Chapt15 Equilibrium

8/10/2019 Chapt15 Equilibrium

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Chemical EquilibriumChapter 15

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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Chapter 15; Equilibrium

I. Dynamic Equilibrium

II. Equilibrium Constant; Kc and Kp

III. Homogenous vs. Heterogeneous

Equilibrium

IV. Calculation of K from Equilibrium

Concentrations


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Chapter 15; Equilibrium

V. Predicting Reaction Direction for

Nonequilibria Mixtures

Reaction Quotient, Q

VI. Le Chatliers Principle

VII. Calculating Equilibrium Concentrations

from Initial Concentrations and K Square Root

Quadratic Formula


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Equi l ibr iumis a state in which there are no observable

changes as time goes by.

Chemical equi l ibr iumis achieved when:

the rates of the forward and reverse reactions are equal and

they are not zero.

the concentrations of the reactants and products remain

constantChemical equilibrium

N2O4(g)

14.1

2NO2(g)


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N2O4(g) 2NO2(g)

Start with NO2 Start with N2O4 Start with NO2& N2O4

equilibrium

equilibrium

equilibrium

14.1

Equilibrium favors the reactant side


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14.1

constant


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N2O4(g) 2NO2(g)

= 4.63 x 10-3

K=

[NO2]2

[N2O4]

aA + bB cC + dD

K=[C]c[D]d

[A]a[B]b

K>> 1

K


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Homogenous equi l ibr ium applies to reactions in which all

reacting species are in the same phase.

N2O4(g) 2NO2(g)

Kc=[NO2]

2

[N2O4]Kp=

NO2P2

N2O4P

In most cases

KcKp

aA (g)+ bB (g) cC (g)+ dD (g)

14.2

Kp= Kc(RT)Dn

Dn = moles of gaseous productsmoles of gaseous reactants

= (c+ d)(a+ b)


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Heterogenous equi l ibr ium 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] = Kcx[CaCO3]

[CaO]Kp= PCO2

The concentration of solidsand pure liquidsare notincluded in the expression for the equilibrium constant.

14.2


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PCO2 = Kp

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

PCO2 does not depend on the amount of CaCO3or CaO

14.2


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Ways Different States of Matter

Can Appear in the Equilibrium

Constant, K

Molarity Partial Pressure

gas, (g) YES YES

aqueous, (aq) YES ---

liquid, (l) --- ---

solid, (s) --- ---


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Writing Equilibrium Constant Expressions

The concentrations of the reacting species in thecondensed phase are expressed in M. In the gaseous

phase, the concentrations can be expressed in Mor in atm.

The concentrations of pure solids, pure liquids and solvents

do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity.

In quoting a value for the equilibrium constant, you must

specify the balanced equation and the temperature.

14.2


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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 Kcand Kp.

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

Kc=

[COCl2]

[CO][Cl2] =

0.14

0.012 x 0.054 = 220

Kp= Kc(RT)Dn

Dn= 12 = -1 R= 0.0821 T= 273 + 74 = 347 K

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

14.2


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The equilibrium constant Kpfor the reaction

is 158 at 1000K. What is the equilibrium pressure of O2if

the PNO = 0.400 atm and PNO= 0.270 atm?2

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

14.2

Kp=2

PNO PO2

PNO2

2

PO2 = KpPNO

2

2

PNO2

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


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Predicting Reaction Direction

from a Nonequilibrium Mixture

Calculate Reaction Quotient, Q

The expression for Q is the same as K

BUT nonequilibrium concentrations

rather than equilibrium concentrations

are put into the expression

Compare the value of Q to K.


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The react ion quo tient (Qc)is calculated by substituting the

initial concentrations of the reactants and products into the

equilibrium constant (Kc) expression.

IF

Qc> Kcsystem proceeds from right to left to reach equilibrium

Qc= Kc the system is at equilibrium

Qc< Kcsystem proceeds from left to right to reach equilibrium

14.4


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Le Chteliers Principle

1. System startsat equilibrium.2. A change/stress is then made to system at

equilibrium.

Change in concentration Change in volume

Change in pressure

Change in Temperature

Add Catalyst3. System responds by shifting to reactant or

product side to restoreequilibrium.


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If an external stress is applied to a system at equilibrium, the

system adjusts in such a way that the stress is partially offset

as the system reaches a new equilibrium position.


Changes in Concentration

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

Add

NH3

Equilibrium

shifts left to

offset stress

14.5


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Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) leftDecrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

14.5

aA + bB cC + dD

AddAddRemove Remove


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Le Chatliers Principle

Removing SO2, O2


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Le Chatliers Principle (cont)


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Changes in Volume and Pressure

(Only a factor with gases)

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


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Chemistry; The Science

in Context;by Thomas

R. Gilbert, Rein V.Kirss, and Geoffrey

Davies, Norton

Publisher, 2004, p 752


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Changes in TemperatureOnly factor that can change value of K

Change Exothermic Rx

Increase temperature Kdecreases

Decrease temperature Kincreases

Endothermic Rx

Kincreases

Kdecreases

14.5


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Le Chteliers Principle; Temperature Effect

Chemistry; The Science

in Context;by Thomas

R. Gilbert, Rein V.Kirss, and Geoffrey

Davies, Norton

Publisher, 2004, p 764


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Adding a Catalyst

does not change Kdoes not shift the position of an equilibrium system

system will reach equilibrium sooner



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uncatalyzed catalyzed

14.5

Catalyst lowers Ea

for bothforward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

t li P i i l


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Change Shift Equilibrium

Change Equilibrium

Constant

Concentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Catalyst no no

14.5

Animation of Le Chateliers Principle
http://localhost/var/www/apps/conversion/animations/CHEM1212/2Equilibrium.exehttp://localhost/var/www/apps/conversion/animations/CHEM1212/2Equilibrium.exe


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Calculating Equilibrium Concentrations

from K and Initial Concentrations

1. Balance chemical equation.

2. Determine Q to see which way reaction will proceed to get

to equilibrium

3. Define changes needed to get to equilibrium. Use x torepresent the change in concentration to get to equilibrium.

Determine expression for equilibrium concentration.

4. Write the expression for equilibrium constant, K.

5. Substitute expressions defined in step 3 into the equilibrium

constant (step 4).

6. Solve for x and plug value back to table made in step 3.

14.4

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


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

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063

Mand [Br] = 0.012 M, calculate the concentrations of thesespecies at equilibrium.

Br2(g) 2Br (g)

Br2(g) 2Br (g)

Letxbe 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 forx

14.4

(0 012 + 2 )2


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Kc=(0.012 + 2x)2

0.063 -x= 1.1 x 10-3

4x2+ 0.048x+ 0.000144 = 0.00006930.0011x

4x2

+ 0.0491x+ 0.0000747 = 0

ax2+ bx+ c=0-b b24ac

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 Mor 0.00844 M

At equilibrium, [Br2] = 0.062x= 0.0648 M

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Chapt15 Equilibrium