Chapter 15 Chemical Equilibrium Lecture Presentation © 2012 Pearson Education, Inc

Chapter 15

Chemical Equilibrium

Lecture Presentation

© 2012 Pearson Education, Inc.

Equilibrium

The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.


Equilibrium

The Concept of Equilibrium• As a system

approaches equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.


Once equilibrium is achieved, the amount of each reactant and product remains constant

Equilibrium

The Equilibrium Constant

• Consider the generalized reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD


Equilibrium



N2O4(g) 2NO2(g)

Keq =[NO2]2

[N2O4]

Equilibrium


Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Kp =(PC

c) (PDd)

(PAa) (PB

b)


Equilibrium

The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2

2NO2 (g) 2NO (g) + O2 (g)

14.2

Kp = 2PNO PO

2

PNO2

2

PO2 = Kp

PNO2

2

PNO2

PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

Equilibrium

Relationship Between Kc and Kp

where

Kp = Kc (RT)n

n = (moles of gaseous product) (moles of gaseous reactant)


Equilibrium

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)

Kc = [COCl2]

[CO][Cl2]=

0.140.012 x 0.054

= 220

Kp = Kc(RT)n

n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

Kp = 220 x (0.0821 x 347)-1 = 7.7

14.2

Equilibrium

Equilibrium Can Be Reached from Either Direction

It doesn’t matter whether we start with N2 and H2 or whether we start with NH3, we will have the same proportions of all three substances at equilibrium.


Equilibrium

What Does the Value of K Mean?

• If K>>1, the reaction is product-favored; product predominates at equilibrium.

• If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.


Equilibrium

Manipulating Equilibrium Constants

The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction:

Kc = = 0.212 at 100 C[NO2]2

[N2O4]

Kc = = 4.72 at 100 C[N2O4][NO2]2

N2O4(g)2NO2(g)

N2O4(g) 2NO2(g)


Equilibrium

Manipulating Equilibrium Constants

The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number:

Kc = = 0.212 at 100 C[NO2]2

[N2O4]

Kc = = (0.212)2 at 100 C[NO2]4

[N2O4]24NO2(g)2N2O4(g)

N2O4(g) 2NO2(g)


Equilibrium

A + B C + D

C + D E + F

A + B E + F

Kc =‘[C][D][A][B]

Kc =‘‘[E][F][C][D]

[E][F][A][B]

Kc =

Kc ‘Kc ‘‘Kc

Kc = Kc ‘‘Kc ‘ x

If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

Equilibrium

Sample Exercise 15.5 Combining Equilibrium Expressions

Given the reactions

determine the value of Kc for the reaction

SolutionAnalyze We are given two equilibrium equations and the corresponding equilibrium constants and areasked to determine the equilibrium constant for athird equation, which is related to the first two.

Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation.

Equilibrium

Sample Exercise 15.5 Combining Equilibrium ExpressionsContinued

SolveIf we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get

Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives

Now we have two equations that sum to give the net equation, and we can multiply the individualKc values to get the desired equilibrium constant.

Equilibrium

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

CaCO3 (s) CaO (s) + CO2 (g)

Kc =‘[CaO][CO2]

[CaCO3][CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kc x‘[CaCO3][CaO]

Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

14.2

Equilibrium

Kc = [Pb2+] [Cl]2

PbCl2(s) Pb2+(aq) + 2Cl(aq)


Example

Equilibrium

At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

[Br]2

[Br2]Kc = Kc =

(0.012 + 2x)2

0.063 - x= 1.1 x 10-3 Solve for x

14.4

Equilibrium

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.0648 M14.4

Equilibrium

Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations

SolutionAnalyze We are given the initial concentrations of H2 and l2 and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant Kc for

Solve First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table.

Second, we calculate the change in HI concentration, which is the difference Between the equilibrium and initial values:

A closed system initially containing 1.000 103 M H2 and 2.000 103 M I2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 103 M. Calculate Kc at 448 C for the reaction taking place, which is

Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant.

Change in [HI] = 1.87 103 M 0 = 1.87 103 M

Equilibrium

Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H2] and [I2]:

Fourth, we calculate the equilibrium concentrations of H2 and I2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed:

Our table now looks like this (with equilibrium concentrations in blue for emphasis):

Continued

[H2] = 1.000 103 M 0.935 103 M = 0.065 103 M[I2] = 2.000 103 M 0.935 103 M = 1.065 103 M

Notice that the entries for the changes are negative when a reactant is consumed and

positive when a product is formed.


Equilibrium

Finally, we use the equilibrium-constant expression to calculate the equilibrium constant:

Continued


Equilibrium

The Reaction Quotient (Q)• Q gives the same ratio the equilibrium

expression gives, but for a system that is not at equilibrium.

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.


Q = [Cinitial]c[Dinitial]d

[Ainitial]a[Binitial]b

aA + bB cC + dD

Equilibrium

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

14.4

Equilibrium

Sample Exercise 15.10 Predicting the Direction of Approach to Equilibrium

SolutionAnalyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium.

Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed.

SolveThe initial concentrations are

The reaction quotient is therefore

At 448 C the equilibrium constant Kc for the reaction

is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 102 mol of HI, 1.0 102 mol of H2, and 3.0 102 of I2 in a 2.00-L container.

Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to reachequilibrium; the reaction as written proceeds left to right to attain equilibrium.

Equilibrium

Le Châtelier’s Principle

“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”


Equilibrium


• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left14.5

aA + bB cC + dD

AddAddRemove Remove

Equilibrium


• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

14.5

Equilibrium

uncatalyzed catalyzed

14.5Catalyst does not change equilibrium constant or shift equilibrium.

• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner


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Chapter 15 Chemical Equilibrium Lecture Presentation © 2012 Pearson Education, Inc