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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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**Properties of an Equilibrium**

Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction

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**Properties of an Equilibrium**

Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction Pink to blue Co(H2O)6Cl2 ---> Co(H2O)4Cl H2O

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**Properties of an Equilibrium**

Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction Pink to blue Co(H2O)6Cl2 ---> Co(H2O)4Cl H2O Blue to pink Co(H2O)4Cl H2O ---> Co(H2O)6Cl2

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**Chemical Equilibrium Fe3+ + SCN- FeSCN2+**

Fe(H2O)63+ + SCN- Fe(SCN)(H2O)53+ + H2O

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**Chemical Equilibrium Fe3+ + SCN- FeSCN2+**

After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.

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**Examples of Chemical Equilibria**

Phase changes such as H2O(s) H2O(liq)

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**Examples of Chemical Equilibria**

Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

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Chemical Equilibria CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq) At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

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**THE EQUILIBRIUM CONSTANT**

For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T)

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**THE EQUILIBRIUM CONSTANT**

For any type of chemical equilibrium of the type 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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**Determining K 2 NOCl(g) 2 NO(g) + Cl2(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 Change Equilibrium

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**Determining K 2 NOCl(g) 2 NO(g) + Cl2(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 Change Equilibrium

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**Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2]**

Before Change Equilibrium

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**Determining K 2 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2]**

Before Change Equilibrium

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**Writing and Manipulating K Expressions**

Solids and liquids NEVER appear in equilibrium expressions. S(s) + O2(g) SO2(g)

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**Writing and Manipulating K Expressions**

Solids and liquids NEVER appear in equilibrium expressions. S(s) + O2(g) SO2(g)

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**Writing and Manipulating K Expressions**

Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)

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**Writing and Manipulating K Expressions**

Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)

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**Writing and Manipulating K Expressions**

Adding equations for reactions S(s) + O2(g) SO2(g) K1 = [SO2] / [O2] SO2(g) + 1/2 O2(g) SO3(g) K2 = [SO3] / [SO2][O2]1/2 NET EQUATION S(s) /2 O2(g) SO3(g)

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**Writing and Manipulating K Expressions**

Adding equations for reactions S(s) + O2(g) SO2(g) K1 = [SO2] / [O2] SO2(g) + 1/2 O2(g) SO3(g) K2 = [SO3] / [SO2][O2]1/2 NET EQUATION S(s) /2 O2(g) SO3(g)

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**Writing and Manipulating K Expressions**

Changing coefficients S(s) /2 O2(g) SO3(g) 2 S(s) O2(g) SO3(g)

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**Writing and Manipulating K Expressions**

Changing coefficients S(s) /2 O2(g) SO3(g) 2 S(s) O2(g) SO3(g)

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**Writing and Manipulating K Expressions**

Changing coefficients S(s) /2 O2(g) SO3(g) 2 S(s) O2(g) SO3(g)

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**Writing and Manipulating K Expressions**

Changing direction S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g)

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**Writing and Manipulating K Expressions**

Changing direction S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g)

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**Writing and Manipulating K Expressions**

Changing direction S(s) + O2(g) SO2(g) SO2(g) S(s) + O2(g)

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**Writing and Manipulating K Expressions**

Concentration Units We have been writing K in terms of mol/L. These are designated by Kc 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 Kp. Kc and Kp may or may not be the same.

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The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N2(g) H2(g) NH3(g)

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The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N2(g) H2(g) NH3(g)

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The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N2(g) H2(g) NH3(g) Conc. of products is much greater than that of reactants at equilibrium.

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The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N2(g) H2(g) NH3(g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

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**The Meaning of K For AgCl(s) Ag+(aq) + Cl-(aq)**

Kc = [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) is product-favored.

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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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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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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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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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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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**Typical Calculations H2(g) + I2(g) ¸ 2 HI(g)**

PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calc. equilibrium concentrations. H2(g) + I2(g) ¸ 2 HI(g)

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial Change Equilib

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial Change -x -x +2x Equilib 1.00-x 1.00-x 2x where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 2. Put equilibrium concentrations into Kc expression.

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides.

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides.

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides. x = 0.79

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H2] = [I2] = x = M

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H2(g) + I2(g) HI(g), Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H2] = [I2] = x = M [HI] = 2x = M

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] Initial Change Equilib

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N2O4] [NO2] Initial Change -x +2x Equilib x 2x

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

Step 2. Substitute into Kc expression and solve. Rearrange: ( x) = 4x2 x = 4x2 4x x = 0 This is a QUADRATIC EQUATION ax bx c = 0 a = 4 b = c =

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

Solve the quadratic equation for x. ax bx c = 0 a = 4 b = c =

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

Solve the quadratic equation for x. ax bx c = 0 a = 4 b = c =

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

Solve the quadratic equation for x. ax bx c = 0 a = 4 b = c = x = ± 1/8(0.046)1/2 = ± 0.027

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**Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)**

x = or But a negative value is not reasonable. Conclusion [N2O4] = x = M [NO2] = 2x = M

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