Problem Set Sulfuryl Chloride Equilibria. Gaseous Equilibrium Edward A. Mottel Department of...

Problem Set

Sulfuryl Chloride Equilibria

Gaseous Equilibrium

Edward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

04/18/23

Gaseous Equilibria

• Reading Assignment: • Zumdahl Chapter 6.3, 6.6-6.8

• This lecture continues the topic of dynamic equilibrium with examples drawn from gaseous systems.

• Changes to the system can be predicted by Le Châtelier's Principle.

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Water-Gas Shift Reaction

• The water-gas shift reaction is a useful industrial process to generate hydrogen gas.

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

What will a graph of the steam pressure as a function of time look like?

Suppose 1 atm of CO and 1 atm of steam are allowedto react in the presence of a catalyst.

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Water-Gas Shift ReactionCO2(g) + H2(g)CO(g) + H2O(g)

1.00

0.00

0.20

0.40

0.60

0.80

0 50 100 150

Time

Pre

ssu

re (

atm

)

CO

H2O

CO2

H2

What will the graph of the other gas pressureslook like?

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

• At 605 K, the equilibrium pressures are

• PCO = 0.67 atm

• PH2O = 0.67 atm

• PCO2 = 0.33 atm

• PH2 = 0.33 atm

• What is the numeric value of the equilibrium constant for the reaction?


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

• Suppose 1 atm of CO2 and 1 atm of H2 are allowed to react in the presence of a catalyst.


What will a graph of the gas pressures look like?

0.00

0.20

0.40

0.60

0.80

1.00

0 50 100 150

Time

Pre

ssu

re (

atm

) CO

H2O

CO2

H2

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


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0.00

0.20

0.40

0.60

0.80

1.00

0 50 100 150

Time

Pre

ssu

re (

atm

)

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

CO

H2O

CO2

H2


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

• Suppose 0.5 atm each of CO, H2O, CO2 and H2 are allowed to react in the presence of a catalyst.


What will a graph of the gas pressures look like?

0.00

0.20

0.40

0.60

0.80

1.00

0 50 100 150

Time

Pre

ssu

re (

atm

) CO

H2O

CO2

H2

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


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0.00

0.20

0.40

0.60

0.80

1.00

0 50 100 150

Time

Pre

ssu

re (

atm

) CO

H2O

CO2

H2

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


04/18/23

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

• Observations• Three different starting combinations of

reactants and products give the same final results.

• Equilibrium can be approached from reactants, products or a combination of both.

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


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Sulfur Dioxide Oxidation

• Sulfur dioxide gas reacts with oxygen to give sulfur trioxide gas.

• The equilibrium constant for this reaction is 3.46 atm-1.

If PSO2 is 4.00 atm and PO2 = 3.00 atm initially,what will be the total pressure of the system

at equilibrium?

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• Process• Write a balanced chemical equation.• Write the mass-action expression.• Set up an “accounting system” that allows

you to determine the pressures of the reactants and products as the system attains equilibrium.

• Determine the final pressure of the system at equilibrium.

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Sulfur Dioxide Oxidation2 SO3(g)2 SO2(g) + O2(g)

( )P2

SO3

( ) ( )P P2

SO2 O2

K = = 3.46 atm-1

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Sulfur Dioxide Oxidation2 SO3(g)2 SO2(g) + O2(g)

4.00 atm 3.00 atm 0 atm

thishas to

increase

thishas to

decrease

thishas to

decrease

-x-2x +2x

04/18/23

ICE Table2 SO3(g)2 SO2(g) + O2(g)

SO2 O2 SO3

Initial

Change

Equilibrium

4.00 3.00 0

-x-2x +2x

4.00-2x 3.00-x 2x

Substitute into the mass-action expressionand solve

04/18/23


K =(2x)2

(4.00-2x)2 (3.00-x)= 3.46 atm-1

solve(((2*x)^2)/(((4-2*x)^2)*(3-x))=3.46,x);

1.403091360, 2.653945650 - 1.228460854 I, 2.653945650 + 1.228460854 I

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

ICE Table

SO2 O2 SO3

Initial

Change

Equilibrium

4.00 3.00 0

-2x -x +2x

4.00-2x 3.00-x 2x

4.00 - 2*1.40= 1.20 atm

3.00 - 1.40= 1.60 atm

2*1.40= 2.80 atm

2 SO3(g)2 SO2(g) + O2(g)

04/18/23


K =(2.80 atm)2

(1.20 atm)2 (1.60 atm)= 3.46 atm-1

= 3.40 atm-1

2 SO3(g)2 SO2(g) + O2(g)

04/18/23


Total Pressure = PSO2 + PO2 + PSO3

=(1.20 + 1.60 + 2.80 ) atm = 5.60 atm

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Equivalent Starting Conditions

• Example:• In the previous example, how much sulfur

trioxide gas could have been placed in an evacuated container to end up with the same equilibrium conditions?

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4.00 atm 3.00 atm 0 atm

2 SO3(g)2 SO2(g) + O2(g)

Suppose 2 atm of SO2 reacts,what will be the resulting pressures?

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4.00 atm 3.00 atm 0 atm

2.00 atm 2.00 atm 2.00 atm

0 atm 1.00 atm 4.00 atm

Equilibrium isshifted as farleft as possible

Equilibrium isshifted as farright as possible

2 SO3(g)2 SO2(g) + O2(g)

Suppose 2 atm of SO2 reacts,what will be the resulting pressures?

Suppose it shifts as far to the right as possible,what will be the resulting pressures?

04/18/23

ICE Table

SO2 O2 SO3

Initial

Change

Equilibrium

0 1.00 4.00

+y+2y -2y

2y 1.00+y 4.00-2y

2*0.60= 1.20 atm

1.00 + 0.60= 1.60 atm

4.00 -2*0.60= 2.80 atm

2 SO3(g)2 SO2(g) + O2(g)

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Perturbing a System at Equilibrium

• Addition or removal of reactant• Addition or removal of product• Addition of a non-reacting component with no

change in volume• Volume change• Temperature change• Le Châtelier's Principle

04/18/23

Le Châtelier's Principle2 SO3(g)2 SO2(g) + O2(g)

• What will be the effect on the reactants and products as each of the following changes are made?

• Additional SO2 is added to the system.

SO2 O2 SO3initially upthen down

down up

04/18/23

Le Châtelier's Principle


• O2 is removed from the system.

SO2 O2 SO3initially down

then upup down

2 SO3(g)2 SO2(g) + O2(g)

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• Additional SO3 is added to the system.

SO2 O2 SO3initially upthen down

up up

2 SO3(g)2 SO2(g) + O2(g)

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• Inert N2 is added to the system with no change in volume or temperature.

SO2 O2 SO3

nochange

nochange

nochange

2 SO3(g)2 SO2(g) + O2(g)

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• What will be the effect on the reactants and products as each of the following changes are made?• The volume of the system is increased

with no change in temperature.

SO2 O2 SO3initiallydown

then up

initiallydown

then up

initiallydown

then down

2 SO3(g)2 SO2(g) + O2(g)

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Changing the Volume of aGaseous Equilibrium System

high pressuresmall volumes

2 SO3(g)2 SO2(g) + O2(g)

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less atlarger volumes

more atlarger volumes

2 SO3(g)2 SO2(g) + O2(g)

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low pressurelarge volumes

2 SO3(g)2 SO2(g) + O2(g)

Pressure Effects

Nitrogen Dioxide-Dinitrogen Tetroxide System

Nitrogen Dioxide - Dinitrogen Tetroxide

NO2(g) + NO2(g) N2O4(g)

NO

ON

O

ON

O

ON

O

O


0 sec 1 sec 10 sec

NO2(g) + NO2(g) N2O4(g)

[reactants]

[products]


Mass-Action Expression

Q =

Why does the gas initially get darker and then lighten?

PNO2( )2

PN2O4

NO2(g) + NO2(g) N2O4(g)

04/18/23

Gas Pressure Analysis

VolumeL

PSO2

atmPO2

atmPSO3

atmPtotal

atmKp

atm-1 % Reaction

1.00 2.00 1.00 3.72 6.72 3.46 3.72/5.72 = 65.0%

10.00 0.326 0.163 0.246 0.735 3.49 0.246/0.572 = 43.0%

100.00 0.0447 0.0223 0.0125 0.0795 3.51 0.0125/0.0572 = 21.9%

Expanding the volume of the system causes the partial pressures of all the gases to decrease.

2 SO3(g)2 SO2(g) + O2(g)

04/18/23


VolumeL

PSO2

atmPO2

atmPSO3

atmPtotal

atmKp

atm-1 % Reaction

1.00 2.00 1.00 3.72 6.72 3.46 3.72/5.72 = 65.0%

10.00 0.326 0.163 0.246 0.735 3.49 0.246/0.572 = 43.0%

100.00 0.0447 0.0223 0.0125 0.0795 3.51 0.0125/0.0572 = 21.9%

As the volume increases, the percentage reactionshifts toward the side with more total moles of gas.

2 SO3(g)2 SO2(g) + O2(g)

04/18/23


VolumeL

PSO2

atmPO2

atmPSO3

atmPtotal

atmKp

atm-1 % Reaction

1.00 2.00 1.00 3.72 6.72 3.46 3.72/5.72 = 65.0%

10.00 0.326 0.163 0.246 0.735 3.49 0.246/0.572 = 43.0%

100.00 0.0447 0.0223 0.0125 0.0795 3.51 0.0125/0.0572 = 21.9%

If the equilibrium is shifted all the wayRIGHT or LEFT

what would be the initial starting conditions?

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Equivalent Starting Conditions

• This system could have been prepared by initially adding 5.72 atm of SO3 gas to the evacuated 1.00 L cylinder, or

• by adding 5.72 atm of SO2 gas and 2.86 atm of O2 gas to the evacuated 1.00 L cylinder, or

• several other possibilities.

04/18/23


VolumeL

PSO2

atmPO2

atmPSO3

atmPtotal

atmKp

atm-1 % Reaction

1.00 2.00 1.00 3.72 6.72 3.46 3.72/5.72 = 65.0%

10.00 0.326 0.163 0.246 0.735 3.49 0.246/0.572 = 43.0%

100.00 0.0447 0.0223 0.0125 0.0795 3.51 0.0125/0.0572 = 21.9%

The stoichiometric relationships remain true:SO2 and O2 ratio

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Moles of Gas Analysis

The stoichiometric relationships remain true:total moles of sulfur containing compounds

is constant.

VolumeL

1.00

10.00

100.00

Totalmoles

0.0643

0.0704

0.0762

MolesSO2

0.0191

0.0312

0.0428

MolesO2

0.0096

0.0156

0.0214

MolesSO3

0.0356

0.0236

0.0120

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Moles of Gas Analysis

As the volume expands, the reactions shifts lefttowards the side with more moles of gas.

VolumeL

1.00

10.00

100.00

Totalmoles

0.0643

0.0704

0.0762

MolesSO2

0.0191

0.0312

0.0428

MolesO2

0.0096

0.0156

0.0214

MolesSO3

0.0356

0.0236

0.0120

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Mole Fraction Analysis

VolumeL

1.00

10.00

100.00

2.00/6.72 = 0.298

0.444

0.562

SO2

0.149

0.222

0.281

O2

0.553

0.334

0.157

SO3

As the volume expands, the reactions shifts left.

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Mole Fraction Analysis

VolumeL

1.00

10.00

100.00

2.00/6.72 = 0.298

0.444

0.562

SO2

0.149

0.222

0.281

O2

0.553

0.334

0.157

SO3

The stoichiometric relationships remain true:SO2 and O2 ratio

2 SO3(g)2 SO2(g) + O2(g)

04/18/23

Problem Set Sulfuryl Chloride Equilibria. Gaseous Equilibrium Edward A. Mottel Department of...

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