condensed matter 2 physical chemistry and applications of liquids and solutions

Condensed Matter 2Condensed Matter 2

Physical Chemistry and Physical Chemistry and Applications of Applications of

Liquids and SolutionsLiquids and Solutions

Course AimsCourse Aims

The course aims to apply the principles of The course aims to apply the principles of thermodynamics and kinetics to the study of the thermodynamics and kinetics to the study of the physical properties of gases, liquids and solids physical properties of gases, liquids and solids on both microscopic and macroscopic scales. on both microscopic and macroscopic scales.

There will be particular emphasis on the There will be particular emphasis on the properties of liquids and solutions. properties of liquids and solutions.

This course provides a bridge between the This course provides a bridge between the elementary study of condensed matter phases elementary study of condensed matter phases and the advanced study of solid-state physics. and the advanced study of solid-state physics.

Learning ObjectivesLearning Objectives

After completion of this course the student After completion of this course the student is expected to have a firm understanding is expected to have a firm understanding the relationship between pressure, the relationship between pressure, temperature, and volume on the behaviour temperature, and volume on the behaviour of solids liquids and gasses. of solids liquids and gasses.

They should be able to apply these They should be able to apply these relationships to understand phenomena relationships to understand phenomena such as vapour pressure, surface tension, such as vapour pressure, surface tension, liquid crystals, etc. liquid crystals, etc.

The student should understand the properties of The student should understand the properties of liquids and solutions (both ideal and real) on liquids and solutions (both ideal and real) on microscopic and macroscopic scales, including microscopic and macroscopic scales, including electrolyte solutions, solvation, and equilibrium electrolyte solutions, solvation, and equilibrium in solution, and appreciate the application of in solution, and appreciate the application of these principles to devices such as batteries and these principles to devices such as batteries and fuel cells. fuel cells.

The student should have some understanding of The student should have some understanding of intramolecular (chemical) and intermolecular intramolecular (chemical) and intermolecular bonding; including the bonding and structure of bonding; including the bonding and structure of crystals.crystals.

The student should be able to quantitatively The student should be able to quantitatively solve problems involving all of the above.solve problems involving all of the above.

Course ContentCourse Content The Relative Stability of Gases Liquids and The Relative Stability of Gases Liquids and

SolidsSolids (4 lectures): (4 lectures):

Including vapour pressure; surface tension; Including vapour pressure; surface tension; supercritical fluids; liquid crystals.supercritical fluids; liquid crystals.

Ideal and real solutionsIdeal and real solutions (4 lectures): (4 lectures):

Binary solution model; Gibbs-Duhem equation; Binary solution model; Gibbs-Duhem equation; suppression and elevation of freezing and suppression and elevation of freezing and boiling temperatures; osmotic pressure; dilute boiling temperatures; osmotic pressure; dilute solutions; equilibrium in solution.solutions; equilibrium in solution.

Electrolyte SolutionsElectrolyte Solutions (4 lectures): (4 lectures): HH, , SS, and , and GG of ion formation; solvation; of ion formation; solvation;

calculating activity coefficients by Debye-Huckel calculating activity coefficients by Debye-Huckel theory; equilibrium of electrolyte solutions.theory; equilibrium of electrolyte solutions.

Electrochemical cellsElectrochemical cells (4 lectures): (4 lectures): Electrode potentials and cell potential; cell EMF Electrode potentials and cell potential; cell EMF

and the equilibrium constant; electrochemical and the equilibrium constant; electrochemical cells; batteries; fuel cells, atomic-scale cells; batteries; fuel cells, atomic-scale electrochemistry and nano-machining; half-cell electrochemistry and nano-machining; half-cell potentials. potentials.

Introductory Solid State PhysicsIntroductory Solid State Physics (4 lectures): (4 lectures): Including intermolecular and intramolecular Including intermolecular and intramolecular

bonding; and heat capacities of solids.bonding; and heat capacities of solids.

Course StructureCourse Structure

Lectures: Fridays 0900 and 1100Lectures: Fridays 0900 and 1100

The course will consist of approximately The course will consist of approximately 20 lectures20 lectures. .

All lectures are compulsory and an All lectures are compulsory and an attendance register will be taken. attendance register will be taken.

Failure to attend lectures at least 75 % of Failure to attend lectures at least 75 % of lectures may result in deregistration from lectures may result in deregistration from the course.the course.

88 Weekly Assignments (best 6 count)Weekly Assignments (best 6 count)(Weeks, 4-12, save Reading Week)(Weeks, 4-12, save Reading Week). .

These sets are compulsory. Failure to These sets are compulsory. Failure to submit at least 75 % of the assignments submit at least 75 % of the assignments may result in deregistration.may result in deregistration.

Deadline:1700 on the Wednesday Deadline:1700 on the Wednesday preceding the problem class.preceding the problem class.

The assignments will be available on the The assignments will be available on the Module’s homepage (Module’s homepage (www.ph.qmul.ac.uk/~phy226www.ph.qmul.ac.uk/~phy226))on the Monday of the week preceding the on the Monday of the week preceding the deadline.deadline.

Problem Classes: Thursdays 1000Problem Classes: Thursdays 1000

8 Weekly Problem Classes. (Weeks, 4-12, 8 Weekly Problem Classes. (Weeks, 4-12, save Reading Week)save Reading Week)

All problem classes are compulsory and All problem classes are compulsory and an attendance register will be taken. an attendance register will be taken.

Failure to attend lectures at least 75 % of Failure to attend lectures at least 75 % of lectures may result in deregistration from lectures may result in deregistration from the course.the course.

Course AssessmentCourse Assessment

One Examination paper: 2 hours 15 One Examination paper: 2 hours 15 minutes (80%).minutes (80%).

Eight assessed problem sets, of which the Eight assessed problem sets, of which the best six sets contribute to the mark (12%).best six sets contribute to the mark (12%).

Two in-class tests (8 % in total) to be run Two in-class tests (8 % in total) to be run during the last lecture before reading week during the last lecture before reading week and the last lecture of term.and the last lecture of term.

Suggested ReadingSuggested Reading

Physical Chemistry, P. W. Atkins.Physical Chemistry, P. W. Atkins. Properties of Liquids and Solutions, J.N. Properties of Liquids and Solutions, J.N.

Murrell.Murrell. Introduction to Solid-State Physics, C. Introduction to Solid-State Physics, C.

Kittel. Kittel.

EquilibriumEquilibriumThings in BalanceThings in Balance

Mechanical EquilibriumMechanical Equilibrium

Forces in balance.Forces in balance. e.ge.g., two equal masses, a rope and a ., two equal masses, a rope and a

pulley.pulley. From Newton II, the net force is zero.From Newton II, the net force is zero. Note the system is reversible.Note the system is reversible.

Thermal EquilibriumThermal Equilibrium

Two objects in thermal contact.Two objects in thermal contact. Heat flows from the warmer to the cooler. Heat flows from the warmer to the cooler.

Until they reach the same temperature.Until they reach the same temperature. (metal objects at room temperature initially (metal objects at room temperature initially

feel cold – sensation diminished and feel cold – sensation diminished and vanishes at equilibrium)vanishes at equilibrium)

Chemical EquilibriumChemical Equilibrium

In a closed system, reactants turn into In a closed system, reactants turn into products.products.

Eventually this process stops Eventually this process stops A chemical reaction is in equilibrium when A chemical reaction is in equilibrium when

there is no tendency for the quantities of there is no tendency for the quantities of reactants and products to changereactants and products to change. .


HH22 +I +I22 →→ 2HI (synthesis)2HI (synthesis)

2HI 2HI → H→ H22 + I + I22 (dissociation) (dissociation)

The same chemical reaction system, but The same chemical reaction system, but the roles are reversed.the roles are reversed.

Both yield the same mixture when the Both yield the same mixture when the change is completed (equilibrium mixture). change is completed (equilibrium mixture).


composition of a chemical reaction composition of a chemical reaction system will tend to change in a system will tend to change in a direction that brings it closer to its direction that brings it closer to its equilibrium composition.equilibrium composition.


The end is the same no mater where you The end is the same no mater where you start. start.

Complete ReactionsComplete Reactions

A + B A + B → C + D→ C + D A reacts with B to form C and D.A reacts with B to form C and D. Implies that at the end of the reaction Implies that at the end of the reaction

there is only C and D.there is only C and D. Implies that C + D Implies that C + D → A + B does not → A + B does not

occur.occur. The reaction is complete (the reactants The reaction is complete (the reactants

turn completely into products).turn completely into products).

Reversible ReactionsReversible Reactions

If at equilibrium there are significant quantities of If at equilibrium there are significant quantities of all reactants, the reaction is not complete, and all reactants, the reaction is not complete, and the reaction is reversible.the reaction is reversible.

A + B = C + DA + B = C + D In principle, all chemical reactions are reversible, In principle, all chemical reactions are reversible,

but this reversibility may not be observable if the but this reversibility may not be observable if the fraction of products in the equilibrium mixture is fraction of products in the equilibrium mixture is very small, or if the reverse reaction is very small, or if the reverse reaction is kinetically kinetically inhibitedinhibited (very slow.) (very slow.)

Law of ‘Mass Action’Law of ‘Mass Action’

The law of mass action states that any chemical The law of mass action states that any chemical change is a competition between a forward and a change is a competition between a forward and a reverse reaction. reverse reaction.

The rate of each of these processes is governed The rate of each of these processes is governed by the concentrations of the substances reacting; by the concentrations of the substances reacting; as the reaction proceeds.as the reaction proceeds.

These rates approach each other, and at These rates approach each other, and at equilibrium they become identical.equilibrium they become identical.

Law of ‘Mass Action’Law of ‘Mass Action’

Rate of forward reaction is:Rate of forward reaction is:

Rate of reverse reaction is:Rate of reverse reaction is:

Thus, at equilibrium,Thus, at equilibrium,

baf BAk

dcr DCk

dcr

baf DCkBAk

How do we know a reaction is at How do we know a reaction is at equilibrium?equilibrium?

Consider.Consider. 2H2H22 + O + O22 = 2H = 2H22O.O.

We add the required amounts of HWe add the required amounts of H22 and O and O22

to a reaction vessel and leave them for a to a reaction vessel and leave them for a month.month.

Is the mixture in equilibrium?Is the mixture in equilibrium?


[set of a spark in the vessel to find out].[set of a spark in the vessel to find out]. The reaction now rapidly proceeds to The reaction now rapidly proceeds to

equilibrium [provided the reaction vessel equilibrium [provided the reaction vessel survives].survives].

This mixture was not at equilibrium.This mixture was not at equilibrium. Although the reaction is Although the reaction is

thermodynamically favoured it was thermodynamically favoured it was kinetically inhibited.kinetically inhibited.


Test for equilibrium:Test for equilibrium: Change P, or T, or an amount of one Change P, or T, or an amount of one

reactant (discussed later). reactant (discussed later). If there is a change in the system we were If there is a change in the system we were

at equilibrium.at equilibrium. If no change is observed we were not at If no change is observed we were not at

equilibrium (the reaction is kinetically equilibrium (the reaction is kinetically inhibited).inhibited).

The Le Chatelier PrincipleThe Le Chatelier Principle


If a system at equilibrium is subjected to a If a system at equilibrium is subjected to a change of pressure, temperature, or the change of pressure, temperature, or the number of moles of a component, there number of moles of a component, there will be a tendency for a net reaction in the will be a tendency for a net reaction in the direction that reduces the effect of this direction that reduces the effect of this change. change.


Example (amount of a reactant change)Example (amount of a reactant change) 2HI 2HI → H→ H22 + I + I2 2 (in an arbitrary eq. mix) (in an arbitrary eq. mix)

Now inject more HNow inject more H22: :

H2 now exceeds eq. concentration.H2 now exceeds eq. concentration. No longer in equilibrium.No longer in equilibrium. A net reaction ensues until a new A net reaction ensues until a new

equilibrium is reached.equilibrium is reached.


This happens in a direction that reduces This happens in a direction that reduces the effect of the added Hthe effect of the added H22..

This can happen by some of the excess HThis can happen by some of the excess H22

reacting with some of the Ireacting with some of the I22 to form more to form more

HI.HI. Chemists would say "the equilibrium shifts Chemists would say "the equilibrium shifts

to the left." to the left."


Example (pressure change )Example (pressure change ) 2NH2NH33 → 3H→ 3H22 + N + N22 (in an arbitrary eq. (in an arbitrary eq.

mix, in a closed reaction vessel)mix, in a closed reaction vessel) Now increase the size of the vessel.Now increase the size of the vessel. Effect of increasing the size is that the Effect of increasing the size is that the

pressure inside the vessel falls.pressure inside the vessel falls. How does the system react?How does the system react?


2NH2NH33 → 3H→ 3H22 + N + N22

If only 2NH3 present, the pressure is 2 If only 2NH3 present, the pressure is 2 arbitrary units.arbitrary units.

If only 3HIf only 3H22 + N + N22 present, the pressure is 4 present, the pressure is 4 arbitrary units.arbitrary units.

Increasing volume decreases the Increasing volume decreases the pressure.pressure.

Therefore system reacts to increase Therefore system reacts to increase pressurepressure


2NH2NH33 → 3H→ 3H22 + N + N22

Pressure in increased by forming more Pressure in increased by forming more products (and decomposing more NH3).products (and decomposing more NH3).

Equilibrium shifts to the right.Equilibrium shifts to the right.


Example (temperature change)Example (temperature change) heat + Nheat + N22 + O + O22 →→ 2 NO (endothermic) 2 NO (endothermic)

Regard Regard heatheat as a "reactant" or "product" in as a "reactant" or "product" in an endothermic or exothermic reaction an endothermic or exothermic reaction respectively.respectively.

Apply the Principle.Apply the Principle.


Suppose this reaction is at equilibrium at Suppose this reaction is at equilibrium at some temperature some temperature TT11 . .

Increase temperature to Increase temperature to TT22..

The system reacts in a way that absorbs The system reacts in a way that absorbs heat.heat.

Equilibrium shifts to the right (product)Equilibrium shifts to the right (product) [by analogy, equilibrium shifts to the left for [by analogy, equilibrium shifts to the left for

exothermic reactions].exothermic reactions].

Equilibrium QuotientEquilibrium Quotient

Earlier condition for the reaction Earlier condition for the reaction

aA + bB aA + bB →→ cC + dD cC + dD

can be expressed as a quotientcan be expressed as a quotient

cba

dc

r

f QBA

DC

k

k

Equilibrium ConstantEquilibrium Constant

If expressed in terms of the equilibrium If expressed in terms of the equilibrium concentrations, this becomes:concentrations, this becomes:

cba

dc

r

f KBA

DC

k

k


KKcc is the value of is the value of QQcc when the reaction is when the reaction is

at equilibrium.at equilibrium. The ratio KThe ratio Kcc/Q/Qcc s serves as an index how the erves as an index how the

composition of the reaction system composition of the reaction system compares to that of the equilibrium state.compares to that of the equilibrium state.

Thus, it indicates the direction in which Thus, it indicates the direction in which any net reaction must proceed. any net reaction must proceed.


QQcc/K/Kcc > 1 > 1

Product concentration too high for Product concentration too high for equilibrium;equilibrium;

net reaction proceeds to left.net reaction proceeds to left.


KKcc/Q/Qcc = 1 = 1

System is at equilibriumSystem is at equilibrium No net change will occur.No net change will occur. < 1 < 1Product Product

concentration too low for equilibrium;concentration too low for equilibrium;net reaction proceeds to right.net reaction proceeds to right.


KKcc/Q/Qcc < 1 < 1

Product concentration too low for Product concentration too low for equilibrium.equilibrium.

Net reaction proceeds to right.Net reaction proceeds to right.

Relating Q and K to ReactionsRelating Q and K to Reactions

Each tiny dot on the graph represents a Each tiny dot on the graph represents a possible combination of NOpossible combination of NO22 and N and N22OO44

concentrations that produce a certain concentrations that produce a certain value of value of QQcc for N for N22OO44 →→ 2 NO 2 NO22. .


Only those dots that fall on the red line Only those dots that fall on the red line correspond to equilibrium states of this correspond to equilibrium states of this system (those for which system (those for which QQcc = K = Kcc ). ).


If the system is initially in a non-equilibrium If the system is initially in a non-equilibrium state, its composition will change in a state, its composition will change in a direction that moves it to one on the line. direction that moves it to one on the line.


Solid/vapour equilibria.Solid/vapour equilibria. E.g.,: Sublimation of iodine IE.g.,: Sublimation of iodine I22(s)(s) = I = I22(g)(g)

The possible equilibrium states of the The possible equilibrium states of the system are limited to those in which at system are limited to those in which at least some solid is present (shaded) least some solid is present (shaded)


However, within this region, the quantity of However, within this region, the quantity of iodine vapor (red line) is constant [iodine vapor (red line) is constant [as long as long as the temperature is unchangedas the temperature is unchanged]. ].


The arrow shows the succession of states The arrow shows the succession of states the system passes through when 0.28 the system passes through when 0.28 mole of solid iodine is placed in a 1-L mole of solid iodine is placed in a 1-L sealed container.sealed container.


The unit slope of this line reflects the fact The unit slope of this line reflects the fact that each mole of Ithat each mole of I22 removed from the removed from the

solid ends up in the vapor. solid ends up in the vapor.


The decomposition of ammonium chlorideThe decomposition of ammonium chlorideNHNH44ClCl(s)(s) = NH = NH33(g)(g) + HCl + HCl(g)(g) is another is another

example of a solid-gas equilibrium.example of a solid-gas equilibrium.


Arrow Arrow 11 traces the states the system traces the states the system passes through when solid NHpasses through when solid NH44Cl is Cl is

placed in a closed container. placed in a closed container.


Arrow Arrow 22 represents the addition of represents the addition of ammonia to the equilibrium mixture; ammonia to the equilibrium mixture;


the system responds by following the path the system responds by following the path 33 back to a new equilibrium state. back to a new equilibrium state.


This state contains a smaller quantity of This state contains a smaller quantity of ammonia than was added (Consistent with ammonia than was added (Consistent with the Le Châtelier principle). the Le Châtelier principle).

Does the reaction stop at Does the reaction stop at equilibrium?equilibrium?

At equilibrium At equilibrium QQcc = = KKcc..

There is no change in the concentrations There is no change in the concentrations of the reactants/products.of the reactants/products.

The absence of any The absence of any net changenet change does not does not mean that nothing is happening.mean that nothing is happening.

Why?Why?


As any reaction is reversible, it can be As any reaction is reversible, it can be expressed at the sum of the forward and the expressed at the sum of the forward and the reverse reactions.reverse reactions.

A = B A = B A A → B→ B rate = rate = kkff[A].[A].

B → AB → A rate = rate = kkrr[B].[B].

This impliesThis implies

r

fc k

k

A

BK


This implies that the rates of the forward and This implies that the rates of the forward and backward reactions are equal: not that they are backward reactions are equal: not that they are zero.zero.

I.eI.e., at equilibrium As are turning into Bs as fast ., at equilibrium As are turning into Bs as fast as Bs turn into As.as Bs turn into As.

Equilibrium is a dynamic process.Equilibrium is a dynamic process.

ExampleExample

The commercial production of hydrogen is The commercial production of hydrogen is carried out by treating natural gas (mainly carried out by treating natural gas (mainly methane) with steam at high temperatures methane) with steam at high temperatures and in the presence of a catalyst (“steam and in the presence of a catalyst (“steam reforming of methane”):reforming of methane”):

CHCH44 + H + H22O O CH CH33OH + HOH + H22

ExampleExample

Given the following boiling points: Given the following boiling points: CHCH44 (methane) = –161°C (methane) = –161°C HH22O = 100°C O = 100°C

CHCH33OH = 65°OH = 65°

HH22 = –253°C. = –253°C.

predict the effects of an increase in the total predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and pressure on this equilibrium at 50°, 75° and 120°C.120°C.

ExampleExample Given the following boiling points: Given the following boiling points: CHCH44 (methane) = –161°C (methane) = –161°C HH22O = 100°C O = 100°C CHCH33OH = 65°OH = 65° HH22 = –253°C. = –253°C. predict the effects of an increase in the total predict the effects of an increase in the total

pressure on this equilibrium at 50°, 75° and pressure on this equilibrium at 50°, 75° and 120°C.120°C.

Solution:Solution: Calculate the change in the moles of Calculate the change in the moles of

gas for each process:gas for each process:

ExampleExample

50°C 50°C

CHCH4 4 (g) + H(g) + H22O (l) CHO (l) CH33OH (l) + HOH (l) + H22(g)(g)

Solution:Solution: Calculate the change in the Calculate the change in the

moles of gas for each process:moles of gas for each process:

One mole gas One mole gas one mole of gas one mole of gasNO CHANGENO CHANGE

ExampleExample

75°C 75°C

CHCH4 4 (g) + H(g) + H22O (l) CHO (l) CH33OH (g) + HOH (g) + H22(g)(g)

Solution:Solution: Calculate the change in the moles of Calculate the change in the moles of

gas for each process:gas for each process:

One mole gas One mole gas two mole of gas two mole of gas

To counter the increase in pressure, equilibrium To counter the increase in pressure, equilibrium shifts to the left.shifts to the left.

ExampleExample

120°C 120°C

CHCH4 4 (g) + H(g) + H22O (g) CHO (g) CH33OH (g) + HOH (g) + H22(g)(g)

Solution:Solution: Calculate the change in the Calculate the change in the

moles of gas for each process:moles of gas for each process:

two mole gas two mole gas two mole of gas two mole of gasNO CHANGENO CHANGE

How to find the equilibrium constant How to find the equilibrium constant for a series of reactions for a series of reactions

Many chemical changes can be regarded Many chemical changes can be regarded as the sum or difference of two or more as the sum or difference of two or more other reactions. other reactions.

If we know the equilibrium constants of the If we know the equilibrium constants of the individual processes, we can easily individual processes, we can easily calculate that for the overall reaction calculate that for the overall reaction according to the following rule: according to the following rule:

How to find the equilibrium constant How to find the equilibrium constant for a series of reactions for a series of reactions

The equilibrium constant for the sum of The equilibrium constant for the sum of two or more reactions is the product of the two or more reactions is the product of the equilibrium constants for each of the equilibrium constants for each of the steps. steps.

ExampleExample

Given the following equilibrium constants:Given the following equilibrium constants:

CaCOCaCO3 3 (s)(s) → Ca → Ca22+ + (aq)(aq) + CO + CO33

2–2– (aq)(aq)

KK11 = 10 = 10–6.3–6.3

HCOHCO3–3– (aq) (aq) → H→ H++(aq) (aq) + CO32–+ CO32–(aq)(aq)KK22 = 10 = 10–10.3–10.3

Calculate the value of Calculate the value of KK for the reaction for the reaction

CaCOCaCO33 (s)(s) + H + H++(aq)(aq) → Ca → Ca2+2+ (aq)(aq) + HCO + HCO3–3– (aq)(aq)

ExampleExample

The net reaction The net reaction CaCOCaCO33 (s)(s) + H + H++(aq)(aq) → Ca → Ca2+2+ (aq)(aq) + HCO + HCO3–3– (aq)(aq)

is the sum of Reaction 1 is the sum of Reaction 1

CaCOCaCO3 3 (s)(s) → Ca → Ca22+ + (aq)(aq) + CO + CO33

2–2– (aq) (aq) KK11 = 10 = 10–6.3–6.3

And the reverse of Reaction 2And the reverse of Reaction 2

HCOHCO3–3– (aq) (aq) ←← H+ H+(aq) (aq) + CO+ CO3 3 2–2– (aq) (aq) KK-2 -2 = 10= 10-(-10.3)-(-10.3)

ExampleExample

Therefore,Therefore,

KK = = KK11 KK-2-2 = 1 = 100(-8.4+10.3)(-8.4+10.3) = = 1010+1.9+1.9

The Harber processThe Harber process

The Haber process for the synthesis of The Haber process for the synthesis of ammonia is based on the exothermic ammonia is based on the exothermic reaction reaction

NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(g)(g)

Apply the le Châtelier principle in order to Apply the le Châtelier principle in order to maximize the amount of product in the maximize the amount of product in the reaction mixture.reaction mixture.


NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(g)(g)

Would prefer Would prefer NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(l)(l)


NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(g)(g)

Would prefer Would prefer NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(l)(l)

it should be carried out at high pressure it should be carried out at high pressure and low temperatureand low temperature. .


NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(g)(g)

But lower temperature slows the reaction.But lower temperature slows the reaction.

A choice has to be made.A choice has to be made.

But But Haber solved the first problem by developing Haber solved the first problem by developing a a catalystcatalyst that would greatly speed up the that would greatly speed up the reaction at lower temperatures. reaction at lower temperatures.


NN22(g) + 3 H(g) + 3 H22(g) = 2 NH(g) = 2 NH33(g)(g)

A catalyst is a substance that lowers the A catalyst is a substance that lowers the energy barrier for a reaction. energy barrier for a reaction.

A catalyst plays no role in the net reaction.A catalyst plays no role in the net reaction.

It is regenerated at the end of the reaction.It is regenerated at the end of the reaction.

Thus, they are only needed in trace Thus, they are only needed in trace quantities.quantities.

condensed matter 2 physical chemistry and applications of liquids and solutions

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