Homework_8chemical Equilibrium FINAL

7/29/2019 Homework_8chemical Equilibrium FINAL

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CHEMICAL EQUILIBRIUM

Reversible Reactions

Reversible Reactions:

A chemical reaction in which the products can react to re-form the reactants as indicat

below with arrows pointing in both directions.Chemical Equilibrium:

When the rate of the forward reaction equals the rate of the reverse reaction and the

concentration of products and reactants remains unchanged

2HgO(s) 2Hg(l) + O2(g)Arrows going both directions ( ) indicates equilibrium in a chemical equation

Reactions which are not reversible (irreversible) have the usual complete arrow only pointin

the right.

Note:Reversing the reaction conditions reverses the direction of chemical change, typica

a reversible reaction.

Example 1.

Thermal decomposition means using 'heat' to 'break down' a molecule into smaller ones. The

decomposition of NH4Cl is endothermic, H +ve (heat absorbed, taken in from the

surroundings) and the formation of NH4Cl is exothermic, H -ve (heat released, given out

the surroundings).

This means if the direction of chemical change is reversed, the energy change sign mustbe reversed but its numerical value stays the same.

Example 2

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

anhydrous copper(II) sulphate is formed. When the white solid is cooled and water added, bl

hydrated copper(II) sulphate is reformed.

CuSO4.5H2O(s) --------- CuSO4(s) + 5H2O(g)

Note: The crystal structure is broken down on heating and the water of crystallisation is giv

off. Thermal decomposition is endothermic (H +ve) as heat is absorbed to drive off the waThe reverse reaction is exothermic (H +ve) i.e. on adding water to cold white anhydrous

copper(II) sulphate the mixture heats up as the blue crystals reform.

These are typical examples you encounter at an earlier study level, but it begs the question, possible to have a situation, under suitable conditions, in which the reaction does not

completely go in one direction or the other and both reactants and products co-exist?",


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the answer is yes! and the situation is called a dynamic chemical equilibrium. The word dynam

used because the 'forward' (L to R) and 'backward' (R to L) reactions do not cease but matc

each other in rate so the concentrations of reactants and products are constant when the

equilibrium is established.

Reversible reactions and the concept of a dynamic chemical equilibrium

Although most reactions you have encountered at an earlier academic level did go to 100%

completion, it is a fact that many reactions do NOT go to completion i.e. 100% yield from

forward reaction.

If ammonium chloride were heated in a closed system, over a certain temperature range, som

the NH4Cl will be sublimed into the gases NH3 and HCl and some of the solid salt remains.

A closed system means nothing can enter or leave the system.When a reversible reaction occurs in a closed system, depending on conditions, a chemica

equilibrium is formed, in which the original reactants and products formed coexist. In othe

words the reaction (i.e. from left to right as the equation is written) never goes to complet Eventually the 'system settles down' and the net concentrations of the reactants and

products remain constant i.e. a state of concentration balance exists. BUT the reactions don't stop! Reactants are continually forming products, and the products

continually re-forming the original reactants, hence the term dynamic equilibrium.

In terms of kinetics ('rates of reaction'), it means that the

rate of formation of product = rate of re-formation of reactants,

or

the rate of the forward reaction = rate of the backward reaction

Example 2 The formation/decomposition of hydrogen iodide.

H2(g) + I2(g) ---------- 2HI(g)(all gases above 200oC)

L to R forward reaction: If you start with pure hydrogen and pure iodine, so much of themcombine to form hydrogen iodide.

R to L backward reaction: If you start with pure hydrogen iodide, some, but not all of it, w

decompose into hydrogen and iodine.

Starting with the same total number of moles of either H2 + I2 or HI, the final equilibriumconcentrations will be the same at the same temperature, volume and pressure. This is furth

illustrated in the diagram below.

Put in figure 15.4 on page 262 of Ababio


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Note:

The final equilibrium composition is the same in each case no matter which direction you sta

from for the same total moles of of substance.

Systems A and B show that the same equilibrium position can be reached irrespective of

whether we dissolve the iodine in trichloromethane or in aqueous potassium iodide.

Some important outcomes from experimentally studying dynamic equilibrium reactions

It doesn't matter whether you start with the 'reactants' or the 'products', either way, if

conditions are suitable; both are present when a state of equilibrium exists.

Eventually the net concentrations of the reactants and products remain the same BUT

forward and backward reactions don't stop.

When a dynamic equilibrium is achieved there is a state of balance between the constant

concentrations of the reactants and products because the rate at which the reactants cha

into products is exactly equal to the rate at which the products change back to the origreactants.

However, the actual relative amounts of the original reactants left, and products form

at equilibrium, depend on the particular reaction and reaction conditions e.g. the initial

concentrations, temperature and pressure (if gaseous reactants or products are involved) an

value of the equilibrium constant

A catalyst does not affect the position of the equilibrium, i.e. the final constant

concentrations are the same with or without a catalyst, you simply get to the equilibrium poin

faster with a catalyst!

In some cases you can adjust reaction conditions sufficiently to make the reaction go

virtually 100% in one direction.

At a given constant temperature, all the final equilibrium concentrations are

mathematically governed by the equilibrium expression and the equilibrium constant.

Equilibrium constant

The effect of concentration on reaction rate was first stated by Gulberg and Waage. T

law is known as the law of mass action. It states that at constant temperature, the rate o

reaction is proportional to the active masses of each of the reactants.

The active mass is given as the concentration of the substance raised to an appropriate powe

moles per dm3 and if the substance is a gas, its partial pressure may be used instead of

concentration.

For a reaction in which nmoles of a substance A reacts with mmoles of substance B. Let the

rate of the reaction be r. Then


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mA + nB ---r--- Products

r & (concentration of A)m and r & (concentration of B)n

r & (concentration of A)m x (concentration of B)n

i.e r & [A]m x [B]n

Introducing a constant,

r = k.[A]m. [B]n Where r is a constant called the Velocity Constant.

For a particular equation,

mA + nB pC + qD

The rate of the left-to-right reaction, r1 = k1 . [A]m. [B]n

The rate of the right-to-left reaction, r2 = k2 . [C]p. [D]q

Where k1and k2are the velocity constants of the forward and backward reactions respectiv

At equilibrium, r1= r2

then k1.[A]m. [B]n = k2. [C]

p . [D]q

therefore

k1 = [C]p . [D]q = K

2 k2 [A]m . [B]n

Where K is the equilibrium constant of the reaction at that temperature.

It is important to note the following:

* Kis the ratio of the two velocity constants k1and k2

* Since the velocity constant usually vary with temperature, then it follows that K will vary w

temperature.

* The addition of a catalyst will however not cause any change in the value of K

* Varying the concentration of any of the substances A,B,C or D will also not affect the valu


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K. (this is because the reaction will automatically adjust itself to attain equilibrium again to g

the same value of K) To do this, the position of the equilibrium shifts to the right when the

concentration of A and B are increased favouring the formation of C and D thereby increasin

their concentrations.

Le Chatelier's Principle

As we can see, every reaction reaches its own specific equilibrium under a given set

conditions. This equilibrium state is dependent on:

* temperature of the reacting system

* pressure of the reacting system(for gases)

* concentration of the reacting system.

A change in any one of these factors will upset the balance of the system and result

shift in the position of the equilibrium. These factors and their effects on chemical sys

in equilibrium were studied by Le Chatelier (1850-1936) who formulated the Le Chateli

principle.

Le Chateliers's Principle states that if an 'instantaneous' change is imposed on an

equilibrium, the position of the equilibrium will further change to minimize the 'enforced

change.

Le Chatelier Translated:When you take something away from a system at equilibrium, the system shifts in such a way

to replace what youve taken away.

When you add something to a system at equilibrium, the system shifts in such a way as to use

what youve added.

In other words, if a change is 'instantaneously' imposed, the equilibrium attempts to restore

original situation, but it cannot do this completely BUT the change 'trend' can be predicted.

when considering the equilibrium rules outlined below, any change affects BOTH the rates of

forward and backward reactions.

* This principle is of great importance in the chemical industry because it can help to:- define the optimum conditions for the chemical processes employed in industry;

- reduce undesirable reversibility;

- predict the effect of an altered factor of the equilibrium position of an untried reaction

Rule 1 - Temperature and energy changes (H)

1a. Raising the temperature favours the endothermic direction (H +ve). The system abs


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the heat energy from the surroundings to try to minimize the temperature increase.

1b.Decreasing the temperature favours the exothermic direction (H -ve). The system

releases heat energy to the surroundings to try to minimize the temperature decrease.

Rule 2 - Gas pressure changes at constant temperature (V)

2a. Increasing the pressure favours the side of the equilibrium with the least number ofgaseous molecules as indicated by the balanced symbol equation. The system attempts to red

the number of gas molecules present to reduce the pressure increase.2b.Decreasing the pressure favours the side of the equilibrium with the most number o

gaseous molecules as indicated by the balanced symbol equation. The system attempts to

increase the number of gas molecules to minimize the pressure decrease.

NOTE:

States symbols (g/l/s/aq) are particularly important when considering equilibrium equat

if no (g) the pressure rule doesn't apply since solids and liquids are virtually in-compressible.

Rule 2 ONLY applies to a reaction with one or more gaseous reactants or products becpressure has no real effect on the 'concentration' on the virtually incompressible liquids or

solids.

If there is NO net change in the number of gas molecules, gas pressure has NO effec

the position of the equilibrium, though pressure increase effectively increases gas

concentration so both the forward and backward reactions will be speed-ed up.

Over and above rule 2, all the individual partial pressures of the gases, must comply with t

mathematics of the Kp equilibrium expression.

The rest of the rules 1, 3 and 4 apply to ANY reaction, whatever the physical states o

reactants and products.

Rule 3 - Concentration changes at constant temperature

3a. If the concentration of a reactant (on the left) is increased, then some of it must chang

the products (on the right) to maintain a balanced equilibrium position.3b. If the concentration of a reactant (on the left) is decreased, then some of the product

the right) must change back to reactants to maintain a balanced equilibrium position.

This means if you change ANY concentration, all the other concentrations must change t

Also, any net concentration changes must comply with the Kc equilibrium expression

Rule 4 - Using a catalyst

A catalyst does NOT affect the position of an equilibrium.

What it does is to enable the reaction get to the point of equilibrium faster!

A catalyst speeds up both the forward and reverse reactions by providing a mechanistic

pathway with alower activation energy, but there is no way it can influence the final 'balanc

concentration ratios. The importance of a catalyst lies with economics of chemical production e.g. bringing ab


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reactions with high activation energies at lower temperatures and so reducing energy

requirements and time, and both reductions save money!

Applying Le Chatelier's Principle and the equilibrium rules

Example 1. The thermal decomposition of calcium carbonate (limestone) to make calcium oxid

(quicklime):

CaCO3(s) CaO(s) + CO2(g) (H = +178 kJ mol-1

)

Note: By convention, the H value quoted corresponds to the forward reaction (L to R),

reversing the sign gives the H for the backward reaction (R to L).

Rule 1- temperature and energy change (H)

Increasing temperature favours the endothermic direction (RHS) so more quicklime is forme

Rule 2 - gas pressure (V)

Decreasing the partial pressure of carbon dioxide increases the yield of quicklime (0 ==> 1 mo

gas).

Rule 3 - concentration

Not applicable to the reactants because you can't decrease or increase the concentration of

solid but you can reduce the concentration of carbon dioxide by venting the gases to increas

yield of quicklime, a common strategy used in industries.

Rule 4 - catalyst: Not applicable.

Example 2. The synthesis of ammonia: nitrogen + hydrogen ammonia

N2(g) + 3H2(g) 2NH3(g) (H = -92 kJ mol-1

)

1mole 3 moles 2 moles

Rule 1- temperature and energy change (H)

The forward, and desirable reaction, to form ammonia, is exothermic, so lowering the

temperature favours its formation.


Increase in pressure favours ammonia formation since 4 mol of gaseous reactants ==> 2 mol


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gaseous products.

Rule 3 - concentration: In terms of enforced change => system response

If the nitrogen or hydrogen concentration was increased, some of this 'extra' gases would

change to ammonia. Likewise, if the nitrogen or hydrogen concentration was decreased, some

ammonia would change to nitrogen and hydrogen.

Rule 4 - catalyst: An iron oxide catalyst is used, time = money for industrial chemicalproduction!

3. The formation of hydrogen iodide from hydrogen and iodine:

H2(g) + I2(g)----- 2HI(g) (H = -10 kJ mol-1

, iodine gaseous above 200oC)

Rule 1 - temperature and energy change (H)

Increasing temperature favours the LHS, i.e. increases the endothermic decomposition of

hydrogen iodide.

Rule 2 - gas pressure (V):

No effect on position of equilibrium, 2 mol gas ==> 2 mol gas, no net change in gas moles.


e.g. if more iodine was added to a constant volume container, the hydrogen concentration or

partial pressure would decrease as some reacts with added iodine to give more hydrogen iodithe system tries to minimise the iodine increase. Please note that there would still be an ove

increase in iodine at the new equilibrium point.

Rule 4 - catalyst: Not applicable.

4. The formation of nitrogen (II) oxide.

N2(g) + O2(g) -------- 2NO(g) (H = +181 kJ mol-1

)


Increase in temperature favours the endothermic formation of NO.

This reaction does not happen at room temperature but is formed at the high temperatures

engines. Unfortunately when released through the car exhaust, it cools to normal temperatur

when NO irreversibly reacts with oxygen in air to form nitrogen(IV) oxide, NO2, which is aci

lung irritant and a reactive free radical molecule involved in the chemistry of photochemical

not good!


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Since 2 mol gaseous reactants gives 2 mol gaseous products, pressure does not affect the

position of the equilibrium.


The concentration of nitrogen is high from air, but although the concentration of oxygen is lothe exhaust gases, there is sufficient present in the combustion process to ensure a small %

NO is formed.Rule 4 - catalyst: Not applicable.

5. One way to producehydrogen for the Haber synthesis of ammoniais to react methane ga

with steam.

CH4(g) + H2O(g) --------- 3H2(g) + CO(g) (H = +206 kJ mol-1

)Rule 1 - temperature and energy change (H)

Increase in temperature favours the endothermic formation of hydrogen (and carbon monoxRule 2 - gas pressure (V)

For the desired forward reaction, 2 mol of reactant gases ==> 4 mol of product gases, so the

increase in product volume is favoured by lower pressure.


Theoretically increase in methane and steam concentrations will increase the hydrogen

concentration, but this essentially means increasing pressure favouring the LHS, so you migh

gain as much hydrogen as you like to!Rule 4 - catalyst

A nickel catalyst is used, but cannot affect the yield.

e.g. nitrogen + hydrogen ammonia

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

If the nitrogen or hydrogen concentration was increased, some of this extra gas would chan

ammonia.

If the nitrogen or hydrogen concentration was decreased, some of ammonia would change ba

to nitrogen and hydrogen.

So in terms of enforced change ==> system response:Increasing nitrogen ==> decreases hydrogen and increases ammonia.

Increasing hydrogen ==> decreases nitrogen and increases ammonia.

Increasing ammonia ==> increases nitrogen and hydrogen.

Decreasing ammonia ==> decreases nitrogen and hydrogen.

Decreasing nitrogen ==> increases hydrogen and decreases ammonia.

Decreasing hydrogen ==> increases nitrogen and decreases ammonia.
http://www.docbrown.info/page07/equilibria3.htm#Synthesishttp://www.docbrown.info/page07/equilibria3.htm#Synthesishttp://www.docbrown.info/page07/equilibria3.htm#Synthesishttp://www.docbrown.info/page07/equilibria3.htm#Synthesis


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Rule 4: A catalyst does NOT affect the position of equilibrium, you just get there faster

catalyst usually speeds up both the forward and reverse reaction but there is no way it can

influence the final 'balanced' concentrations. However, the importance of a catalyst lies with

economics e.g. (i) bringing about reactions with high activation energies at lower temperature

and so saving energy or (ii) saving time is saving money.

LeChatelier Example #1

A closed container of ice and water at equilibrium. The temperature is rais

Ice + Energy Water

The equilibrium of the system shifts to the _______ to use up the added

energy.

right


A closed container of N2O4 and NO2 at equilibrium. NO2 is added to the

container.

N2O4 (g) + Energy 2 NO2(g)

The equilibrium of the system shifts to the _______ to use up the addedNO2.

left


A closed container of water and its vapor at equilibrium. Vapor is removed

from the system.

water + Energy vapor

The equilibrium of the system shifts to the _______ to replace the vapor.

right


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A closed container of N2O4 and NO2 at equilibrium. The pressure is increas

N2O4 (g) + Energy 2 NO2(g)The equilibrium of the system shifts to the _______ to lower the pressure

because there are fewer moles of gas on that side of the equation.

leftActivity for practice ( solve the following on a separate piece of paper )

Show the effect of the four(4) rules on the following reactions

1. N2O4(g) 2NO2(g) 2. 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H

3. The effect of change in concentration on Iron(III) thiocyanate system and potassi

heptaoxochromate (VI) system

4. Include the summary tables of Table 15.1; 15.2 and 15.3

EQUILIBRIUM IN PRACTICE

Chemical equilibrium and Le Chatelier's Principle, are applied to decide on the optimum

conditions of operation. These conditions, temperature, Pressure, Concentration and

Catalysts, are chosen by industrial chemists with the aim of:

* Minimizing cost of production by ensuring

[a] the capital cost of the plant is not too high.

[b] the starting materials are cheap.


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* maximizing yield of product by:

[a] shifting the equilibrium position in the desired direction.

[b] increasing the value of the equilibrium constant K, for the process concerned.

The Haber Process and the Contact Process are two important industrial processes that uses

Chaterlier's Principle among others to determine their optimum operating condition.

3. The Synthesis of ammonia - The Haber Process

Ammonia gas is synthesized in the chemical industry by reacting nitrogen gas with hydrog

gas in what is known as the Haber-Bosch Process, named after two highly inventive and

subsequently famous chemists.

The nitrogen is obtained from liquified air (80% N2). Air is cooled and compressed under h

pressure to form liquid air (liquefaction). The liquid air is fractionally distilled at low temperto separate oxygen (used in welding, hospitals etc.), nitrogen (for making ammonia), Noble Ga

e.g. argon for light bulbs, helium for balloons).

The hydrogen is made by reacting methane (natural gas) and water or from cracking

hydrocarbons (both reactions are done at high temperature with a catalyst).

CH4 + H2O ==> 3H2 + CO

The synthesis equation for this reversible reaction is ...

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

.. which means an equilibrium will form, so there is no chance of 100% yield even if we use,

you actually do, the theoretical reactant ratio of nitrogen : hydrogen of 1 : 3 !

In forming ammonia 92kJ of heat energy is given out (i.e. exothermic, 46kJ of heat release

mole of ammonia formed).

N2 + 3H2 2NH3 H = -92JK/mol

Four moles of 'reactant' gas form two moles of 'product' gas, so there is net decrease in ga

molecules on forming ammonia.

N2 + 3H2 2NH3

1 mole 3 moles 2 moles

So applying the equilibrium rules from section 2 above, the formation of ammonia is favour

by

(a) Using high pressure because you are going from 4 to 2 gas molecules (the high pressure

speeds up the reaction because it effectively increases the concentration of the gas moleculbut higher pressure means more dangerous and more costly engineering.

(b) Carrying out the reaction at a low temperature, because it is an exothermic reaction is

favoured by low temperature, but this may produce too slow a rate of reaction,


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So, the idea is to use a set of optimum conditions to get the most efficient yield of ammo

and this involves getting a low % yield (e.g. 8% conversion) but fast. Described below are th

conditions to give the most economic production of ammonia.

these arguments make the point that the yield* of an equilibrium reaction depends on the

conditions used.

* The word 'yield' means how much product you get compared to the theoretical maximpossible if the reaction goes 100%.

In industry pressures of 200 - 300 times normal atmospheric pressure are used in line w

the theory.

Theoretically a low temperature would give a high yield of ammonia BUT...

Nitrogen is very stable molecule and not very reactive i.e. chemically inert, so the rate of

reaction is too slow at low temperatures.

To speed up the reaction an iron catalyst is used as well as a higher temperature (e.g. 40

450oC).

The higher temperature is an economic compromise, i.e. it is more economical to get a low yifast, than a high yield slowly!

Note: a catalyst does NOT affect the yield of a reaction, i.e. the equilibrium position BU

you do get there faster!At the end of the process, when the gases emerge from the iron catalyst reaction

chamber, thegas mixture is cooled under high pressure, when only the ammonia liquefie

is so can be removed and stored in cylinders.

Any unreacted nitrogen or hydrogen (NOT liquified), is recycled back through the reac

chamber, nothing is wasted! [nitrogen (-196

o

C) and hydrogen (-252

o

C) have much lower boilpoints than ammonia (-33

oC). Boiling points increase with pressure, but these normal atmosph

pressure values offer a fair comparison].To sum up: A low % yield of ammonia is produced quickly at moderate temperatures and

pressure, and is more economic than getting a higher % equilibrium yield of ammonia at

more costly high pressure and a slower lower temperature reaction.

Include Fig 15.8 on page 270 and table 15.4. in Ababio.


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CONTACT PROCESS ( industrial production of sulphuric acid )

The oxidation of sulphur dioxide to sulphur trioxide

e.g. in the Contact Process for manufacturing sulphuric acid.

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

)


The exothermic formation of sulphur trioxide is favoured by low temperature.


Higher pressure favours a higher yield of sulphur trioxide as 3 gas moles ==> 2 gas moles, tho

1-2 only atm is used in practice because the equilibrium is already so far to the right (about 9


Air is used as the source of oxygen and despite its dilution with nitrogen the concentration, t

oxygen concentration is high enough to move the equilibrium very much to the RHS.

Rule 4 - catalystA vanadium(V) oxide, V2O5, catalyst ensures the high yield of 99% SO3 is attained fas

but no more!

Include table 15.5 page 271 in Ababio.

EQUILIBRIUM CONSTANT FREE ENERGY AND ELECTRODE POTENTIAL

At equilibrium, the free energy change of a closed system at constant temperature and press

is at its minimum {meaning that the system is at its utmost stability} At this stage, there is nnet flow of heat or species occurs and no work can be done by the system as such we can str

simple relationship between Gof a reaction and the equilibrium constant, K.

Go = -RT InK

G--- Standard free energy change of the system; K ---- is the equilibrium constant;

R ----- is the gas constant; T ---- is the temperature of the system in kelvin


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Since redox reactions occur in an electrochemical cell, at equilibrium, a reaction such as thi

given as;

Zn(s) + Pb2+(aq) Zn2+(aq) + Pb(s)

a relationship between Eoand K is given as,

Eo = RT InKnF

Eo is standard electrode potential of the reaction

Ris the gas constant; Tis the temperature in kelvin; Fis the Faraday's constant

Kis the equilibrium constant; nis the number of moles of electrons

The value ofKusually used to effect electroplating of metals. For the reaction above, a higvalue of K would favour the forward reaction resulting in a thicker coating of lead on the ob

in question.

Go= -nFEo

This is the relationship between electrode potential and free energy change of anelectrochemical cell. A negative value of delta G shows that work is obtained from the

electrochemical cell.

ACIDBASE EQUILIBRIUM

This is a study of the chemical equilibrium that exists between acids and their associated ba

Two theories have been put forward on acid-base action. The earliest being that proposed by

Arrhenius in the nineteenth century and the other by Bronsted-Lowry.

According toArrhenius, an acid is defined as a substance which dissociates in water to prod

hydrogen ions, H+ e.g HCl H++ Cl-

In the same vein, a base is defined as a substance which gives hydroxide ions in solution

NaOH(s) Na+(aq) + OH

-

Chemist discovered that the hydrogen ion cannot exist on its own in solution and that each proto

associates with a water molecule to form hydroxonium or Oxonium ion.Hence the dissociation of an

in water is represented more accurately as;

HCl(aq) + H2O(l) H3O+(aq) + Cl

-(aq)


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With these arguments, the Arrhenius theory recognized

1. Therole of the Solvent water molecules in the dissociation of acids. 2. Why substance exhibit a

properties in the presence of water. 3. that acidity is not an inherent property of the substance but

rather the mode of behaviour of a substance under certain conditions.

Bronsted-Lowry Theory

According to the theory of Bronsted-Lowry, an acid is a substance which donates a proton, while a ba

a molecule or ion which accepts the proton. This describes an acid-base reaction in a non-aqueous sol

CH3COOH(aq) + OH-(aq) CH3COO

-(aq) + H2O(l)

acid base base acid

In the forward reaction, an ethanoic acid molecule donates a proton to a hydroxide ion which

accepts it. The ethanoic acid behaves as an acid while the hydroxide ion as a base. In the revreaction, the ethanoate ion accepts a proton from a water molecule acting as a base, and the

water acting as an acid.

The ethanoic acid and the ethanoate ion are aCONJUGATEacid-base pair similarly the wa

molecule and the hydroxide ion are also aCONJUGATE acid-base pair in the reverse reac

Thus a Conjugate acid-base pair is an acid-base combination in which one is related to

gain or loss of protons.

Other examples are

HCl(g) + NH3(g) NH4+ Cl-(s)

acid base acid base

By the arrows, the equilibrium shifts to the right to favour the forward reaction because hydrogen chloride is a stronger acid t

ammonium ion.

HCl(aq) + H2O(l) Cl-(aq) + H3O

+(aq

acid base base acid

By the arrows it shows that equilibrium shifts to the right to favour the forward reaction because

hydrogen chloride is a stronger acid than hydroxonium ion.

H2O(l) + H2O(l) H3O+(aq) + OH

-(aq)

acid base acid base


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The equilibrium is balanced because the conjugate acid and base are equal.

NOTE: The lengths of the arrows indicate the favour direction of the reaction.

HYDROGEN ION CONCENTRATION

Dissociation of Water

At 250 C, pure water ionizes as follows;

[H+] = [OH-] = (10-7 x 10-7) moldm -3

Here the concentration of both hydrogen and hydroxide ions are both equal to 10-7 mold

The ionic product of water is represented as

Kw = [H+][OH-]

= (10-7 x 10-7) (moldm-3)2

= 10-14 moldm-3 at 250C

Kw is kept constant under all circumstances at 250C. Thus in all neutral, Acidic (i.e acid

added to the solution), and alkaline solutions (i.e alkaline added to the solution), the

concentrations of both hydrogen and hydroxide ions are jas reflected in the table belo

Table 15.6 Hydrogen ion concentration of solutions [Ababio page 273]

The increase and decrease in hydrogen and hydroxide concentrations conversely is in the

attempt to maintain Kw at 10-14 mol dm-3.


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pH Scale

Sorensen devised the logarithmic pH scale in 1909 in order to:

Avoid the clumsy ness in the handling of negative indices.

Accommodate the wide range of H+ and OH- concentrations commonly encountered in acid-b

reactions. He defined the pH of a solution as, as the negative logarithm of the hydrogen ion

concentration to the base 10.

Example; If the hydrogen ion concentration of a given aqueous medium is 10-5 moldm-3, then t

acidity of the solution can be written as;

[H+] = 10-5

Log[H+

] = log 10-5

= -5

pH = -log [H+]

= -(-5)

= 5

Thus if [H+

] = 10-x

, then pH = x

Since concentration of [H+] is inversely proportional to [OH-], then the value of pH would ind

both acidity and alkalinity of a solution.

[H+][OH-] = 10-14

Therefore, pH + pOH = 14 where pOH is the hydroxide ion concentration

Then pOH = 14 pH

Since [H+] = [OH-] = 10-7, then a high pH value would indicate a low hydrogen ion

concentration i.e weak acidity and a high hydroxide ion concentration i.e strong alkalinit

Likewise, a low pH value indicates a high hydrogen ion concentration i.e strong acidity a

low hydroxide ion concentration i.e weak alkalinity.


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Include examples 1 to 3 from pages 273 to 275

STRENGTHS OF ACIDS AND BASES

Take for instance,

HA(aq) + H2O H3O+(aq) + A

-(aq)

The position of equilibrium indicates the strength of an acid.

In a strong acid the reaction goes virtually to completion as below

HCl(aq) + H2O(l) H3O+(aq) + Cl

-(aq)

In a weak acid, most of the acid molecules are dissociated.

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H2O(aq)

Note that the reverse arrow is supposed to be longer than the forward arrow

Read and follow the line of class discussion to make notes on:

hydrolysis of salts * Indicators * Buffer solutions


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Homework_8chemical Equilibrium FINAL

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