AE2_LeChatelier 2008

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THEORY BOOKLET NAME: HSC CHEMISTRY ACIDIC ENVIRONMENT 2 (Le Chatelier’s Principle)

Transcript of AE2_LeChatelier 2008

Page 1: AE2_LeChatelier 2008

THEORY BOOKLETNAME:

HSCCHEMISTRY

ACIDIC ENVIRONMENT 2(Le Chatelier’s Principle)

1300 008 008

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HSC Chemistry Acidic Environment 2[Le Chatelier’s Principle]

matrixeducation.com.auLevel 8, Strathfield Plaza, 11 The Boulevarde, STRATHFIELD, NSW 2135

1. Factors affecting the rate of reaction

Nature of the reactants

Every reaction has its own rate and its own activation energy. In general,

reactions that take place between ions in aqueous solution are extremely

rapid, almost instantaneous.

The rate of a reaction is defined as the change in concentration of reactants

or products over time. In a reaction with a slow rate, the reactants react to

form products over a long period of time.

Activation energies for these reactions are very low because usually

no covalent bonds need to be broken. Define the term ‘activation energy’.

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As we might expect, reactions between covalent molecules, whether in

aqueous solution or not, are much slower. Many of these reactions require

anywhere from 15 minutes to 24 hours or more for most of the starting

compounds to be converted to the products. There are, of course, reactions

that take a good deal longer, but such reactions are seldom useful.

Concentration

Consider a reaction A + B C + D.

In most cases, a reaction rate increases when the concentration of either or

both reactants (A or B) is increased.

For many reactions, though by no means all, there is a ………………….1

relationship between concentration and rate; that is, when the

concentration of a reactant is doubled, the rate also doubles. All of this is

easily understandable on the basis of the collision theory.

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If we double the concentration of A, there are twice as many molecules of A

in the same volume. The molecules of B in that volume now collide with

twice as many A molecules per second as before. Since the reaction rate

depends on the number of collisions per second, the rate is doubled.

Rate of reaction between gaseous H2 and I2 increases with increasing

concentration:

Source: Tro, N. J., Introductory Chemistry, Prentice-Hall Inc, 2003.

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Temperature

In virtually all cases, reaction rate increases with increasing temperature. An

approximate rule for many reactions is that:

As the temperature goes up by 10oC, the rate of reaction doubles.

This rule is far from exact, but it is not far from the truth in many cases. This

is quite a large effect and says, for example, that if we run a reaction at

90 °C instead of at room temperature (20 °C), the reaction will go about 128

times faster.

Increasing temperature provides more energy that the reactants need to

change into products. The reaction of the mixture in the diagram c) below is

therefore the fastest as the reaction mixture is at the highest temperature.

Source: Tro, N. J., Introductory Chemistry, Prentice-Hall Inc, 2003.

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If it takes 20 hours to convert 100g of A to product C at 20 °C, then it would

take only about 10 minutes at 90 °C. Temperature is therefore a powerful

tool that lets us increase the rates of reactions that are inconveniently slow.

It also lets us decrease the rates of reactions that are inconveniently fast. In

some cases we choose to run reactions at low temperatures because

explosions would result (the reaction would run out of control) at room

temperature.

Presence of a Catalyst

Any substance that ………………….2 the rate of a reaction without itself

being used up is called a catalyst.

Many catalysts are known. Some increase the rate of only one reaction

while others can affect several reactions. Although we have seen that

reactions can be sped up by increasing the temperature, in some cases they

are still too slow even at the highest temperatures we can conveniently

reach.

In other cases it is not feasible to increase the temperature, because other,

unwanted reactions would also be sped up. In such cases a catalyst, if we

can find the right one for a given reaction, can prove very valuable.

Many important industrial processes rely on catalysts and virtually all

reactions that take place inside living organisms are catalysed

by…………………….3

How do catalysts increase the rate of reaction?

Catalysts work by allowing the reaction to take a different pathway, one

with lower activation energy.

Without the catalyst, the reactants would have to get over the energy hill.

The catalyst provides a lower hill. As we have seen, lower activation energy

means a higher reaction rate.

2 increases3 enzymesCopyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 5 of 30

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An analogy:

Consider the following analogy which can be used to explain the effect of

catalyst.

There are two ways in which a ball can be pushed over a hill.

1. One way is to simply push them directly over the hill. This is analogous

to an increase in temperature for a chemical reaction.

2. The other way is find a path that goes around the hill. This is analogous

to the role of a catalyst for a chemical reaction.

Source: Tro, N. J., Introductory Chemistry, Prentice-Hall Inc, 2003.

The second method obviously requires less energy to achieve, just try it! The

use of a catalyst therefore increases the rate of reaction whilst improving

energy efficiency of the process.

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By lowering the activation energy using a catalyst, less energy is required for

the reactants to undergo reaction. This effect is shown on a reaction profile

as:

Source: Tro, N. J., Introductory Chemistry, Prentice-Hall Inc, 2003.

The reaction between ethene gas, C2H4, and hydrogen gas, H2, to give

ethane gas, C2H6:

C2H4(g) + H2(g) C2H6(g)

goes so slowly without a catalyst that is not practical even if we increase the

temperature to any reasonable level.

If however the mixture of gases is shaken with finely divided solid platinum,

the reaction takes place at a convenient rate. The C2H4 molecules and the

H2 molecules meet each other on the surface of the platinum where the

proper bonds can be broken and the reaction can proceed.

Each catalyst increases the rate of reaction by providing a surface on which

the reactants can meet. In this way bonds within a molecule are weakened,

or interactions between molecules strengthened, facilitating ease of reaction.

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The sucrose molecule fits perfectly within a pocket on the surface of the

sucrase enzyme. Once held in place, the enzyme weakens the bond

between glucose and fructose, thus lowering the activation energy of the

reaction.

Source: http://www.maxanim.com/biochemistry/index.htm

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2. Le Chatelier’s PrincipleStudents learn to

define Le Chatelier’s principle identify factors which can affect the equilibrium in a reversible reaction

When a reaction has reached equilibrium, the forward and reverse reactions

are taking place at the same rate, and the concentrations of all the

components do not change as long as we don’t do anything to the system.

In 1888 Henri Le Chatelier (1850-1936) put forth the statement known as

Le Chatelier’s principle:

If a system at equilibrium is disturbed, then the system adjusts itself so as to minimise the disturbance.

Addition of Reaction Components

Suppose that this reaction has reached equilibrium:

CH3COOH + C2H5OH CH3COOC2H5 + H2O

Acetic acid Ethanol Ethyl acetate

This means that the reaction flask contains all these substances and that the

concentrations no longer change. We now disturb the system by adding

some acetic acid from outside.

The result is that the concentration of acetic acid suddenly increases, which

increases the rate of the forward reaction. As a consequence, the

concentrations of products (ethyl acetate and water) begin to increase. At

the time the concentrations of reactants are decreasing.

Now, an increase in the concentrations of products causes the rate of the

reverse reaction to increase, but the rate of the forward reaction is

decreasing, so eventually the two rates will be equal again and a new

equilibrium is established.

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When that happens, the concentrations once again remain constant, but

they are not the same as they were before. The concentrations of ethyl

acetate and water are higher now, and the concentration of ethyl alcohol

lower. The concentration of acetic acid is higher because we added some,

but it is less than it was immediately after we made the addition.

This always happens when we added more of any component to system

equilibrium. The addition constitutes a stress; the system relieves the stress

by increasing the concentrations of the components on the other side. We

say that the equilibrium shifts towards the right:

Adding CH3COOH + C2H5OH CH3COOC2H5 + H2O CH3COOH

The addition of acetic acid causes the reaction to move toward the right:

more CH3COOC2H5 and H2O are formed and some of the CH3COOH and

C2H5OH are used up. The same thing happens if we add C2H5OH.

On the other hand, if we add H2O or CH3COOC2H5, the reaction shifts to the

left:

CH3COOH + C2H5OH CH3COOC2H5 + H2O AddingCH3COOC2H5

We can summarise by saying that the addition of any component causes

the equilibrium to shift to the opposite side.

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Removal of a Reaction Component

It is not always as easy to remove a component from a reaction mixture as it

is to add one, but there are often ways to do it.

The removal of a component, or even a decrease in its concentration, lowers

the corresponding reaction rate and changes the position of the equilibrium.

In this case, the reaction is shifted toward the side from which the reactant

was removed.

In the case of the acetic acid-ethyl alcohol equilibrium, ethyl acetate has the

lowest boiling point of the four components and can be removed by

distillation. If this is done, the equilibrium shifts to that side:

CH3COOH + C2H5OH CH3COOC2H5 + H2O Removing CH3COOC2H5

The concentrations of CH3COOH and C2H5OH decrease and that of H2O

increases. The effect of removing a component is thus the opposite of

adding one: the removal of a component causes the equilibrium to shift to the side the component was removed from.

No matter what happens to the individual concentrations, the value of the

equilibrium constant remains unchanged.

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Concept Check 3.1

(i) When dinitrogen tetroxide, N2O4, a colourless gas, is enclosed in a vessel a colour will appear indicating the formation of brown nitrogen dioxide, NO2. The intensity of the brown colour indicates the amount of nitrogen dioxide formed, the equilibrium reaction is

N2O4(g) 2NO2(g)

When more N2O4 is added to the equilibrium mixture, the brownish colour becomes darker. Explain what happened.4

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(ii) What happens to the equilibrium reaction

2NOBr(g) 2NO(g) + Br2(g)

When Br2 gas is added to the equilibrium mixture?

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4 The darker colour indicates that more nitrogen dioxide was formed. This happened because the addition of the reactant shifted the equilibrium to the right, forming more products, NO2.Copyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 12 of 30

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Concept Check 3.2

(i) When acid rain containing sulfuric acid attacks a marble statue made of calcium

carbonate, the following equilibrium reaction can be written

CaCO3(s) + H2SO4(aq) CaSO4(s) + CO2(g) + H2O(l)

How does the fact that CO2 is a gas influence the equilibrium?5

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(ii) In a power plant using coal, SO2 is generated from pyrite, FeS2, small amounts of

which are present in the coal, according to the reaction:

4FeS2(s) + 11O2(g) 2Fe2O3(s) + 8SO2(g)

SO2 is much more soluble in water than O2 and reacts with water to form H2SO3.

What is the effect on the equilibrium if the above reaction is performed in a moist

atmosphere?6

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5 The gaseous CO2 diffuses away from the reaction site, thus removing the product from the equilibrium mixture. The rate of forward reaction is greater than the rate of reverse reaction. Therefore the equilibrium shifts to the right.6 Since the reaction is taken in a moist environment, the SO2 which is more soluble in water than O2, will be removed away from the products. Therefore in order to restore its equilibrium, FeS2 and O2 will react to produce more of SO2. This means that the rate of forward reaction has increased and is greater than the rate of reverse reaction. Hence, the equilibrium shifts to the right.Copyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 13 of 30

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Change in Temperature

The effect of a change in temperature on a reaction that has reached

equilibrium depends on whether the reaction is exothermic (gives off heat) or

endothermic (requires heat). Let us look at an exothermic reaction:

2H2(g) + O2(g) 2H2O(l) + 137 000 cal

This equation tells us that 2 moles of H2 react with 1 mole of O2 to give 2

moles of H2O and heat.

If we look upon the heat as a product of this reaction, we can use the same

type of reasoning as we did before.

An increase in temperature means that we are adding heat. Since heat is a

product, its addition pushes the equilibrium to the opposite side. We can

therefore say that, if the reaction is at equilibrium and we increase the

temperature, the reaction goes to the left - the concentrations of H2 and O2

increase and that of H2O decrease. This is true of all exothermic reactions.

For an exothermic reaction

An increase in temperature drives an exothermic reaction toward the

reactants (to the left).

A decrease in temperature drives an exothermic reaction toward the

products (to the right).

For an endothermic reaction:

An increase in temperature drives an endothermic reaction toward

the products.

A decrease in temperature drives an endothermic reaction toward the

reactants.

Remember that a change in temperature changes not only the position of

equilibrium but the value of K as well.

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Concept Check 3.3(i) The conversion of nitrogen dioxide to dinitrogen tetraoxide is an exothermic reaction:

2NO2(g) N2O4(g) + 57 kJ mol-1

NO2 is a brown gas while N2O4 is colourless.To observe this equilibrium we can contain this reaction mixture within a sealed test tube. See figure below.

Source: Petrucci, Harwood & Herring, General Chemistry: Principles and Modern Applications, 8th Ed, Prentice-Hall, 2002.

When the test tube containing the reaction mixture is taken from an ice-water mix at 0 °C (left) and warmed in a cup containing hot water at 50 °C (right) the brown colour is darker; hence, the NO2 concentration is higher than at 0 °C. Why?

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(ii) In a reaction A B, the amount of B has increased when the temperature of the reaction was raised. Was this an exothermic or an endothermic reaction? Explain.

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Molecular Workbench Simulation

Chemical Equilibrium and Le Chatelier's Principle

http://molo.concord.org/database/activities/193.html

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4. Carbon Dioxide and CarbonatesStudents learn to

describe the solubility of carbon dioxide in water under various conditions as an equilibrium process and explain in terms of Le Chatelier’s principle

What is a Carbon dioxide?

Carbon dioxide (CO2) is a c……………….., non-toxic gas with a “faintly

pungent” odor.

The carbon dioxide molecule is l……………. with double bonds between the

carbon and oxygen atoms.

Draw the structure of the CO2 molecule

Physical and Chemical Properties

Gaseous CO2 is readily condensed to the liquid state by cooling and

compression and upon further cooling the liquid freezes to a white solid

(called “dry ice”).

The substance does not melt upon warming, but sublimes at 78.5 °C. Solid

CO2 leaves no trace when it evaporates (except for the water that

condenses on its surface). It is very convenient for use in cooling ice cream

(it is often used in Baskin Robins Ice cream shops), reactive chemicals, and

many other things.

At high temperatures (above 1700 °C) carbon dioxide decomposes to give

carbon monoxide and oxygen.

2CO2(g) 2CO(g) + O2(g)

This is an endothermic reaction.

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The atmosphere contains only 0.0325% carbon dioxide by volume, but

atmospheric carbon dioxide plays an important role in photosynthesis and

the cycling of carbon and oxygen through the environment. You have

learned this in The Identification and Production of Materials.

On the other hand, life is impossible in an atmosphere that contains too

much carbon dioxide. A CO2 concentration of 1% by volume in air can

cause headaches, 10% can cause severe distress and over 30% causes

unconsciousness and death.

Some uses of Carbon Dioxide

Carbon dioxide is inert, does not support combustion and is denser than air,

allowing it to blanket a fire, it is therefore an excellent …………………….

…………………………7 agent. Liquid carbon dioxide (held in tanks at

elevated pressure) is used in fire – extinguishing systems in airplanes, ships,

chemical plants and other industrial installations. Hand – held extinguishers

also often contain carbon dioxide under pressure. When the pressure is

released by opening a valve, the liquid carbon dioxide escapes and

immediately evaporates. Expansion of the gas causes a dramatic drop in

temperature. As a result, the CO2 freezes and forms a blanket of CO2

“snow”.

………………………….8 in production of organic chemicals.

Recovery of oil from depleted wells.

Refrigeration and food preservation.

………………………….9 carbonation

7 fire extinguishing8 Raw material9 BeverageCopyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 18 of 30

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Solubility of Carbon Dioxide in water

Carbon dioxide dissolves readily in water and reacts to form

………………..10 acid, H2CO3. This acid cannot be isolated, and carbon

dioxide is usually shown reacting directly with water to form hydrogen ions

and hydrogen carbonate ions:

CO2(g) CO2(aq) Equation 1CO2(aq) + H2O(l) H+

(aq) + HCO3(aq) Equation 2

In pure water a saturated CO2 solution at 1 atm pressure and 25 °C contains

about 0.034 mol/L of CO2 gas, most of which is simply dissolved in the

water. Only approximately one out of every 400 CO2 molecules in solution

reacts with a water molecule to give carbonic acid (H2CO3). This acid cannot

be isolated but does exist in small concentrations in aqueous CO2 solutions.

Carbon Dioxide in soda water

Why do carbonated soft drinks have a sharp acidic taste?

Coke and other effervescent soft drinks contain carbon dioxide gas

under pressure. As CO2(g) is dissolved in water (see Equation 1), the

concentration of CO2(aq) in equation 2 is ……………………….11

Since the concentration of reactants has ………..…………….12 , the rate

of forward reaction will also increase while the rate of reverse reaction

remains ………….………………….13 Since the rate of forward reaction

is greater than the rate of reverse reaction, the concentration of

products will ……………….14 The concentration of H+ and HCO3

…………….……15

Hence the position of the equilibrium has shifted towards the r….

………….

Therefore the soft drink has a sharp acidic taste (pH = 3.8).

10 carbonic11 increased12 increased13 unchanged14 increase15 increaseCopyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 19 of 30

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Why does a soft drink undergo effervescence when the cap is removed?

CO2(aq) + H2O(l) H+(aq) + HCO3(aq)

When the cap is opened, the system at equilibrium is opened. As

a result CO2(g) in the system escapes.

To restore equilibrium, CO2(g) must be reproduced. Therefore, CO2(aq) is

converted to CO2(g).

This represents a removal of reactants as CO2(aq) is converted to CO2(g).

Therefore, the rate of forward reaction …..…………….16 as the

concentration of the reactant has decreased.

Since the rate of reverse reaction is ………….…….17 than the rate of

forward reaction, the equilibrium position shifts towards the

……………..18

As a result more CO2(aq) is produced to restore equilibrium.

Hence effervescence is observed.

Why does a soft drink taste “flat” when it loses its carbon dioxide gas?

Loss of CO2(g) represents a removal of reactants as CO2(aq) is converted to

CO2(g).

Therefore, the rate of forward reaction …………………19 as the

concentration of the reactant has decreased.

Since the rate of reverse reaction is …….……………..20 than the rate of

forward reaction, the equilibrium position shifts towards the ……………..21

This represents a decrease in concentration of H+ and HCO3 since more

CO2(aq) is being produced to restore equilibrium.

Therefore, a decrease in concentration of H+ means a decrease in acidity

and loss of sharp acidic taste.

16 decreases17 greater18 left19 decreases20 greater21 leftCopyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 20 of 30

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Solubility of carbon dioxide in water under various conditions

Solubility of CO2 in water:

CO2(g) + H2O(l) H2CO3(aq)

Disturbance Effect on equilibrium (right means increased solubility)

Increase in pressure Favours reaction with fewer moles of gas in product,

reducing pressure, therefore shifts right, CO2 dissolves

more readily.

Increase in concentration

of CO2 in system

System aims to reduce concentration of CO2 and thus it

reacts, i.e., shifts equilibrium right

Increase in temperature Favours endothermic reaction, i.e. reverse reaction,

equilibrium shifts left, and CO2 is less soluble

Concept Check 4.1An equilibrium exists between gaseous and dissolved carbon dioxide in water as shown by the following equation:

CO2(g) CO2(aq)

With reference to Le Chatelier’s principle explain the following:

(a) Fizzing occurs when a bottle of a carbonated drink is opened. 2

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(b) It is observed that the fizzing is less if the bottle is kept under refrigeration rather than at room temperature. Deduce whether the dissolving process is exothermic or endothermic. Explain your reasoning. 2

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Concept Check 4.2

Carbon dioxide is poorly soluble. Small amounts of carbon dioxide can be dissolved in water and like nearly all gases it is more soluble in cold water.

(a) Write an equation showing carbon dioxide in water. Explain in terms of Le Chatelier’s Principle why high pressure is used to increase the amount of carbon dioxide which dissolves. 2

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(b) Suggest a reason why more gas dissolves at lower temperature. 1

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Concept Check 4.3 [2005 NEAP Q23]

Use Le Chatelier’s Principle to relate the increase in burning fossil fuels to a possible increase in the acidity of the oceans.22 4

COURSE OUTCOMES: H4, H8, H13MARKING GUIDELINES

CRITERIA MARKS

Relates increase in atmospheric CO2 to increases in dissolved CO2 and reaction with water to produce H+

(aq), through application of Le Chatelier’s Principle.3 ~ 4

Describes increases in acidity due to more dissolving of CO2 or increased production of H+

(aq)

2

Describes Le Chatelier’s Principle using CO2 in some way. 1

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22 As the concentration of CO2 in the atmosphere rises, more of it will dissolve in the oceans. This is a consequence of Le Chatelier’s Principle, which states that when an equilibrium is disturbed the reaction will shift to minimize the changes. In this case the first equilibrium system affected is CO2(g) CO2(aq). Then CO2(aq) + H2O(l) H+

(aq) + HCO3(aq) This increase in concentration of H+ means the acidity has increased.Copyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 23 of 30

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Concept Check 4.4 [2005 CSSA Q24]

Two identical bottles of soda water (carbonated water), one at room temperature (25 °C) and one just out of the refrigerator, had their pH determined using a probe and data logger. The results are tabulated below.

Soda Water pH Temperature (C)

Bottle A 5.21 25

Bottle B 4.63 4

Account for the difference in pH in terms of Le Chatelier’s Principle.23

COURSE OUTCOMES: H2, H8, H14MARKING GUIDELINES

CRITERIA MARKS

Demonstrates a thorough knowledge of Le Chatelier’s Principle with regard to the solubility of carbon dioxide in water

4

Demonstrates a sound knowledge of Le Chatelier’s Principle Outline changes in the CO2/H2CO3 equilibrium

2 ~ 3

Identifies Le Chatelier’s Principle ORIdentifies that at lower temperatures, [H+] increases

1

23 The solubility of CO2(g) in water is an equilibrium reaction : CO2(aq) + H2O(l) H+

(aq) + HCO3(aq) where H is negative.For this equilibrium, if you increase the temperature then the equilibrium will shift to the LHS reducing the solubility of the carbon dioxide. Since the CO2(g) when dissolved, reacts with water to produce H+ ions, the lower the solubility then the higher the pH. Hence, in bottle A at the higher temperature, we would expect a higher pH, indicating less H+ ions in solution as the solubility of the CO2(g) is lower than bottle B which is at a lower temperature and hence, would have more CO2(g) dissolved and hence a lower pH according to Le Chatelier’s Principle.Copyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 24 of 30

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HSC Chemistry Acidic Environment 2[Le Chatelier’s Principle]

5. First Hand Investigation: Decarbonation of Soda Water

Students identify data, plan and perform a first-hand investigation to

decarbonate soft drink and gather data to measure the mass changes involved and calculate the volume of gas released at 25 °C and 100 kPa

Soda water or carbonated water is a solution of carbon dioxide in water.

Soft drinks are carbonated water with added flavours and sweeteners.

Materials: Weighing equipment accurate to at least the nearest gram.

An unopened small bottle (or can) of soda water with liquid level as low as

possible; 250 mL or 300 mL size is adequate.

Either a source of dry heat such as an electric hotplate or a saucepan in

which the soda water can be gently warmed. A dry towel and a thermometer

(the warming method) OR 1g of table salt per 50 mL of soda water (the

salting method)

Method:1. Weigh the unopened bottle of soda water: ……………..g

2. Slowly remove the cap controlling the release of bubbles so that the soda water

does not foam out the top and you lose soda water. If this happens you will need

to buy another bottle and start all over again. Observe what happens to the

release of bubbles when you retighten the cap.

3. Reweigh the bottle including the cap: ………………….g

4. Calculate the mass of carbon dioxide lost when the cap was removed.

5. Now choose either the warming method or the salting method.

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WARMING METHOD: Uses reduced water solubility of gas with temperature rise.

6. Carefully stand the open bottle on an electric hotplate or in a saucepan of heated

water. Insert the thermometer. Stirring encourages release of gas bubbles.

7. Heat the soda water bottle to about 38C (Human body temperature)

8. Remove the soda water bottle, dry it well with a towel, then reweigh the bottle

and its cap.

Weight of bottle and cap after warming: …………………….g

Calculate the change in weight due to loss of CO2 gas to the air: see calculations

below.

SALTING METHOD: Uses the addition of ions which attract water molecules, thereby

reducing the ability of water to dissolve gases.

6. Weigh about 1g of table salt for each 50 mL of soda water accurately:

…………..g

7. Add the salt very slowly and carefully to the soda water so that the water does

not degas too quickly. This is especially important in the early stages. If soda

water foams over the top of the container you will lose weight and need to start all

over again. Control spraying so that spray does not leave the container and

affect weighing.

8. Reweigh the bottle and cap after salting: ………………….. g

Allowing for the weight of salt added to the soda water, calculate the change in

weight due to loss of CO2 gas to the air:

Calculations To calculate the volume that this mass of CO2 gas would occupy, you need

to convert the mass in g to mole using the equation:

The molar mass of CO2 = 12 + 2 16 = 44g/mole. Therefore, number of

moles of CO2, n = m/44

The molar volume of a gas at 25 °C and 100 kPa pressure is 24.8 L The

volume of gas can be calculated.

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For 1g of CO2

at 25C and 100kPa

Conclusions

Compare the volume of gas released with the volume of liquid soda water which contained that amount of gas.24

………………………………………………………………………………………

………………………………………………………………………………………

What do these volumes illustrate about the distance between CO2 particles in the gas phase compared with the distance between CO2 particles in solution?25

………………………………………………………………………………………

………………………………………………………………………………………

Explain why bubbles of CO2 gas escape from solution in a bottle when the cap is undone.26

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

24 Volume of gas released is a few times greater than the volume of soda water which contained that gas.25 These volumes indicate that CO2 molecules are further apart when in the gas phase than when dissolved in soda water.26 When the cap is undone the pressure of CO2 acting on the water surface decreases. The number of CO2 molecules entering the water is now much less than the number of CO2 molecules leaving the water. Equilibrium shifts to oppose the change by CO2 passing from solution to gas phase.Copyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 27 of 30

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Concept Check 5.1 [2004 NEAP Q25]

As part of your practical work you decarbonated a beverage. A student decarbonated a sample of soda water by opening the bottle it was in and leaving it for a period of time, weighing it at regular intervals. She also used a non-carbonated sample of water as a control, recording its mass at the same intervals.

Mass (g)

Initial After 12 hours 24 hours 36 hours 48 hours 60 hours

Soda water 385.0 382.6 381.1 380.7 380.3 380.0

Plain water 385.0 384.7 384.2 383.7 383.4 383.0

(a) Graph the information shown for each water sample on the same graph.

379

380

381

382

383

384

385

0 12 24 36 48 60 72Time / hours

Mas

s / g

ram

s

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(b) Interpret the trends shown in the graph.27

Marking Guidelines

Criteria MarksInterprets each graph and forms relationship between them 2

…………………………………………………………………………………………………

……………………………………………………………………………………………

…………………………………………………………………………………………………

……………………………………………………………………………………………

(c) Use the graph to determine the volume of CO2 gas produced at 25oC and 100 kPa. Show your working.28

…………………………………………………………………………………………………

……………………………………………………………………………………………

…………………………………………………………………………………………………

……………………………………………………………………………………………

27 The plain water graph shows the loss of mass due to evaporation. The soda water graph shows the loss of mass due to water evaporation AND carbon dioxide gas. When no more CO2 remains the line parallels the water graph.28 From the graph, the mass of carbon dioxide gas lost is 3.0g. 1.69LCopyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 29 of 30

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ANSWERS

1 directCopyright © Matrix Education 2001-2009 Results ABOVE Expectations! Page 30 of 30