(a) Advanced Calculations using the Equilibrium Constant; (b) Le

Chatelier’s Principle; (c) Equilibria of Real Gases

Chemistry 142 B

Autumn Quarter 2004

J. B. Callis, Instructor

Lecture #19

Solving Equilibrium Problems

• Write the balanced equation for the reaction.• Write the equilibrium expression.• List the initial concentrations.• Calculate Q and determine the direction of shift to

equilibrium.• Define the change needed to reach equilibrium and define

the equilibrium concentrations. • Substitute the equilibrium concentrations into the

equilibrium expression and solve for the unknown. • Check the solution by calculating K and making sure it is

identical to the original K.

Problem 19-1 – Graphical Solution to the Quadratic Equation

Let us begin by considering the equilibrium between NO2, a red-brown gaseous pollutant formed by automobiles, and its dimer, N2O4. This equilibrium can be expressed by the chemical equation:

2 NO2(g) = N2O4(g)

The equilibrium constant for this reaction is

where PN2O4 and PNO2 are the equilibrium partial pressures of the two gases. KP has the numerical value of 8.8 at T = 25 oC when the partial pressures are expressed in atmospheres.

Problem 19-1 – Graphical Solution cont.Let us now consider the problem of finding the equilibrium partial pressures of NO2 and N2O4, given the value of KP, and the initial pressures of NO2 and N2O4 ( (PNO2)0 and (PN2O4)0 ). Then if x atm is the additional amount of N2O4 formed by the equilibrium, the partial pressure of NO2 must decrease by 2x atm.

Inserting the equilibrium partial pressures into the equilibrium expression results in the equation:

:yield toedrationaliz becan which

K

Finding the Solution to an Equilibrium Problem Using a Graphical Method

From Graph:

x = and

x =

The later is chosen because

it gives all positive concentrations.

(PNO2)eq = 0.726 atm

(PN2O4)eq = 4.637 atm

x, atm f(x)4.0 31.204.1 24.414.2 18.334.3 12.954.4 8.274.5 4.304.6 1.034.7 -1.534.8 -3.394.9 -4.555.0 -5.005.1 -4.755.2 -3.795.3 -2.135.4 0.235.5 3.305.6 7.075.7 11.555.8 16.735.9 22.616.0 29.20

=4*K*x 2̂-(4*K*PNO2o+1)*x+K*PNO2o 2̂-PN2O4o

Graphical Solution to the Equilibrium Problem

-10.00

-5.00

0.00

5.00

10.00

15.00

20.00

25.00

30.00

35.00

4.0 4.5 5.0 5.5 6.0 6.5

x, atm

f(x)

Problem 19-2: Using Exact Solution of Quadratic Equation

K.179at mequilibriu

at species theall of pressure partial thecalculate

initially, L10 in alcohol isopropyl of g 10For

K)179at 0.444(K

)()()(

:follows as writtenis gas hydorgen and acetone

toalcohol isopropyl of ondissociati The

22323

gHgCOCHgCHOHCH

Problem 19-2: Solution(a)

: tablereaction theConstruct

right the toproceeds reaction that theobvious isit - calculatednot is Q

atm 0.00 P Pproducts theof pressure partial Initial

atm, 0.617 = Plisopropano theof pressure partial Initial0H2

0Ac

0Iso

Pressure (atm)

Isopropyl alcohol, IPA Acetone, Ac Hydrogen

H2(g)

Init.

Change

Equil.

CHOH(g)CH 23 CO(g)CH 23

Problem 19-2: Continued (b)

:formula quadratic thefrom obtained be can roots (two) thewhere

form general theof quadratica is expression This

0

:right theon zero withsideleft theonx

of powers descending in rearrange and

tion,multiplica indicated theperform wex,for solve To

:expression mequilibriu theinto

table thefrom ionsconcentrat mequilibriu thengSubstituti

K

Problem 19-2: Continued (c)

444.0

:Check

atm

atm

:ionsconcentrat mequilibriu gCalculatin

atm :root negative theignore we

ions,concentrat positive all toleadsroot positive only the Since

x and x

solutions candidate twoobtain wequadratic, theFrom

2

2

IPA

HAC

IPA

HAc

P

PPK

P

xPP

x

Problem 19-3: Problems Involving Higher Order Polynomials

m.equilibriuat result that willgH

of pressure partial theDetermine ly.respective atm, 1.60 and

atm 2.30 atm, .40 1 are reaction) (before pressures partial

initial their andK, 600at chamber

evacuatedan into introduced are CO and OH ,CH Gaseous

K) 600at 108.1(K

)(3)()(

:reaction following heConsider t

2

24

7P

224

gHgCOOHgCH

Problem 19-3: Solution(a)

: tablereaction theConstruct

right the toproceeds reaction that theobvious isit - calculatednot is Q

Pressure (atm)

CH4(g) H2O(g) CO(g) 3 H2(g)

Init.

Change

Equil.

Problem 19-3: Continued (b)

.polynomialorder fourth a is expression This

:right on the zero with sideleft on they

of powers descendingin rearrange and

tion,multiplica indicated theperform wefor x, solve To

108.1

:expression mequilibriu theinto

table thefrom ionsconcentrat mequilibriu thengSubstituti

7PK

Problem 3- Continued (c)

atm P

thenm,equilibriuAt

pressures. starting with thecompared small indeed is valueThis

y

:gives sidesboth ofroot cube theTaking

y

yieldsy for Solving 108.1

equation eapproximat theuse weThus

small. isy that imagine wei.e. right, themuch to

shift thatnot doesreaction thesmall, isK sincey Fortunatel

H2

3

7

Le Chatelier’s Principle

If a change in conditions (a ‘stress’) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions.

Henri Le Chatelier, 1884

The Effect of a Change in Concentration

• If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to to reduce the concentration of the added component.

• If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to to increase the concentration of the removed component.

Problem 19-4: The Effect of a Change in Concentration

Consider the following reaction:

2 H2S(g) + O2(g) = 2 S(s) + 2 H2O(g)

What happens to:

(a) [H2O] if O2 is added?

Ans:

(b) [H2S] if O2 is added?

Ans:

(c) [O2] if H2S is removed?

Ans:

(d) [H2S] if S is added?

Ans:

Effect of a Change in Pressure

1. Add or remove a gaseous product at constant volume.

2. Add an inert gas (one not involved in the reaction) at constant volume.

3. Change the volume of the container.

There are three ways to change the pressure of a reaction system involving gaseous components at a given temperature:

Problem 19-5: The Effect of a Change in Pressure

How would you change the total pressure of each of the following reactions to increase the yield of the products:

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

Ans:

(b) S(s) + 3 F2 (g) = SF6 (g)

Ans:

(c) Cl2(g) + I2(g) = 2 ICl (g)

Ans:

The Effect of a Change in Temperature

• Changes in concentration or pressure alter the equilibrium position.

• In contrast, changes in temperature alter the value of the equilibrium constant.

Exothermic Reactions

• Releases heat upon reaction. H is negative.

• Addition of heat to an exothermic reaction shifts the equilibrium to the left.

• The value of K decreases in consequence.

Endothermic Reactions

• Absorbs heat upon reaction. H is positive.

• Addition of heat to an endothermic reaction shifts the equilibrium to the right.

• The value of K increases in consequence.

Using Le Chatelier’s Principle to Describe the Effect of a Temperature Change on a System in Equilibrium

• Treat the energy as a reactant (in an exothermic process) or as a product (in an endothermic process).

• Predict the direction of the shift as if an actual reactant or product has been added or removed.

Problem 19-6: The Effect of a Change in Temperature on the Position of Equilibrium

How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions:

(a) CaO(s) + H2O (l) = Ca(OH)2 (aq) H0 =-82 kJ

Ans:

(b) (a) CaCO3 (s) = + CaO(s) + CO2 (g) H0 = 178 kJ

Ans:

(c) SO2 (g) = S(s) + O2(g) H0 = 297 kJ

Ans:

Equilibria Involving Real Gases

• We have thus far assumed that gas phase equilibria involve gases that behave as ideal gases.

• The effect of non-ideal behavior is to cause real equilibrium constants to become non-constant under differing conditions of total pressure.

The Activity Coefficient

The activity of the ith gaseous component of the equilibrium system is represented as:

ref

obsii

i P

Pa

Where i is called the activity coefficient for correcting Pi

obs to the ideal value.

For equilibrium pressures of 1 atm or less, the value of Kp calculated from the observed pressures is expected to be within about 1% of the true value

Answers to Problems in Lecture 19

1. From graph: x = 5.391 and x = 4.637. The later is chosen because it gives all positive concentrations.

2.

3. 4. (a) The reaction proceeds to the right so H2O increases. (b) Some H2S reacts with the

added O2 to move the reaction to the right, so [H2S] decreases. (c) The reaction proceeds to the left to re-form H2S, more O2 is formed as well, O2 increases. (d) S is a solid, so its concentration does not change. Thus, [H2S] is unchanged.

5. (a) The only gas is the product CO2. To move the reaction to the right decrease the pressure. (b)With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF6. (c) The number of moles of gas is the

same on both sides of the equation, so a change in pressure will have no effect.

6. (a) Add heat to the right side. Adding heat shifts the system to the left. [Ca(OH)2] and K will decrease. (b) Add heat to the left side. Adding heat shifts the system to the right. [CO2] and K will increase. (c) Add heat to the left side. Adding heat shifts the system to the right. [SO2] will decrease and K will increase.

atm 101.7P 3H2

atm 270.0atm 347.02 IPAHAc PPP