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How Far? How Fast? - Chemical Equilibria

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Chemical Equilibria- Introduction

The vast majority of organic chemical reactions are reversible - that is the products decompose to the reactants as the reactants are forming products. A general equation would look like this -

aA + bB cC + dD

At the position of equilibrium the rate of the forwards reaction is equal to the rate of the reverse reaction. At equilibrium a certain percentage of the reaction mixture will be products -

aA + bB cC + dD

The position of equilibrium lies somewhere between 0% and 100% products.

The position of equilibrium is represented by the equilibrium constant, given the symbol Kc -

and [ ]eqm = the concentration of a molecule at equilibrium.

N.B.: A catalyst does not effect the value of Kc at a given temperature. It increases the rate of the forwards and backwards reactions.

The value of Kc is only affected by a change in temperature.

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Chemical Equilibria- Le Chatelier's Principle

 

In summary,

Whatever is done to an equilibrium, the equilibrium will undo.

(1) Change in concentration :

The value of Kc is a constant for a particular reaction at a constant temperature. Therefore, if the concentration of one of the components in a reaction is changed the other values must change to keep the value of Kc constant.

e.g. if a reactant concentration is increased the reaction will form more products. The values on the top of the equation must increase to balance the change on the bottom of the equation.

(2) Change in pressure :

When the pressure of a gas phase reaction is increased then the reaction tries to reduce the pressure by moving to the side of the equation with the fewest molecules -

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

an increase in pressure forms more products in this reaction from the Contact process.

(3) Change in temperature :

The affect of a temperature change on an equilibrium depends on the enthalpy change associated with that reaction.

An increase in temperature will lead to a reaction using up the excess energy by following an endothermic route and a decrease in temperature leads to an exothermic path being followed.

e.g. for an exothermic reaction an increase in temperature will lead to more reactants being formed and an overall decrease in the value of the equilibrium constant.

and,

for an endothermic reaction an increase in temperature will lead to more products being formed and an overall increase in the equilibrium constant.

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Chemical Equilibria- Kp

The value of the equilibrium is normally expressed in terms of concentrations as explained before. However, for a reaction involving gaseous components an alternative expression for the equilibrium constant is possible. This method expresses the equilibrium in terms of the partial pressures of each gaseous component -

aA + bB cC + dD

where,

Px = the partial pressure of the gas X
= pressure of gas X/total pressure
= the mole fraction of X x the total pressure
= cX x P

N.B.:

(i) the sum of all the mole fractions of the gaseous components is equal to 1.

(ii) the sum of the partial pressures must be equal to the total pressure.

If a reaction involves a mixture of gaseous and liquid/solid components then only the gaseous components are used in the calculation of Kp. All components would be used to determine Kc for a reaction.

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Chemical Equilibria- Kc and Kp Calculations

(1) Calculation of an equilibrium constant :

For the preparation of ethyl ethanoate the equilibrium is -

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

An experiment is conducted and the data obtained for the equilibrium concentrations are as follows -

Component Equilibrium concentration
CH3COOH 0.1 mol dm-3
CH3CH2OH 0.1 mol dm-3
CH3CO2C2H5 0.2 mol dm-3
H2O 0.2 mol dm-3
(2) Calculation of equilibrium concentrations from Kc values :

For a reaction involving the hydrolysis of ethyl ethanoate a mixture was found to contain 0.1 mol dm-3 ethanoic acid by titration with standard sodium hydroxide(aq). What is the concentration of the ethyl ethanoate present in the equilibrium mixture?

Let Kc = 0.25 for the reaction,

CH3COOCH2CH3 + H2O CH3COOH + CH3CH2OH

If 0.1 mol dm-3 ethanoic acid is formed then 0.1 mol dm-3 ethanol must have been formed as well, since the equation shows a 1:1 ratio of the products.

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Chemical Equilibria - Industrial Concerns

Haber process

- the manufacture of ammonia

- conditions

- uses of ammonia and nitrogen containing compounds

Contact process

- the manufacture of sulphuric acid

- conditions

- uses of sulphuric acid

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Chemical Equilibria - Acid / Base Equilibria

(1) Br°nsted-Lowry theory :

There are a number of ways of defining an acid. The simplest if to consider hydrogen ion (proton) transfer between molecules.

An acid is a molecule that donates protons to water and a base accepts protons from water. This can be written as a chemical equilibrium,

Exemplar equations -

CH3COOH + H2O CH3COO- + H3O+
NH3 + H2O NH4+ + -OH

The strength of an acid (or a base) can be inferred by the value of Kc for these equilibria i.e. a high Kc value means more products formed and therefore a stronger acid or base and a low value for Kc means less products formed and therefore a weaker acid or base.

The strength of an acid or base is also quantified by a pH value for the aqueous solution.

(2) Conjugate pairs :

In the previous examples of Br°nsted acid/base behaviour the reactants and products occur in pairs of acids and bases, so-called conjugate pairs -

This pairing behaviour will always be seen in the reaction between an acid and a base.

(3) Water equilibrium constant, Kw :

Water is involved in an equilibrium reaction with itself -

For this equilibrium,

If both sides of the equation are multiplied by [H2O]2eqm then,

Where Kw is the equilibrium constant for water and Kw = 1x10-14 mol2 dm-6 at 25 ░C and 1 atm pressure.

(4) Acid equilibrium constant, Ka :

For the equilibrium,

A-H + H2O H3O+ + A-

where A-H is a protonated acid and A- is its conjugate base,

If both sides of the equation are multiplied by [H2O]eqm then the expression becomes,

where Ka is the acid equilibrium constant and always has units of mol dm-3.

The value of pKa for an acid gives a more useful scale than the simple pH values and achieves the same aim - categorising a molecule by its ability to donate protons.

(5) Acid/base Indicators :

An acid/base indicator can be thought of as being a weak acid, with the conjugate acid (HIn) and base (In-) differing in colour.

At the end-point of the titration the concentrations of the conjugate acid and base are the same i.e. [HIn]eqm = [In-]eqm. By putting this into the previous expression for Ka,

Ka = [H3O+]eqm

=> pH = pKa , called pKIn for indicators.

Therefore, the pH at the colour change point for an indicator is the pKIn value for that indicator.

e.g. for methyl red pKIn = 5.1 and for phenolphthalein pKIn = 9.3.

The range over which the indicator acts is generally about ▒1 unit from the pKIn value i.e. pH 4.1-6.1 for methyl red and pH 8.3-10.3 for phenolphthalein.

(6) pH changes during titration :

The end-point of a titration is accompanied by a rapid increase in the pH. This can be seen for a number of different acid/base reactions in the graph below -

The strong acid/strong base combination has the highest rise in pH value, with the mid-point being at about pH 7.

Any indicator that changes colour during this rapid rise in pH will detect the end-point of the reaction and therefore be able to show the volume of base required to neutralise the acid.

Just about any indicator will suffice for strong acid/strong base reactions, whilst an indicator that changes at a basic pH is needed for the weak acid/strong base combination (e.g. phenolphthalein) and an indicator that changes at an acidic pH is needed for the strong acid/weak base combination (e.g. methyl orange).

For a multibasic acid there is more than one end-point for the titration. This can be seen in the graph below, taken from data produced by one of my sets -

Each rapid rise in pH is a sign that one of the hydrogens is being lost from the acid molecule.

The rise at about pH 4.6 is due to this reaction :

H3PO4 H+ + H2PO4-

and the rise at about pH 9.4 is due to this reaction :

H2PO4- H+ + HPO42-

(7) Buffer solutions :

A buffer solution is a solution that can maintain a constant pH value no matter how much acid or base is added to it.

A buffer solution is prepared by taking a weak acid, e.g. ethanoic acid, and dissolving a salt of its conjugate base in it, e.g. sodium ethanoate.

The pH of the resulting solution can be calculated by using the following equation (which is a rearrangement of the previous Ka expression) -

and using the previous expression for pH, the equation becomes -

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Chemical Equilibria - Acid / Base Equilibria Calculations

(1) Calculating the pH of water :

At the neutral point of water [H3O+]eqm = [-OH]eqm i.e. the concentration of protons is equal to the concentration of hydroxide ions.

Kw = [H3O+]2eqm = 1 x 10-14 mol2 dm-6

= [H3O+]eqm = 1 x 10-7 mol dm-3

= pH = -log10(1 x 10-7) = 7

However, at 50 ░C Kw changes to 5.47 x 10-14 mol2 dm-6.

= Kw = [H3O+]2eqm = 5.47 x 10-14 mol2 dm-6

=> [H3O+]eqm = 2.34 x 10-7 mol dm-3

=> pH = -log10(2.34 x 10-7) =6.63

Therefore it can be seen that water becomes more acidic as its temperature increases.

(2) Calculating Ka from pH values :

Given the pH of a 0.1 mol dm-3 solution of carbonic acid, H2CO3, is 4.17 at 25 ░C, calculate the value of Ka for the acid.

pH = 4.17

=> [H3O+]eqm = 10-4.17 = 6.76 x 10-5 mol dm-3

Substituting these numbers into the expression for Ka gives,

Ka = ( 6.76 x 10-5 mol dm-3 )2 / ( 0.1-6.76 x 10-5 mol dm-3 )

=> Ka = 4.60 x 10-7 mol dm-3

(3) Calculating the pH of a weak acid :

At 25 ░C Ka for ethanoic acid is 1.7 x 10-5 mol dm-3. What is the pH of a 0.1 mol dm-3 solution of this acid?

The equilibrium concentration of the acid can be estimated to be 0.1 mol dm-3 as the concentration value x will be very small compared to 0.1.

Substituting these values into the expression for Ka gives,

1.7x10-5 mol dm-3 = x2 / 0.1 mol dm-3

=> x2 = 1.7x10-6 mol2 dm-6

=> [H3O+]eqm = 1.3x10-3 mol dm-3

=> pH = 2.9

(4) Calculating the pH of a buffer solution :

What is the pH of a buffer solution containing 0.5 mol dm-3 methanoic acid and 2.5 mol dm-3 potassium methanoate? Ka for methanoic acid = 1.6 x< 10-4 mol dm-3.

Using the previous expression relating pH and pKa,

pH = -log10( 1.6 x 10-4 ) + log10( 2.5/0.5 ) = -( -3.8 ) + log10( 5 )

=> pH = 3.8 + 0.7 = 4.5

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Chemical Equilibria - Solubility products

(1) General introduction :

Inorganic salts are usually classified as soluble or insoluble in water. The extent to which a salt is soluble is often expressed as the amount or mass that will dissolve in a unit volume of water, e.g. units of mol dm-3 or g dm-3.

However, even salts classified as insoluble in water do dissolve to a certain extent. These salts exist in a state of equilibrium between the solid salt and its dissolved aqueous ions -

e.g. AaBb(s) aA+(aq) + bB-(aq)

For this equilibrium,

If both sides of the equation are multiplied by [AaBb(s)]eqm then the expression becomes,

Kc x [AaBb(s)]eqm = [A+(aq)]aeqm[B-(aq)]beqm = Ksp

Ksp is the solubility product for the salt and has units of (mol dm-3 )a+b.

(2) Common ion effect :

When an 'insoluble' salt is dissolved in a solvent containing one of its ions,e.g. silver chloride in dilute hydrochloric acid, the solubility of the salt is greatly reduced.

AgCl(s) Ag+(aq) + Cl-(aq)

Using Le Chatelier it can be seen that an increase in the concentration of chloride ions leads to the equilibrium moving to the left hand side. This forms more solid silver chloride and therefore the solubility has decreased.

(3) Calculations :

(i) Ksp from solubilties -

The solubility of silver sulphide, Ag2S, is 2.48 x 10-15 mol dm-3. Calculate Ksp for Ag2S.

Ksp = [Ag+]2eqm[S2-]eqm

=> Ksp = (4.96x10-15 mol dm-3)2 x 2.48x10-15 mol dm-3

=> Ksp = 6.10x10-44 mol3 dm-9

(ii) Solubilities from Ksp values -

Calculate the solubility of calcium sulphate, CaSO4, given that Ksp = 8.64x10-8 mol2 dm-6.

Ksp = x2
=> x2 = 8.64 x 10-8 mol2 dm-6
=> x = 2.94 x 10-4 mol dm-3

Therefore, the solubility of calcium sulphate is 2.94 x 10-4 mol dm-3.

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written by Dr Richard Clarkson : © Saturday, 1 November 1997

updated : Sunday, 30th October, 2005

mail to: chemistryrules@rjclarkson.demon.co.uk

created with the aid of ChemWindow«5.1


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