chemistry form 5 chapter 1

Form 5 Chemistry Notes – Ms. R. Buttigieg Pg. 1

Form 5 Chemistry Notes

What you shall be needing

� A4 notebook for practicals � 2 copybooks for HW and CW � A copy of the Chemistry syllabus � SEC exam papers 2004-2006 � JL papers 2003 -2005

(can print out from internet by going on http://www.curriculum.gov.mt/ then choose Examination papers, Select JL papers and the years you need)

Main topics to be discussed during the year:

1. Energetics 2. Organic Chemistry 3. Carbon chemistry and structures


Topic 1 Energetics

Energy changes accompany both physical and chemical changes. Principle of Conservation of Energy: Energy can be changed from one form to another but it can neither be created nor destroyed.

Exothermic and endothermic reactions � Chem 4 You pg. 190-195, GCSE Chem pg.190-191 When a chemical reaction happens, energy is transferred to or from the surroundings and often there is a temperature change. For example, when a bonfire burns it transfers heat energy to the surroundings. Objects near a bonfire become warmer and the temperature rise can be measured with a thermometer. Exothermic reactions These are reactions that transfer energy (liberate energy) to the surroundings. The energy is usually transferred as heat energy, causing the reaction mixture and its surroundings to get hotter. The temperature increase can be detected using a thermometer. Some examples of exothermic reactions are:

• burning • neutralisation reactions between acids and alkalis, and • the reaction between water and calcium oxide

These reactions can be represented by a thermochemical equation like the one below:

2 H2 (g) + O2 (g) 2H2O (l) ∆ H = -575 kJ

� The symbol ∆ H shows the change in energy. � The minus (-) sign shows that energy is lost from the substance and an exothermic

reaction has occurred.

Endothermic reactions These are reactions that take in energy from the surroundings. The energy is usually transferred as heat energy, causing the reaction mixture and its surroundings to get colder. The temperature decrease can be detected using a thermometer. Some examples of endothermic reactions are:

• electrolysis • the reaction between ethanoic acid and sodium carbonate, and • the thermal decomposition of calcium carbonate in a blast furnace

N2 (g) + O2 (g) 2NO (g) ∆ H = + 180.6 kJ (+ energy gained by substance)


� The chemical energy stored in the bonds in a substance gives us a measure of a chemical's energy level.

� The higher the energy level of a substance, the more chemical energy is stored in its bonds.

� The reactants and products in a chemical reaction usually have different energy levels, which are shown in a type of graph called an energy level diagram.

� The vertical axis on this diagram represents the energy level and the horizontal axis represents the progress of the reaction from reactants to products.

Energy level diagrams for exothermic reactions

In an exothermic reaction, reactants have a higher energy level than the products. The difference between these two energy levels is the energy released to the surroundings in the reaction, and an energy level diagram shows this as a vertical drop from a higher to a lower level:

� Usually some extra energy is needed to get the reaction to start.

� This is called the activation energy and is drawn in energy level diagrams as a hump.

� Catalysts reduce the activation energy needed for a reaction to happen - this lower activation energy is shown by the dotted line in the diagram here.

Energy level diagrams for endothermic reactions

� In endothermic reactions the reactants have a lower energy level than the products.

� The difference between these two energy levels is the energy gained from the surroundings in the reaction, represented in an energy level diagram as a vertical jump from a lower to a higher level - the bigger the jump, the more energy is gained.

• State symbols are important as the energy change also depends on the state

• Energy must be supplied to break bonds.

• Energy is released when bonds are formed.

1. STUDY CHEMISTRY FOR YOU PG. 194,

o Work out pg. 198 nos 8,9.

o Work out pg. 257 nos 25,26


In an exothermic reaction, the energy released from forming new bonds is greater than

the energy needed to break existing bonds.

In an endothermic reaction the energy needed to break existing bonds is greater than the

energy released from forming new bonds.

These energy changes can be shown in an energy level diagram.

The products have less energy than

the reactants.

The products have more energy than

the reactants.

30 kJ of energy has been given out. 15 kJ of energy has been taken in.

The reaction is exothermic The reaction is endothermic.

The amount of energy transferred is found by comparing the energy in the reactants with

that in the products.

Reaction type: exothermic endothermic

heat is: released absorbed

reaction vessel temperature: rises falls

enthalpy change is negative positive

net bond: formation breaking

Answer the following:

1. When sodium carbonate and calcium chloride solution are mixed, the temperature of the solution dropped by 4oC. Is the reaction exothermic or endothermic? ______________________

2. Are combustion reactions exothermic or endothermic? ______________________

3. Is energy needed or given out when bonds are broken? ______________________

4. Is energy needed or given out when bonds are formed? ______________________

5. Are ∆H values for an exothermic reaction positive or negative? ______________________

Now let’s have a look at some of the reactions accompanied by energy changes See GCSE pg. 190-191


Energy changes are caused by bond breaking and forming Breaking a bond - requires energy; Forming a bond – releases energy

Energy Changes Standard enthalpy of combustion – the heat change which occurs when 1 mole of the substance burns completely in oxygen at STP (temperature of 25oC, pressure of 1 atm) Example: When carbon burns completely it forms Carbon dioxide. If not completely burnt it forms Carbon monoxide.

C (s) + O2 (g) � CO2 (g) ∆ H = -394 kJ/mol Standard enthalpy of formation – the heat change which occurs when 1 mole of the substance is formed from its elements at STP Example: When 1 mole of Carbon disulphide is formed from its elements.

C (s) + 2 S (g) � CS2 (l) ∆ H = +117 kJ/mol Standard enthalpy of precipitation – the heat change which occurs when 1 mole of the substance is precipitated from solution at STP Example: When 1 mole of Silver Chloride is precipitated

Ag+ (aq) + Cl- (aq) � AgCl (s) ∆ H = - 50.2 kJ/mol Standard enthalpy of neutralization – the heat change which occurs when 1 mole of acid is neutralized by an alkali to form 1 mole of water at STP .

Example: NaOH (aq) + HCl (aq) � NaCl (aq) + H2O (l) ∆ H = - 57.5 kJ/mol


Standard enthalpy of solution– the heat change which occurs when 1 mole of the substance is dissolved in a stated quantity of water (or till further dilution does not involve any energy change).

The standard enthalpy of solution = lattice energy + hydration energy

The lattice energy is the energy needed to separate the ions and break down the lattice. (Endothermic) The hydration energy is the energy released when the ions bond with water molecules (exothermic) Standard enthalpy of reaction – the energy change that occurs when the reactants react in the number of moles as expressed by the equation.

Work these: 1. Give the meaning of:

� Enthalpy � Exothermic � Principle of Conservation of energy � Standard enthalpy of combustion � STP

2. 2 H2 (g) + O2 (g) � 2H2O (l) ∆ H = -575 kJ

a) Explain the reaction in words b) What is the standard enthalpy of reaction for the reaction above? c) What is the standard enthalpy of combustion of Hydrogen? d) What is the standard enthalpy of formation of water?

3a) The standard enthalpy of neutralization of hydrochloric acid and that for nitric acid is the

same. Why? b) The standard enthalpy of neutralization of sulphuric acid is double that for hydrochloric acid.

Why?


Find the standard enthalpy change in each of the following and say whether each reaction is exothermic or endothermic.

1. Hydrogen + Chlorine ==> Hydrogen Chloride

o The symbol equation is: H2(g) + Cl2(g) ==> 2HCl(g) o but think of it as: H-H + Cl-Cl ==> H-Cl + H-Cl o (where - represents the chemical bonds to be broken or formed) o the bond energies in kJ/mol are: H-H 436; Cl-Cl 242; H-Cl 431

2. Hydrogen Bromide ==> Hydrogen + Bromine

o The symbol equation is: 2HBr(g) ==> H2(g) + Br2(g) o but think of it as: H-Br + H-Br ==> H-H + Br-Br o the bond energies in kJ/mol are: H-Br 366; H-H 436; Br-Br 193

3. hydrogen + oxygen ==> water

o 2H2(g) + O2(g) ==> 2H2O(g) o or 2 H-H + O=O ==> 2 H-O-H (where - or = represent the covalent bonds) o bond energies in kJ/mol: H-H is 436, O=O is 496 and O-H is 463

4. nitrogen + hydrogen ==> ammonia

o N2(g) + 3H2(g) ==> 2NH3(g)

o or N N + 3 H-H ==> 2 o bond energies in kJ/mol: NN is 944, H-H is 436 and N-H is 388

SEC CHEMISTRY MAY 2002 – Paper I 5. Sulphur dioxide and oxygen react to form sulphur trioxide according to the equation:

2 SO2 (g) + O2 (g) -------- 2SO3 (g) ∆ H = -197 kJ

a) Is the reaction exothermic or endothermic? b) What do you understand by the symbol ∆ H? c) What is the value of ∆ H for the reverse reaction? d) Use the information given in the equation above to draw a simple energy level diagram for

the formation of sulphur trioxide. Show ∆H on your diagram. 6. Consider the following reactions: i. C (s) + O2 (g) � CO2 (g) ∆ H = -393.5 kJ ii. C (s) + 2S (s) � CS2 (l) ∆ H = +117 kJ a) Say whether the temperature increases, decreases or doesn’t change in each reaction. b) Use the terms endothermic and exothermic to describe each reaction c) Which one of the reactions would you use to heat a room? Give a reason for your answer.


Finding the Standard Enthalpy of Combustion in the Lab � The diagram shows the apparatus to find the heat of

combustion in the laboratory. (draw complete setup) � The calorimeter has a copper spiral so that the hot gases

formed by the combustion of ethanol pass through the spiral and so heat the surrounding water.

� The calorimeter is well lagged on the outside. � The end of the spiral leads to a flask with water and

thermometer. This serves as a control to check whether heat energy is lost from the calorimeter.

� If all the heat energy is transferred to the water in the calorimeter, the temperature in the flask should not rise.

METHOD:

1) A lamp is filled with ethanol, capped and then the lamp and cap is weighed.

2) The lamp is lit and put beneath the calorimeter. 3) The air-supply is controlled to get a blue flame. 4) Water is stirred continuously. 5) After some time, the flame is extinguished; the

lamp is immediately capped and weighed. 6) The maximum temperature of the water is noted. 7) The standard heat of combustion can then be found. Readings you need to take

Mass of water Mass of lamp and ethanol before burning Mass of lamp after lighting Final temperature Initial temperature Mass of Copper Calorimeter

Mass of Ethanol burnt = Mass of lamp before – Mass of lamp after Temperature Rise = Final – Initial temperature Calculation: 1. Heat energy liberated by ethanol = heat energy absorbed by water + copper calorimeter

= mw x cw x ∆θ + mCu x cCu x ∆θ = ___ joules

2. If x moles were burnt during the experiment, x moles liberated ___ joules 1 mole liberates? Joules Sources of Error:

� Heat is lost to the surroundings � Some ethanol may have evaporated


A simpler method

This method is specifically for determining the heat energy released

(given out) for burning fuels. The burner is weighed before and after

combustion to get the mass of fuel burned. The thermometer records

the temperature rise of the known mass of water (1g = 1cm3).

You can use this system to compare the heat output from burning

various fuels. The bigger the temperature rise, the more heat energy is

released. See calculations below for expressing calorific values.

This is a very inaccurate method because of huge losses of heat eg

radiation from the flame and calorimeter, conduction through the

copper calorimeter, convection from the flame gases passing by the

calorimeter etc. BUT, at least using the same burner and set-up, you

can do a reasonable comparison of the heat output of different fuels.

• You need to know the following: o the mass of material reacting in the calorimeter (or their concentrations and

volume), o the mass of water in the calorimeter, o the temperature change o the specific heat capacity of water, (shorthand is SHCwater), and this is 4.2J/g/oC,

� this means it takes 4.2 J of heat energy to change the temperature of 1g of

water by 1oC (or K).

• Example o 100 cm3 of water (100g) was measured into the calorimeter. o The spirit burner contained the fuel ethanol C2H5OH ('alcohol') and weighed 18.62g

at the start. o After burning it weighed 17.14g and the temperature of the water rose from 18

to 89oC. o The temperature rise = 89 - 18 = 71oC (exothermic, heat energy given out). o Mass of fuel burned = 18.62-17.14 = 1.48g. o Heat absorbed by the water = mass of water x SHCwater x temperature

� = 100 x 4.2 x 71 = 29820 J (for 1.48g) � heat energy released per g = energy supplied in J / mass of fuel burned in g � heat energy released on burning = 29820 / 1.48 = 20149 J/g of C2H5OH

o this energy change can be also expressed on a molar basis. � Relative atomic masses Ar: C = 12, H = 1, O = 16 � Mr(C2H5OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g � Heat released (given out) by 1 mole of C2H5OH = 46 x 20149 = 926854

J/mole � At AS level this will be expressed as enthalpy of combustion =

Hcomb = -926.9 kJmol-1 � The data book value for the heat of combustion of ethanol is -1371 kJmol-1,

showing lots of heat loss!


Rates of reactions

How fast (or slow) a reaction occurs depends on 5 important factors:

� Temperature � Surface area � Presence of a catalyst � Concentration � Pressure


Factors, which increase the speed of a reaction, are those that increase the number of

collisions between particles and the energy of these collisions.

Concentration: So increasing the concentration of hydrochloric acid in its reaction with calcium carbonate increases the speed of reaction because there are more acid particles per unit volume. So they collide more with the calcium carbonate and the chance of them reacting is greater

Surface Area: Increasing the surface area of calcium carbonate in the same reaction also increases the speed of reaction as you are exposing more calcium carbonate to be acted upon by the acid. More collisions occur and the chance of reaction is greater.

Effect of temperature: Increasing the temperature increases the rate of reaction as raising the

temperature makes the particles move faster (have more kinetic energy). This means that the

particles collide more frequently with each other and the rate of the reaction increases.

� Also, the faster the particles are traveling, the greater is the proportion of them, which

will have the required activation energy for the reaction to occur. � As a general guide, raising the temperature of a reaction by 10 °C will double the rate of

a reaction. � The gradient of the plot will be twice as steep

Pressure: affects reactions involving gases, since increasing pressure increases the concentration of gas. Catalysts: Light catalyses certain reactions like the decomposition of silver chloride into silver

and chlorine – known as photoreduction of silver halides

In surface catalysts, the reaction takes place on the surface. Catalysts reduce the activation

energy needed for the reaction to begin.


Examples of catalysts

Light - A mixture of Chlorine, Cl2, and Hydrogen, H2, explodes when exposed to sunlight to give Hydrogen Chloride, HCl. In the dark, no reaction occurs, so activation of the reaction by light energy is required.

ClClClCl2 2 2 2 (g) + H (g) + H (g) + H (g) + H2222 (g) ==> 2 HCl (g) (g) ==> 2 HCl (g) (g) ==> 2 HCl (g) (g) ==> 2 HCl (g)

Catalyst Reaction

Manganese Dioxide - MnO2 2 H2O2 (aq) � 2 H2O (l) + O2 (g)

Iron - Fe N2 (g) + 3 H2 (g) � 2 NH3 (g)

Vanadium (V) Oxide 2 SO2 (g) + O2 (g) � 2 SO3 (g)

In a Catalytic Converter 2 NO (g) + 2 CO (g) � N2 (g) + 2 CO2 (g)

Platinum/Rhodium 4 NH3 (g) + 5 O2 (g) � 4 NO (g) + 6 H2O (l)

Experiments to determine the Rate of Reaction: In such reactions it is important that only one factor is changed each time. When the two reactants are mixed the stopwatch is started and the volume/mass are noted at regular time intervals. Graphs can then be plotted.

The effect of temperature on Reaction Rate

� A crucible with zinc is placed floating on the dilute hydrochloric acid in the flask. � The flask is connected to an empty gas syringe. � The two reactants are mixed, while the stopwatch is started. � The volume of gas produced is measured at regular intervals till production stops.

Zn (s) + 2 HCl (aq) � ZnCl2 (aq) + H2 (g)

� The temperature is then changed and the same experiment repeated (keeping mass, volume and acid concentration constant).

� Both graphs are plotted on the same axis. � It is noted that the reaction at a higher temperature has a steeper gradient.


The effect of surface area on Reaction Rate � The same set-up on page 12 is used. � Calcium carbonate chips are reacted with dilute hydrochloric acid. � Calcium carbonate powder is then reacted with dilute hydrochloric acid. � Both reactions are compared. This reaction can also be done on a balance. When the two reactants are mixed, the stopwatch is started and the mass taken at regular intervals.

The effect of concentration on Reaction Rate

A typical reaction used for such experiments is the reaction between sodium thiosulphate solution and dilute hydrochloric acid. The experiment is repeated at different concentrations.

• To follow this reaction in your investigation you can measure how long it takes for a certain

amount of sulphur to form. • You do this by observing the reaction down through a conical flask, viewing a black cross on

white paper (see diagram below). • The X is eventually obscured by the sulphur precipitate and the time noted.

Na2S2O3 (aq) + 2 HCl (aq) � 2 NaCl (aq) + SO2 (g) + H2O (l) + S (s)

Form 5 Chemistry Notes – Ms. R. Buttigieg Pg. 14 The effect of catalyst on Reaction Rate

� Two test tubes are filled with some hydrogen peroxide. � In one flask some manganese dioxide powder, which has been

previously weighed, is placed in one test tube. � As soon as the manganese dioxide is added rapid effervescence

occurs. � The gas produced relights a glowing splint (oxygen) � When the manganese dioxide is filtered, washed, left to dry and

weighed, it is found that the reaction did not use it up. So MnO2 affects the speed of reaction without being used up.

The effect of light on silver chloride

� Some silver nitrate is added onto sodium chloride solution. � A white precipitate of sodium chloride is formed. � One portion is placed in sunlight, the other in the dark. � After some time the one in sunlight goes black, the other remains white.

Silver nitrate ( ) + Sodium chloride ( ) � Silver chloride ( ) + Sodium nitrate ( )

Answer the following: 1. Draw a labeled diagram of the set-up to follow the speed of reaction by measuring the volume

of gas produced when calcium carbonate is reacted with hydrochloric acid.

a. If the experiment is done to find the effect of surface area you would carry out two

experiments. What must be kept the same in both experiments? What must be

changed from one experiment to the next?

b. Both graphs are plotted on the same graph paper, sketch the graphs that are expected.

i. Clearly label each graph

ii. Mark the point on the graph where the speed of reaction is the highest

iii. Mark the point where the reaction stops.

c. What can you conclude from this experiment?

d. Explain what happens in terms of particles.

2. Write the name of the catalyst used in the following reactions:

a. 2 H2O2 (aq) � 2 H2O (l) + O2 (g) b. 4 NH3 (g) + 5 O2 (g) � 4 NO (g) + 6 H2O (l)

3. Give an example and the equation of a reaction that is affected by light.


Reversible reactions and chemical equilibrium

If the direction of a reversible reaction is changed, the energy change is also reversed.

For example: the thermal decomposition of hydrated copper(II) sulphate.

On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of

anhydrous copper(II) sulphate is formed.

o This a thermal decomposition and is endothermic as heat is absorbed (taken in)

o The energy is needed to break down the crystal structure and drive off the water.

When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed.

o The reverse reaction is exothermic as heat is given out.

o i.e. on adding water to white anhydrous copper(II) sulphate the mixture heats up as

the blue crystals reform.

blue hydrated copper(II) sulphate + heat white anhydrous copper(II) sulphate + water

CuSO4.5H2O(s) CuSO4(s) + 5H2O(g) Another example: is the chromate/dichromate equilibrium

Yellow chromate ion is turned orange by addition of acid. The orange dichromate is returned to yellow by the addition of base.

2 CrO42- (aq) + 2 H+(aq) Cr2O7

2- (aq) + H2O (l)

Chromate ions, CrO42- (aq), are yellow. Dichromate ions, Cr2O7

2- (aq), are orange. Since this is an equilibrium, both ions are present in a solution at the same time. The color of the solution indicates which of the ions is present in the greater amount.

Another example: is esterification

Ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid as a catalyst to produce the ester, ethyl ethanoate. The reaction is slow and reversible. To reduce the chances of the reverse reaction happening, the ester is distilled off as soon as it is formed

.


Kinetic picture of dynamic equilibrium

� A dynamic equilibrium occurs when you have a reversible reaction in a closed system.

Nothing can be added to the system or taken away from it apart from energy.

� At equilibrium, the quantities of everything present in the mixture remain constant, although

the reactions are still continuing. This is because the rates of the forward and the back

reactions are equal.

A dynamic equilibrium occurs when two reversible processes occur at the same rate. Many

processes (such as chemical reactions) are reversible.

An example of the process can be imagined if a bucket is filled with water and placed in a small

room. The water from the bucket will evaporate, and the air in the room will start to become

saturated with water vapor. Eventually, the air will be completely saturated with water, and the

level of water in the bucket will stop falling. However, water from the bucket is still evaporating.

What is happening is that molecules of water in the air will occasionally hit the surface of the

water and condense back into the liquid water, and this occurs at the same rate at which water

evaporates from the bucket. This is an example of dynamic equilibrium, because the rate of

evaporation equals the rate of condensation.

Water in a closed system, at constant temperature, is in a dynamic equilibrium.

H2O(l) � H2O(g) forward reaction; H2O(g) � H2O(l) reverse reaction

At equilibrium H2O(l) H2O(g)

If you change the conditions in a way which changes the relative rates of the forward and back

reactions you will change the position of equilibrium - in other words, change the proportions of

the various substances present in the equilibrium mixture.


In a saturated solution of KCl (aq), at constant temperature, with excess KCl (s) the following

reactions occur.

KCl(s) � K+(aq) + Cl-(aq) forward reaction (dissolution)

K+(aq) + Cl-(aq) � KCl(s) reverse reaction (crystallization)

At equilibrium KCl(s) K+(aq) + Cl-(aq)

The diagram shows what happens to the molecules when there is a dynamic equilibrium.

If conditions are changed, the equilibrium shifts to one side rather than the other, depending on which direction is favoured.

Le Chatelier’s Principle says where the equilibrium shifts when the conditions are changed.

It states that: The equilibrium shifts so as to oppose the change disturbing it.

E.g The following reaction is in equilibrium: N2 (g) + 3 H2 (g) 2 NH3 (g) If the reaction proceeds to the right �, it decreases pressure as it reduces the number of particles from 4 molecules (a nitrogen molecule and 3 hydrogen molecules) to 2. If the reaction proceeds to the right �, it produces heat; exothermic. So if from outside the system the pressure is increased this will result in a higher percentage of ammonia. As according to Le Chateliers’ principle:

� We increase pressure � The equilibrium tries to decrease it � Shift to right � producing more ammonia � In industry a pressure of 300 atm is used to produce ammonia.


Temperature - If the mixture is heated from outside the equilibrium will reduce the heat so the equilibrium shifts to the left �, reducing the amount of ammonia.

� So increasing the temperature reduces the amount of ammonia � In industry a balance is found:

o A low temperature produces more ammonia at equilibrium but this is formed very slowly;

o A high temperature produces less ammonia at equilibrium but this is formed quickly.

� So a temperature of 450oC is used. � Less ammonia is formed at equilibrium but it is formed more quickly.

The effect of a catalyst on the rate of attainment of equilibrium and not on the equilibrium position. � A catalyst does not effect the amounts of products at equilibrium as it catalyses both the

forward and backward reaction. It helps the equilibrium point to be reached more quickly. � In industry, iron is used in this reaction.


� In industry ammonia is removed from the mixture as it becomes liquefied under that high pressure.

� This helps the equilibrium to keep shifting to the right � to increase the amount of ammonia. The following is another equilibrium reaction:

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

� The forward reaction reduces pressure and is exothermic � If we perform the reaction at a low temperature, equilibrium shifts to increase

temperature; so it shift to the right � and produces more sulphur trioxide. � In industry they don’t use low temperatures as this reduces rates. � So a temperature of 450oC is used. � Less sulphur trioxide is formed at equilibrium but is formed more quickly.

Oxygen

� The oxygen used is 3 times as theoretically required. � Equilibrium shifts to reduce the amount of oxygen; so it shifts too the right and produces

more sulphur trioxide. Pressure

� If pressure is increased, the equilibrium shift to reduce pressure � So it shifts to the right � to produce more sulphur trioxide � In industry high pressures aren’t used as at atmospheric pressure, 98% sulphur trioxide is

already formed at equilibrium � The catalyst vanadium (V) oxide doesn’t change the amount of sulphur trioxide produced

at equilibrium but it helps to reach the equilibrium point more quickly. JL 2005 Annual

3. a) This question is about the effect of concentration and of temperature on an equilibrium

mixture. The equation for the system is:

i) What colour would the mixture turn to if some concentrated hydrochloric acid is added?

Give a reason for your answer using Le Chatelier’s principle.(3 marks)

ii) The forward reaction is endothermic. If a warm sample of the mixture is cooled, it turns pink.

Explain this observation in terms of Le Chatelier’s principle. (2 marks)

b) Cobalt chloride can act as a catalyst. What effect (if any) does a catalyst have on the

equilibrium position? Explain your answer. (3 marks)

c) Cobalt is a transition metal. Give two properties (apart from acting as a catalyst) that

cobalt would show. (2 marks)

chemistry form 5 chapter 1

Documents

le chateliers principle

dilute hydrochloric acid

blue hydrated copper

energy level diagrams

white anhydrous copper

break existing bonds

activation energy needed

energy level diagram