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CHEMICAL EQUILIBRIUM Chapter 16 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

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2 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction

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3 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O

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4 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Properties of an Equilibrium Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Equilibrium systems are DYNAMIC (in constant motion)DYNAMIC (in constant motion) REVERSIBLEREVERSIBLE can be approached from either directioncan be approached from either direction Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O ---> Co(H 2 O) 6 Cl 2

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5 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Chemical Equilibrium Fe 3+ + SCN - FeSCN 2+ Fe(H 2 O) 6 3+ Fe(SCN)(H 2 O) 5 3+ + SCN - + + H 2 O

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6 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Chemical Equilibrium Fe 3+ + SCN - FeSCN 2+ After a period of time, the concentrations of reactants and products are constant.After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.The forward and reverse reactions continue after equilibrium is attained.

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7 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Examples of Chemical Equilibria Phase changes such as H 2 O(s) H 2 O(liq)

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8 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3 - (aq) CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3 - (aq)

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9 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Chemical Equilibria CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3 - (aq) CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3 - (aq) At a given T and P of CO 2, [Ca 2+ ] and [HCO 3 - ] can be found from the EQUILIBRIUM CONSTANT.

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10 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B c C + d D a A + b B c C + d D the following is a CONSTANT (at a given T)

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11 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B c C + d D a A + b B c C + d D the following is a CONSTANT (at a given T) If K is known, then we can predict concs. of products or reactants.

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12 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Determining K 2 NOCl(g) 2 NO(g) + Cl 2 (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][Cl 2 ] [NOCl][NO][Cl 2 ] Before2.0000 Change Equilibrium0.66

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13 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Determining K 2 NOCl(g) 2 NO(g) + Cl 2 (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][Cl 2 ] [NOCl][NO][Cl 2 ] Before2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

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14 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Determining K 2 NOCl(g) 2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] [NOCl][NO][Cl 2 ] Before2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

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15 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Determining K 2 NOCl(g) 2 NO(g) + Cl 2 (g) [NOCl][NO][Cl 2 ] Before2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33

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16 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g)

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17 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g)

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18 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq)

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19 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq)

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20 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Adding equations for reactions S(s) + O 2 (g) SO 2 (g) K 1 = [SO 2 ] / [O 2 ] S(s) + O 2 (g) SO 2 (g) K 1 = [SO 2 ] / [O 2 ] SO 2 (g) + 1/2 O 2 (g) SO 3 (g) SO 2 (g) + 1/2 O 2 (g) SO 3 (g) K 2 = [SO 3 ] / [SO 2 ][O 2 ] 1/2 K 2 = [SO 3 ] / [SO 2 ][O 2 ] 1/2 NET EQUATION S(s) + 3/2 O 2 (g) SO 3 (g) S(s) + 3/2 O 2 (g) SO 3 (g)

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21 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Adding equations for reactions S(s) + O 2 (g) SO 2 (g) K 1 = [SO 2 ] / [O 2 ] S(s) + O 2 (g) SO 2 (g) K 1 = [SO 2 ] / [O 2 ] SO 2 (g) + 1/2 O 2 (g) SO 3 (g) SO 2 (g) + 1/2 O 2 (g) SO 3 (g) K 2 = [SO 3 ] / [SO 2 ][O 2 ] 1/2 K 2 = [SO 3 ] / [SO 2 ][O 2 ] 1/2 NET EQUATION S(s) + 3/2 O 2 (g) SO 3 (g) S(s) + 3/2 O 2 (g) SO 3 (g)

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22 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O 2 (g) SO 3 (g) S(s) + 3/2 O 2 (g) SO 3 (g) 2 S(s) + 3 O 2 (g) 2 SO 3 (g) 2 S(s) + 3 O 2 (g) 2 SO 3 (g)

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23 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O 2 (g) SO 3 (g) S(s) + 3/2 O 2 (g) SO 3 (g) 2 S(s) + 3 O 2 (g) 2 SO 3 (g) 2 S(s) + 3 O 2 (g) 2 SO 3 (g)

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24 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O 2 (g) SO 3 (g) S(s) + 3/2 O 2 (g) SO 3 (g) 2 S(s) + 3 O 2 (g) 2 SO 3 (g) 2 S(s) + 3 O 2 (g) 2 SO 3 (g)

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25 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Changing direction S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g)

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26 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Changing direction S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g)

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27 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Changing direction S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g) SO 2 (g) S(s) + O 2 (g) SO 2 (g) S(s) + O 2 (g)

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28 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Writing and Manipulating K Expressions Concentration Units We have been writing K in terms of mol/L. These are designated by K c But with gases, P = (n/V)RT = conc RT P is proportional to concentration, so we can write K in terms of P. These are designated by K p. K c and K p may or may not be the same.

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29 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product- favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

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30 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product- favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

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31 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product- favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Conc. of products is much greater than that of reactants at equilibrium.

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32 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product- favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product- favored.

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33 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K For AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 Conc. of products is much less than that of reactants at equilibrium. The reaction is strongly reactant-favored. Ag + (aq) + Cl - (aq) AgCl(s) AgCl(s) is product-favored. Ag + (aq) + Cl - (aq) AgCl(s) AgCl(s) is product-favored.

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34 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

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35 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Which way does the reaction “shift” to approach equilibrium? See Screen 16.9.

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36 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium.

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37 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Q (2.3) < K (2.5).

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38 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. Q (2.33) < K (2.5). Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must ____________.

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39 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Typical Calculations PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calc. equilibrium concentrations. H 2 (g) + I 2 (g) ¸ 2 HI(g)

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40 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H 2 ][I 2 ][HI] [H 2 ][I 2 ][HI] Initial 1.001.000 ChangeEquilib

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41 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H 2 ][I 2 ][HI] [H 2 ][I 2 ][HI] Initial 1.001.000 Change-x-x+2x Equilib1.00-x1.00-x2x where x is defined as am’t of H 2 and I 2 consumed on approaching equilibrium.

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42 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 2. Put equilibrium concentrations into K c expression.

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43 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 3. Solve K c expression - take square root of both sides.

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44 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 3. Solve K c expression - take square root of both sides.

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45 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 3. Solve K c expression - take square root of both sides. x = 0.79

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46 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 3. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M

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47 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 3. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

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48 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g)

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49 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N 2 O 4 ][NO 2 ] [N 2 O 4 ][NO 2 ] Initial0.500 ChangeEquilib

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50 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N 2 O 4 ][NO 2 ] [N 2 O 4 ][NO 2 ] Initial0.500 Change-x+2x Equilib0.50 - x2x

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51 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Step 2. Substitute into K c expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0029 - 0.0059x = 4x 2 0.0029 - 0.0059x = 4x 2 4x 2 + 0.0059x - 0.0029 = 0 4x 2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029

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52 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029

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53 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029

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54 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029 x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027

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55 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) 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 [N 2 O 4 ] = 0.050 - x = 0.47 M [N 2 O 4 ] = 0.050 - x = 0.47 M [NO 2 ] = 2x = 0.052 M [NO 2 ] = 2x = 0.052 M

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