chapter 13 chemical equilibrium: how can things that are moving seem to be standing still?

Chapter 13Chemical Equilibrium:

How can things that are moving seem to be standing

still?

Chapter 13

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

13.1The Equilibrium Condition

13.2 The Equilibrium Constant

13.3 Equilibrium Expressions Involving Pressures

13.4 Heterogeneous Equilibria

13.5 Applications of the Equilibrium Constant

13.6 Solving Equilibrium Problems

13.7 Le Châtelier’s Principle

Section 13.1

The Equilibrium Condition

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Objectives

• To define the Chemical Equilibrium• To describe the equilibrium condition in terms of

reactant and product concentration• To describe the equilibrium condition in terms of

forward and reverse reaction rates

Section 13.1


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A Gaseous Solute C = kP

Section 13.1


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Liquid/Vapor Equilibrium

Section 13.1


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Chemical Equilibrium

• So far, we have seen chemical equations written as:

• H2O(g) + CO(g) H2(g) + CO2(g)

• Or stated as such that No Reaction occurs

• This may be an oversimplification, whereas…

• H2O(g) + CO(g) H2(g) + CO2(g)

• Is a more thorough description of what is happening in virtually every chemical reaction.

• But, what do these symbols mean?

Section 13.1


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Chemical Equilibrium

• The double arrow sign, , represents an equilibrium condition.

• Equilibrium is: The state where the concentrations of all reactants and products remain constant with time.

• On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

• Equilibrium is : Macroscopically static • Microscopically dynamic

Section 13.1


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N2(g) + 3H2(g) 2NH3(g)

Concentrations reach constant levels where the rate of the forward reaction equals the rate of the reverse reaction.

Section 13.1


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The Changes with Time in the Rates of Forward and Reverse Reactions

• Bees wake up and start leaving to collect honey…time passes. • the # of bees leaving = the # of bees entering…equilibrium

Section 13.1


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Concept Check

Consider an equilibrium mixture in a closed vessel reacting according to the equation:

H2O(g) + CO(g) H2(g) + CO2(g)

You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

Section 13.1


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Objectives Review

• To define the Chemical Equilibrium• To describe the equilibrium condition in terms of

reactant and product concentration• To describe the equilibrium condition in terms of

forward and reverse reaction rates

Section 13.2

Atomic MassesThe Equilibrium Constant

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Objectives

• To define the equilibrium constant (K)• To write the equilibrium constant expression for

various chemical reactions

Section 13.2


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Consider the following reaction at equilibrium:

jA + kB lC + mD

• A, B, C, and D = chemical species.• Square brackets = concentrations of species at equilibrium.• j, k, l, and m = coefficients in the balanced equation.• K = equilibrium constant (given without units).

j

l

k

m

[B][A]

[D] [C]K =

Section 13.2


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Write the equilibrium constant expression:

• H2O(g) + CO(g) H2(g) + CO2(g)

• 4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

Section 13.2


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Conclusions About the Equilibrium Expression

• Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse.

• When balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus

Knew = (Koriginal)n.

• K values are usually written without units.

Section 13.2


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• K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially.

• For a reaction, at a given temperature, there are many equilibrium positions but only one value for K. Equilibrium position is a set of equilibrium

concentrations. K = [Products] [Reactants]

More Conclusions About the Equilibrium Expression

Section 13.2


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Objectives Review

• To define the equilibrium constant (K)• To write the equilibrium constant expression for

various chemical reactions • Work Session: Pg 613 # 1, 6, 12, 17& 18 a&b,

19

Section 13.3

The Mole Equilibrium Expressions Involving Pressures

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Objectives

• To write the Equilibrium Expression for gas reactions (Kp)

• To calculate K from Kp • To see how the 5-Step Problem Solving Method

can help you calculate without having to “KNOW” how to do a problem…

Section 13.3


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• K involves concentrations.

• Kp involves pressures.

Coefficient

Coefficient

Pr oducts =

Reac tantsK

Coefficient

Coefficient

(Pr odPr essure) =

(ReactPr essure)Kp

Section 13.3


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Example

N2(g) + 3H2(g) 2NH3(g)

Write the K and Kp expressions:

3

2 2

2

NH

p 3

N H

P =

P PK

2

33

2 2

NH =

N HK

Section 13.3


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Example

N2(g) + 3H2(g) 2NH3(g)

Calculate Kp given the equilibrium pressures at a certain temperature:

3

2

2

2NH

1N

3H

= 2.9 10 atm

= 8.9 10 atm

= 2.9 10 atm

P

P

P

Section 13.3


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Example

N2(g) + 3H2(g) 2NH3(g)

3

2 2

2

NH

p 3

N H

P =

P PK

22

p 31 3

2.9 10 =

8.9 10 2.9 10

K

4p = 3.9 10K

Section 13.3


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The Relationship Between K and Kp

Kp = K(RT)Δn

• Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.

• R = 0.08206 L·atm/mol·K• T = temperature (in kelvin)

• Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C.

Section 13.3


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Example

N2(g) + 3H2(g) 2NH3(g)

Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C.

p

2 44

7

=

3.9 10 = 0.08206 L atm/mol K 308K

= 2.5 10

nK K RT

K

K

Section 13.3


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The Power of the 5 Step!

• Take a look at the AP reference sheet…• Even though you may only see an equation once,

or perhaps never, and you may think that you forget or never learned how to use the equation…..

• The 5 Step is the “HOW” of working with equations…

• Being more familiar is more comfortable, however, take comfort in the Power of the 5 Step!

Section 13.3


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Objectives Review

• To write the Equilibrium Expression for gas reactions (Kp)

• To calculate K from Kp • To see how the 5-Step Problem Solving Method

can help you calculate without having to “KNOW” how to do a problem…

• Work Session pg 615 # 21, 26

Section 13.4

Heterogeneous Equilibria

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Objectives 13.4 and 13.5

• To define homogeneous and heterogeneous equilibria

• To understand why pure solids and liquids are left out of the equilibrium expression

• To evaluate the extent of a reaction based on the numerical value of K

• To use the Reaction Quotient (Q) to determine if a system is at equilibrium and what directional shift will occur to reach EQ

Section 13.4


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Homogeneous Equilibria

• Homogeneous equilibria – involve the same phase:

N2(g) + 3H2(g) 2NH3(g)

HCN(aq) H+(aq) + CN-(aq)

Section 13.4


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• Heterogeneous equilibria – involve more than one phase:

2KClO3(s) 2KCl(s) + 3O2(g)

2H2O(l) 2H2(g) + O2(g)

Section 13.4


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• The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids

are constant. (Think of them as 1)

2KClO3(s) 2KCl(s) + 3O2(g)

3

2 = OK

Section 13.5

Applications of the Equilibrium Constant

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• A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right. Reaction goes essentially to completion. A + B AB

The Extent of a Reaction

Coefficient

Coefficient

Pr oducts =

Reac tantsK

Section 13.5


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• A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left. Reaction does not occur to any

significant extent. A + B No Rx

The Extent of a Reaction

Coefficient

Coefficient

Pr oducts =

Reac tantsK

Section 13.5


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Concept Check

If the equilibrium lies to the right, the value for K is __________.

large (or >1)

If the equilibrium lies to the left, the value for K is ___________.

small (or <1)

Coefficient

Coefficient

Pr oducts =

Reac tantsK

Section 13.5


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Exercise

Consider the reaction represented by the equation:

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

• Trial #1:

6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M.

What is the value for the equilibrium constant for this reaction? Need [at Equilibrium] –Use ICE Table…

2

3

FeSCN =

Fe SCNK

Section 13.5


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Set up ICE Table (Initial, Change, Equil’ Conditions)

Fe3+(aq) + SCN–(aq) FeSCN2+(aq)

Initial 6.00 10.00 0.00

Change - 4.00 – 4.00 +4.00

Equilibrium 2.00 6.00 4.00

K = 0.333 (1:1 Stoich)

2

3

FeSCN 4.00 = =

2.00 6.00 Fe SCN

MK

M M

Section 13.5


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ExerciseConsider the reaction represented by the equation:


• Trial #2:

Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same temperature as Trial #1) (Means you know K)

Equilibrium: ? M FeSCN2+(aq) (ICE TABLE)

(NEXT SLIDE)

Section 13.5


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Fe3+(aq) + SCN-(aq) FeSCN2+(aq)Trial #2:

Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same temperature as Trial #1) (Means you know K)(ICE TABLE)

I

CE

5.00 M FeSCN2+

2

3

FeSCN =

Fe SCNK

Section 13.5


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• Trial #3:

Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq)

Equilibrium: ? M FeSCN2+(aq) (ICE Set Up K)

(Don’t solve)

3.00 M FeSCN2+

2

3

FeSCN =

Fe SCNK

Section 13.5


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Exercise (Start with the ICE, Keq, and Work Through the Algebra, no Quadratic needed)

Consider the reaction represented by the equation (assume K = 0.333) :


Fe3+ SCN- FeSCN2+

Trial #1 9.00 M 5.00 M 1.00 M

Trial #2 3.00 M 2.00 M 5.00 M

Trial #3 2.00 M 9.00 M 6.00 M

Find the equilibrium concentrations for all species.

Section 13.5


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Exercise (Answer)

Trial #1: [Fe3+] = 6.00 M; [SCN-] = 2.00 M; [FeSCN2+] = 4.00 M



Section 13.5


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• To determine if a system is at EQ, apply the law of mass action using initial (instantaneous) concentrations instead of equilibrium concentrations.

• In other words, calculate the Q expression and see if the conditions are the same as the K expression…

Reaction Quotient, Q

Coefficient

Coefficient

Pr oductsQ = K ?

Reac tants

Section 13.5


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• Q = K; The system is at equilibrium. No shift will occur.

• Q > K; Consuming products and forming reactants, until

equilibrium is achieved. The system shifts to the …… ….left.


Coefficient

Coefficient

Pr oductsQ = K ?

Reac tants

Section 13.5


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• Q < K; Consuming reactants and forming products, to

attain equilibrium. The system shifts to the ………… ………right.


Coefficient

Coefficient

Pr oductsQ = K ?

Reac tants

Section 13.5


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For the synthesis of ammonia at 500o C, Keq = 6.0 X 10-2. Predict if the following system is in equilibrium. If not, indicate the direction in which the system will shift in order to reach eq’.

a. [NH3] = 1.0 X 10-3M; [N2] = 1.0 X 10 -5M; [H2] = 2.0 X 10 -3 M

a. 1.25 X 107 Left shift

b. [NH3] = 2.0 X 10-4M; [N2] = 1.5 X 10 -5M; [H2] = 3.54 X 10 -1 M

@ EQ’M

c. [NH3] = 1.0 X 10-4M; [N2] = 5.0M; [H2] = 1.0 X 10 -2 M

2.0 X 10 -3 Right Shift

Section 13.5


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http://www.usca.edu/chemistry/genchem/sigfig.htm

Section 13.5


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Objectives Review 13.4 and 13.5

• To define homogeneous and heterogeneous equilibria

• To understand why pure solids and liquids are left out of the equilibrium expression

• To evaluate the extent of a reaction based on the numerical value of K

• To use the Reaction Quotient (Q) to determine if a system is at equilibrium and what directional shift will occur to reach EQ

• Work Session: Pg 615 # 29, 31, 33, 37

Section 13.6

Solving Equilibrium Problems

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Objectives

• To practice using the ICE method for solving Equilibrium Problems

Section 13.6


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1) Write the balanced equation for the reaction.

2) Write the equilibrium expression using the law of mass action.

3) List the initial concentrations.

4) Calculate Q, and determine the direction of the shift to equilibrium.


Section 13.6


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5) Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations.

6) Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown.

7) Check your calculated equilibrium concentrations by making sure they give the correct value of K.


Section 13.6


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Concept Check

A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation:

NH3(g) N2(g) + H2(g)

At equilibrium 1.00 mol of ammonia remains.

Calculate the value for K. (Balance)(ICE)

K = 1.69

Section 13.6


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Concept Check

Section 13.6


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Concept Check

A 1.00 mol sample of N2O4(g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation:

N2O4(g) 2NO2(g)

K = 4.00 x 10-4

Calculate the equilibrium concentrations of: N2O4(g) and NO2(g). (ICE,

Concentration of N2O4 = 0.097 M

Concentration of NO2 = 6.32 x 10-3 M

Section 13.6


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Concept Check

Section 13.6


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Objectives Review

• To practice using the ICE method for solving Equilibrium Problems

• Work Session: page 616 39, 41, 43*, 45 (setup only), 47, 51(setup only), 53 (setup only)

Section 13.7

Le Châtelier’s Principle

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Objectives

• To understand Le Chatelier’s Principle• To predict shifts in equilibrium

• HPhen (aq) H+ (aq) + Phen- (aq)

• CLEAR PINK

Section 13.7


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• Le Chatelier’s Principle

• If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

Section 13.7


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Effects of Changes on the System

1. Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs.

2. Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product).

Section 13.7


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Effects of Changes on the System

3. Pressure:

a) The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs.

b) Addition of inert gas does not affect the equilibrium position.

c) Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas.

Section 13.7


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Section 13.7


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Equilibrium Decomposition of N2O4

Section 13.7


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Predicting Shifts in Equilibria

• Which way will the equilibrium shift if: (don’t forget volume…)

N2(g) + 3H2(g) 2NH3(g)

HCN(aq) H+(aq) + CN-(aq) H = -12 kJ

Section 13.7


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Objectives

• To understand Le Chatelier’s Principle• To predict shifts in equilibrium• Work Session: pg 617 # 57, 59, 63

Section 13.7


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Objectives

chapter 13 chemical equilibrium: how can things that are moving seem to be standing still?

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