in a 1.00 L flask. Calc. equilibrium Calc. concentrations. H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] Initial Change Equilib 1.00 [I2] 1.00 [HI] 0 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] [Iso] must become ________ and [n] must ____________. Q= H2(g) + I 2(g) ¸ 2 HI(g) HI(g) Kc = [HI]2 = 5 5.3 [H2 ][I2 ] ¸ Page 4 25 26 27 H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] Initial Change Equilib 1.00 -x 1.00-x [I2] 1.00 -x 1.00-x [HI] 0 +2x 2x H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c Step 2. Put equilibrium concentrations into K c expression. expression. H2(g) + I2(g) ¸ 2 HI(g), Kc = 55.3 ¸ 2 2 c Step 3. Solve K c expression - take expression square root of both sides. Kc = [2x]2 = 55.3 [1.00-x][1.00-x] 7.44 = 2x 1.00-x x = 0.79 Therefore, at equilibrium [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 where x is defined as am’t of H2 and I2 and consumed on approaching equilibrium. 28 29 30 Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO 2(g) N2O4(g) ¸ 2 NO 2(g) ¸ Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO2(g) NO 2 ¸ 24 Kc = [NO2 ]2 = 0.0059 at 298 K [N2O4 ] Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO2(g) NO Kc = [NO2 ]2 = 0.0059 at 298 K [N2O4 ] If initial concentration of N 2O4 is 0.50 M, what are is the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] Initial 0.50 0 Change Equilib If initial concentration of N 2O4 is 0.50 M, what are is the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] Initial 0.50 0 Change -x +2x Equilib 0.50 - x 2x Page 5 31 32 33 Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO2(g) NO 2 ¸ 24 Step 2. Substitute into Kc expression and solve. expression [NO2 ]2 (2x) 2 K c = 0 .0059 = = [N2O 4 ] (0.50 - x) Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO2(g) NO 2 ¸ 24 Solve the quadratic equation for x. ax2 + bx + c = 0 bx a=4 b = 0.0059 c = -0.0029 2 - 4 ac -b b Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO2(g) NO 2 ¸ 24 x= -0.0059 (0.0059) 2 - 4 (4)(-0.0029) 2(4) Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0029 - 0.0059x = 4x 2 0.0029 4x2 + 0.0059x - 0.0029 = 0 4x 0.0059x x= 2a (0.0059) 2 - 4 (4)(-0.0029) 2(4) This is a QUADRATIC EQUATION QUADRATIC ax2 + bx + c = 0 ax bx a=4 b = 0.0059 c = -0.0029 x= -0.0059 x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 -0.00074 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 x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 -0.00074 Page 6 ...
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This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.

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