Equilibrium Power Point At

8/3/2019 Equilibrium Power Point At

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Chemical Equilibrium


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When and ONLY when a system reaches

equilibriumdoes

Kc >>1 products dominate at equilibrium

Kc


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An equilibrium can be disrupted in several ways:

an addition or removal ofreagents (reactants or

products)

a change in volume orpressure

change of temperature


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JUST AFTER the disturbance the system is

NO LONGER AT EQUILIBRIUM

The new system EQUILIBRATES so that the

ORIGINAL Kc is respected (except in the case

where the temperature is changed)

Kc

[C]y x [D]z

[A]w x [B]x= Kc


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Adding more reactants (symbolic)

reactant

side

product

side

unstable stateinitial state

new equilibrium

established


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reactant

side

product

side

unstable stateinitial state

new equilibrium

established

Removing products (symbolic)


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Systems at equilibrium which are disturbed

react according to Le Chateliers principle.

Le Chateliers Principle

If a system at equilibrium is

subjected to a small change then

the equilibrium will shift so as tominimize the effect of the change

Henry Le Chatelier (1899)


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Applications of Le Chatelier s principle


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How is it that by increasing the pressure of a gas

phase reaction, the value of Kc can be respected?

Pressure

increasex 2

In the compressed mixture the

volume has decreased by 2, so theconcentration factor [A]x[B]2 has

risen x8 while [C] has only gone up

x2. Kc is re-established by decreasing

[A]x[B]2

Consider the following reaction at constant T:

A + 2B C

Initial equilibrium

Disturbed equilibrium

New equilibrium

=A

=B

=C


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The Haber process

Fritz Haber (1868-1934)

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

140 M tons per annum

N2 from air, H2 from natural gas or coal

Challenge: to get a good yield quickly

and cost effectively

80% for fertilizers 20% for polymers and other chemicals


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Pressureincrease:

Temperatureincrease:

Iron catalyst:

An optimum temperature and pressure are needed

which allow a reasonable rate and yield of ammonia

N2(g) + 3H2(g) 2NH3(g) H = -92KJ/mol

rate

yield

cost

rate

yield

cost

rate


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Pressure dependence at different temperatures


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Temperature dependence at different pressures


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How it is done

For modern developments see:

http://www.ias.ac.in/resonance/Sept2002/pdf/Sept2002p69-77.pdf


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The Contact Process

3. Converts the sulphur trioxide into

concentrated sulphuric acid

1. Makes sulphur dioxide

2. Converts the sulphur dioxide into sulphur

trioxide (the reversible reaction at the

heart of the process)


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Conver t ing the su lphu r d iox ide i n tosu lphur t r iox ideThis is a reversible reaction, and the

formation of the sulphur trioxide is

exothermic.

2SO2(g) + O2(g) 2SO3(g) H = -196KJ mol-1

Catalysts have no effect on the position

of the equilibrium.

Their only function is to speed up the

reaction.


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Oxygen Why not add more?

By increasing the proportion of oxygen you can increase the

percentage of the sulphur dioxide converted, but at the

same time decrease the total amount of sulphur trioxide

made each day. The 1 : 1 mixture turns out to give you the

best possible overall yield of sulphur trioxide.

Temperature Why so high?

In order to get as much sulphur trioxide as possible in the

equilibrium mixture, you need as low a temperature aspossible. However, 400 - 450C isn't a low temperature!

The lower the temperature you use, the slower the reaction

becomes. 400 - 450C is a compromise temperature producing

a fairly high proportion of sulphur trioxide in the equilibrium

mixture, but in a very short time

Pressure Why so low?

Even at these relatively low pressures, there is a 99.5%

conversion of sulphur dioxide into sulphur trioxide. The very

small improvement that you could achieve by increasing the

pressure isn't worth the expense of producing those highpressures.


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Equilibriumcalculations

These are best done by:

1. writing an equation for the reaction

2. entering the known concentrations of

reactants and products before and after the

change

3. manipulating the equilibrium expression to

obtain the desired unknown variables


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Examples

1. Liquid phase equilibria

When 46g (1.0 mol) of ethanol was reacted with 30.0g (0.50

mol) of ethanoic acid at 373 K the equilibrium mixture

contained 37.0g (0.42 mol) of ethyl ethanoate. Calculate the

value of Kc at 373K to the nearest integer.

C2H5OH(l) + CH3CO2H(l) CH3CO2C2H5(l) + H2O(l)

initial moles:1.0 0.50 0 0

equilibrium: 1.0-0.42 0.50-0.42 0.42 0.42


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0.58 0.08 0.42 0.42

C2H5OH(l) + CH3CO2H(l) CH3CO2C2H5(l) + H2O(l)

concentrations:

Kc = (0.42)2 /V2

V V V V

(0.58/V) x (0.08/V)

N.B. Volume will have no

influence on Kc this time

but could if coefficients

were different!

Kc = 4


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2. Gas phase equilibria with no overall volume change

Examples

H2(g) + I2(g) 2HI(g)

Ifamoles of hydrogen and bmoles of iodine are

reacted together and ifxmoles of hydrogen react

with xmoles of iodine then:

initial amount: a b 0

at equilibrium: a-x b-x 2x

concentrations: a-x b-x 2x

Kc =(2x)2 / V2

(a-x)/V x (b-x)/V

V V V

Volume will have

no influence on Kc

here.


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Examples

3. Gas phase equilibria with increase in volume

PCl5(g) PCl3(g) + Cl2(g)

Initial amount: a 0 0

At equilibrium: a-x x x

Concentrations: a-x x x V V V

Kc = (a-x)/V

x2/V2

Kc =x2

(a-x) x V

Here a volume increase or

pressure decrease will

favour the formation of

products


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Examples

4. Gas phase equilibria with decrease in volume

The Haber process (1)

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

initial: a 3a 0

eqm.: a-x 3a-x 2x

Concn.: (a-x) (3a-x) 2x

V V V

Kc =(4x2)/V2

((a-x)/V) x ((3a-x)/V)3


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Examples

Kc =(4x2)/V2

((a-x)/V) x ((3a-x)/V)3

Kc =4x2V2

27 x (a-x)4

Note that as volume decreasesor pressure increases, xmust increase

to maintain the equilibrium.

The Haber process (2)


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IB question

For the reaction:

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

1.00M of ammonia was placed in a 1.00 L

container. After some time, [H2] = 1.35 M,

and was not changing. What is the value ofKc?

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