DUNCANRIG SECONDARY ADVANCED HIGHER CHEMISTRY

When a reversible reaction occurs in a closed system an equilibrium is formed, in which the original reactants and products formed coexist.

UNIT ONE BOOKLET 4

Equilibrium

At equilibrium there is a state of balance between the concentrations

of the reactants and products and once a state of chemical equilibrium is

reached there is no further change in concentration. However, the reactions

do not stop! – reaction is said to be DYNAMIC!

At equilibrium the rate at which the reactants change into products is

exactly equal to the rate at which the products change back to the original reactants.

Remember forward rate = backward rate

The final relative equilibrium amounts of the reactants and products depend on the reaction conditions e.g. the temperature and pressure.

Le Chatelier - any system at equilibrium will counteract an applied change.

Raising the temperature – the equilibrium shifts in the direction of the endothermic reaction.

Raising the pressure – the equilibrium shifts in the direction with fewer moles of gas.

Raising the concentration – the equilibrium shifts to lower the concentration of the reactant/product added.

Using a catalyst – no change in the equilibrium position.

For a reaction at equilibrium it is possible to write what is known as the equilibrium expression and from this calculate a value for a quantity known as the equilibrium constant, K. The equilibrium constant characterises the composition of the equilibrium mixture.

Consider the general reaction:The equilibrium expression for this reaction is given by

The equilibrium constant relates the concentrations of the reactants and the products at a given temperature to a numerical value.This value can therefore give us an idea of the position of equilibrium for any given reaction.

Therefore the size of K will indicate if the reaction is PRODUCT favoured or REACTANT favoured. A HIGH value for K {greater than one} the PRODUCTS are favoured. A LOW value for K {less than one} the REACTANTS are favoured. For the reaction K = 47 at 490oCThe equilibrium expression for this reaction is

The value of K indicates that, at 490oC there will be a 47:1 concentration ratio of HCl to H2 and Cl2

Equilbrium constants have NO UNITS.

Any solids that are present in an equilibrium reaction are NOT included in the equilibrium expression. Solids do not have

a concentration.In heterogeneous equilibria (reactions where chemicals are in

different phases) liquids are considered as solvents and as such their concentration does not alter and so they are NOT included in the equilibrium expression.

Equilibrium constants are affected by temperature.Consider the following equilibrium reaction

As this reaction is exothermic any increase in temperature will shift the equilibrium to the left.

This will lead to [A] increasing and the [B] decreasing.If this happens the value K will DECREASE.Obviously the reverse will be true for an endothermic reaction.The table summarises what you should know.

Reaction ExothermicEndothermicIncrease temperature shifts left shifts right[Product] decreases increases

[Reactant] increases decreases

Value of K decreases increases

The opposite will be true if the reactions are cooled.Probably the easiest way to think of this is to remember the conditions for one type of reaction and work out the rest from this starting point.So you could use….

The value of K will decrease in an exothermic reaction if the reaction temperature is increased.

1. Write the equilibrium expression for each of the following reactionsa. CH3NH3

+(aq) + OH–(aq) CH3NH2(aq) + H2O(l) b. H+(aq) + OH–(aq) H2O(l) c. C(s) + 2F2(g) CF4(g) d. Fe2S3(s) 2Fe3+(aq) + 3S2–(aq) e. 2N2(g) + O2(g) N2O(g) f. N2(g) + 3H2(g) 2NH3(g)

2. Write the equilibrium expression for reactions c and e if they were written in reverse.

3. Which of the following reactions are product favoured and which are reactant favoured?a. Reaction A --- K = 200 b. Reaction B --- K = 25c. Reaction C --- K = 0.13 d. Reaction d --- K = 0.78

4. The table shows values for the equilibrium constant of a reaction carried out at different temperatures.

a. What is the relationship between K and temperature for this reaction?b. Which temperature would produce the highest yield of product?c. Use the data to explain whether the forward reaction is exothermic or endothermic.

Temperature is the only factor which alters the value of the equilibrium constant.

Concentration, pressure and catalysts DO NOT change the value of K.

This point can be difficult to grasp with respect to concentration and pressure as these factors do shift the position of equilibrium and therefore the concentrations of the reactants and the products will

also change.

How is this possible without a change in the value of K.Consider the equilibrium reaction between dinitrogen tetroxide (N2O4) and nitrogen dioxide (NO2)

N2O4 2NO2

The equilibrium expression; K = [NO2]2/[N2O4]At the start of the reaction there is 1 mol l-1 N2O4 and no NO2.

As the reaction proceeds it reaches equilibrium when the concentrations are constant (0.75 mol l-1 N2O4 and 0.50 mol l-1 NO2).

At this time more NO2 is added and the concentration increases immediately to around 0.70 mol l-1.

Temp/℃ K25 6.4 × 102

200 4.4 × 10−1

300 4.3 × 10−3

400 1.6 × 10−4

500 1.5 × 10−5

Two different coloured cobalt(II) complex ions, [Co(H2O)6]2+  and [CoCl4]2-, exist together in equilibrium in solution in the presence of chloride ions:

The equation for the reaction is:

[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l) PINK BLUE

This equilibrium can be disturbed by changing the temperature of the reaction mixture. The colour changes accompanying the changes in equilibrium position are as predicted by  Le Chatelier’s principle.

1. Place 15 cm3 of the violet coloured cobalt solution provided into three boiling tubes.

2. Keep one tube as a control.

3. Place the second tube in a beaker containing water at approximately 90oC

4. Place the third tube in a beaker containing an ice/water at approximately 5oC

5. When the colours have developed swap over the boiling tubes in the “hot and cold beakers”

a. Initial colour of equilibrium mixture. _____________________________________

At the start of the reaction there is 1 mol l-1 N2O4 and no NO2.

As the reaction proceeds it reaches equilibrium when the concentrations are constant (0.75 mol l-1 N2O4 and 0.50 mol l-1 NO2).

At this time more NO2 is added and the concentration increases immediately to around 0.70 mol l-1.

b. Colour of equilibrium mixture when temperature is increased ___________________

c. Colour of equilibrium mixture when temperature is decreased __________________

d. Observations when the boiling tubes are swapped over.

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a. State the names of the two cobalt complex ions.

[Co(H2O)6]2+ _____________________________________________

[CoCl4]2- _____________________________________________

b. Write the equilibrium expression for this reaction.

c. Explain why the mixture you were given was violet in colour.

_____________________________________________________________

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d. Which complex ion’s concentration increased when the temperature of the reaction mixture increased?

______________________________________________________________

e. In which direction does the equilibrium move when the temperature of the reaction mixture is increased?

______________________________________________________________

f. Is the forward reaction exothermic or endothermic?

______________________________________________________________

g. What would happen to the [Cl-] if

(i) the temperature is increased ____________________________________

(ii) the concentration of the [CoCl4]2- was increased? ____________________

h. What would happen to the value of the equilibrium constant, K, if

(i) the temperature is increased ____________________________________

(ii) the concentration of the [CoCl4]2- was increased? ____________________

To calculate the value of the equilibrium constant, K, for a particular reaction two factors must be known.

1. The ester ethyl ethanaote can be prepared by reacting ethanol with ethanoic acid.

CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

The equilibrium concentrations of ethanol and ethanoic acid are 0.333 mol l-1 and the equilibrium concentrations of ethyl ethanoate

and water are 0.667 mol l-1.Step 1 – Write the equilibrium expression.

Step 2 - Substitute the values given into the expression and do the arithmetic.

Remember - the equilibrium constant symbol is a capital K and it has no units.

2. In an experiment to determine the equilibrium constant for the reaction between ethanoic acid and ethanol, 0.20 mol of ethanoic acid

All the species are liquid – reaction is homogeneous – all are included in the expression.

this may not always be the case.

were mixed with 0.25 mol of ethanol. The mixture was allowed to react for one week at 50oC. At equilibrium the mixture contained 0.053 mol of ethanoic acid.

CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

Calculate the value of the equilibrium constant, K at 50oC.In this example the equilibrium quantities for all the species are

NOT given and must be worked out. The question does provide the initial quantities of the reactants.Step 1 – Write the equilibrium expression.

Step 2 – Calculate the number of moles of each species at equilibrium.The question tells us that at equilibrium 0.053 mol of ethanoic acid remain in the mixture.

Therefore moles of ethanoic acid reacting must be what was there at the beginning minus what is left at equilibrium.

Moles ethanoic acid reacting = 0.20 – 0.053 = 0.147 mol

In this example the stoichiometry of the reaction is 1:1:1:1 for all reactants and products. Using this fact and the moles of ethanoic acid which reacted, the equilibrium quantities of all the species can be found

CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

Initial mol 1 1 0 0

Equilibrium 0.053 0.25 - 0.147 0.147 0.147 mol = 0.103

Step 3 – Substitute values into the equilibrium expression.

3. For the reaction Fe3+ (aq) + 4Cl- (aq) FeCl4-(aq) the equilibrium constant, K, has a value of 8.1 x 10-2 at 298K

given in question.

Equation mole ratio is 1:1. Moles of product formed = moles of reactant used.

Initial moles ethanol – moles of ethanol reacted.

In an equilibrium mixture at 298 K, the concentration of the chloride ion was 0.20 mol l-1 and the concentration of the iron(III) ion was 0.69 mol l-1. Calculate the concentration of the complex ion, FeCl4-(aq). Step 1 – Write the equilibrium expression.

Step 2 – Rearrange expression for FeCl4-(aq).

Step 3 – Substitute values and do the arithmetic. [FeCl4-(aq)] = 8.1x10-2 x 0.69 x (0.20)4 = 8.9x10-5 mol l-1.

1. PCl5 (g) PCl3(g) + Cl2(g) Calculate the equilibrium constant, K, given the following equilibrium concentration data for the reaction shown above. [PCl3(g)] = 0.015 mol l-1 [Cl2(g)] = 0.015 mol l-1 [PCl5(g)] = 1.18 x 10-3

mol l-1

2. The equilibrium shown below was established in a solvent at 10°C in a one litre container.

N2O4(l) 2NO2(l) The initial concentration of the N2O4(l) was 0.1307 mol l-1 and the equilibrium concentration of the NO2(l) was 0.0014 mo l-1.

a. Calculate the value of the equilibrium constant, K, for this reaction at 10oC.b. What would happen to the value of K if the initial concentration of N2O4(l) was doubled? 3. An experiment was carried out to determine the equilibrium constant, K, for the following reaction.

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

Initially 0.110 mol of hydrogen and 0.110 mol of iodine were mixed in a one litre container at 473K.

The number of moles of hydrogen iodide present at equilibrium was found to be 0.171 mol.a. Calculate the value of the equilibrium constant, K, at 473K.b. What will happen to the value of K if the temperature of the equilibrium mixture is increased?

c.Calculate the value of K’, the equilibrium constant for the reverse reaction.

4. Carbonyl bromide, COBr2, decomposes according to the equation,COBr2(g) CO(g) + Br2(g)

The value of the equilibrium constant,K, for this reaction at 346K is 0.190.a. If the concentration of CO(g) and Br2(g) is 0.046 mol l-1 at 346K,

calculate the concentration of COBr2(g) at 346K .

5. The reaction between carbon monoxide and hydrogen proceeds according to the equilibrium

CO(g) + 2H2(g) CH3OH(g) A one litre flask maintained at 700K contains 0.10 mol of carbon

monoxide. After 0.30 mole of hydrogen is added, 0.06 mol of methanol are formed. Calculate the equilibrium constant, K, at 700K.

6. The ester commonly known as diethyl malonate (DEM) occurs instrawberries and grapes. It can be prepared from acid A according to

the following equilibrium.

a. A mixture of 2.50 mol of A and 10.0 mol of ethanol was left to reach equilibrium in an inert solvent in the presence of a small amount of concentrated sulfuric acid.The equilibrium mixture formed contained 1.80 mol of DEM.

(i) Using A to represent the acid and DEM to represent the ester, write the equilibrium expression for this reaction.(ii) Calculate the number of moles of ethanol, water and A in the equilbrium mixture.(iii) Calculate the value of the equilibrium constant, K.(iv) What would happen to the value of K if the total volume of the equilibrium mixture was doubled by the addition of more of the inert solvent.

(b) In a second experiment, the equilibrium mixture was found to contain 0.85 mol of A, 7.2 mol of ethanol, 2.1 mol of DEM and 3.4 mol of water.(i) Calculate the value for the equilibrium constant, K for these

concentrations.(ii) Suggest a reason why the value for the equilibrium constant, K, in a(iii) is different from the value in b(i).Liquids which do not mix with each other are Immiscible.

A solute is the chemical dissolved in a solvent.Polar solutes will tend to dissolve in polar solvents. Non – polar solutes tend to dissolve in non – polar solvents.Hydrocarbons are non – polar and water is polar.If a substance is added to a mixture which is soluble to a greater or lesser extent in two immiscible liquids, on shaking and then allowing the mixture to settle, the concentrations in each layer become constant.

However, there is continual interchange of solute between the liquid layers via the interface i.e. a dynamic equilibrium is formed. When this happens the system will have an equilibrium constant – this is called the ‘Partition Coefficient’.

For example, iodine which is soluble in both water and hexane, will partition between the two solvents.

1. An aqueous solution of an organic acid, X, was shaken with chloroform until the following equilibrium was established.

a. Write the equilibrium expression for the reaction.

b. Why would the titration of the organic acid in the chloroform layer have been more difficult to carry out in practice than the titration of the organic acid in the aqueous

layer?c. Name this piece of apparatus. d. 25.0cm3 of the upper layer needed 20.0cm3 of 0.050 mol l–1 NaOH(aq) for neutralisation. 25.0cm3 of the lower layer needed 13.3cm3 of 0.050 mol l–1 NaOH(aq) for neutralisation.

The equation for the system on the left is A(Y) A(X)The equilibrium expression is

Partition can used to extract and help purify a desired product from a reaction mixture. This technique is called solvent extraction and employs the use of a separating funnel. The method depends on the desired material being more soluble in one liquid phase than another

If some coffee is dissolved in water and then some dichloromethane is added, the caffeine in the coffee is more soluble in the dichloromethane. Thus the watery coffee will now be caffeine free. As the dichloromethane and water are immiscible, two layers will form which can be separated easily. The caffeine can be recovered by evaporating the dichloromethane. This method can be improved by extracting the solute multiple times, or by using several smaller volumes rather than one large volume of the extracting

solvent.

(i) Assuming the organic acid reacts with NaOH in a one mole to one mole ratio, calculate the concentration of the organic acid in each layer.(ii) Calculate the value of the partition coefficient. (iii) What would happen to the value of the partition coefficient if the volume of chloroform used was doubled.d. The organic acid is soluble in ethanol. Suggest why ethanol would not be a good choice of solvent to extract the organic acid from the water.2. An aqueous solution of an organic acid, X, was shaken with ethoxyethane until the following equilibrium was established.

20.0cm3 of the upper layer needed 15.0cm3 of 0.010 mol l–1 KOH(aq) for neutralisation. 20.0cm3 of the lower layer needed

12.0cm3 of 0.010 mol l–1 KOH(aq) for neutralisation. Calculate the value of the partition coefficient.

A list of “learning outcomes” for the topic is shown below. When the topic is complete you should review each learning outcome.Your teacher will collect your completed notes, mark them, and then decide if any revision work is necessary.

Be able to predict the direction of equilibrium if concentration, pressure or temperature are altered in an equilibrium reaction

Be able to write the equilibrium expression for any equilibrium reaction.

State that solids are not included in the equilibrium expression.

State that, in heterogenous equilibria, liquids considered as solvents are not included in the equilibrium expression.

State that every equilibrium reaction has an equilibrium constant, K, and that this constant has no units.

Need Help

Understand

Revise

State that if the equilibrium constant is greater than 1 the equilibrium reaction is product favoured.

State that if the equilibrium constant is less than 1 the equilibrium is reactant favoured.

State that only temperature will alter the value of the equilibrium constant. Factors such as concentration, volume and pressure do not alter the value of the equilibrium constant.

Be able to calculate a value for the equilibrium constant from given either the equilibrium concentrations or the initial concentrations of all or some of the species in the equilibrium.

State that a partition coefficient is an equilibrium constant for a system which has a substance dissolved in two immiscible solvents.

State that solvent extraction is a process used to remove a dissolved substance from a solution. This process can be used to purify a substance.

State that more of the dissolved substance can be removed from solution if the extraction is carried out more than once or several small volumes of solvent are used instead of one large volume of solvent.

Be able to calculate the partition coefficient given the concentrations of the dissolved substance.

State that the value of a partition coefficient is altered by temperature but not by changing the volume of the solvent of the mass of the dissolved chemical.

I have discussed the learning outcomes with my teacher.

Teacher comments.

Date. __________________________________

Pupil Signature. __________________________

Teacher Signature. _______________________

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