Kamalkant Chem for All Bbsr 1 (14)

7/27/2019 Kamalkant Chem for All Bbsr 1 (14)

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Basic Physical Chemistry by Paul D.I. Fletcher, Department of Chemistry, University of Hull

Basic Physical Chemistry. Prof. Paul D.I. Fletcher

7 lectures and 1 additional slot in semester 2 of year 1 as part of module 06512

Foundations of Chemistry 2)

The aims of this segment of the module are to introduce the key concepts of:

states of matter, intermolecular forces, ideal gas behaviour, equilibrium constants, enthalpy, entropy and free energy, reaction rates. rate constants and activation energy electromagnetic spectrum and spectroscopy.

This module segment will also emphasise key skills in physical chemistry such as

units, powers of 10 and quantitative problem solving.

Synopsis (approximately by lecture)

Lectures 1 & 2. States of matter and intermolecular forces. Ideal gas behaviour

Lectures 3 & 4. Equilibrium constant and its relation to enthalpy, entropy and free

energy.Lectures 5 & 6. Reaction rates, rate constants and rate laws. First and second order

reactions and their integrated rate equations, activation energy and

variation of rate with temperature.

Lecture 7. The electromagnetic spectrum and spectroscopy. Energies and

molecular processes occurring with microwaves, infrared, visible,

UV and X ray photons.

Lect/Sem 8. Completion of lectures and revision of key points

Support and Assessment

This module segment will include several short tests to help you checkwhether you have grasped the main concepts.

Module 06512 will be assessed in a 2 hour exam with 4 questions, one each onthe organic, inorganic, analytical and physical sections of the module. Youwill answer all 4 questions, each of which will have some choice in its parts.

Seminars and tutorials forming part of module 06514 (Chemical &Professional Skills A) will include problem solving exercises related to this

course.

You will also do one session per week of physical chemistry practical inweeks 3-6.

Texts

P.W. Atkins and J. de Paula, Atkins Physical Chemistry OUP, 7th Edition,Oxford

Good choice for your main physical chemistry text if you are OK at maths.

P.W. Atkins, The Elements of Physical Chemistry, OUPSimplified version of the main Atkins text for those who struggle with the

maths.

K.J. Laidler and J.H. Meiser, Physical Chemistry, Houghton Mifflin Co., 3rdEdition, Boston.

Good alternative to Atkinss main text.


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States of Matter

Property Solid Liquid Gas

Size and shape Fixed size and

shape

Fixed size but takes

shape of container

Takes the size and

shape of the

container

Order Long-range orderover 100s of atoms

or molecules

Only short rangeorder over 1 2

neighbour atoms or

molecules

No order at all

Density high densities,

typically

1-10 g cm-3

.

high densities,

typically

1-10 g cm-3

.

low densities,

typically

0.1 to 0.001 g cm-3

Compressibility incompressible

(almost)

incompressible

(almost)

very compressible

Atom or

molecular

separation

atoms or molecules

are touching, i.e.

separation = size.

atoms or molecules

are touching, i.e.

separation = size.

atoms or molecules

are separated by

many atom or

molecular

diameters

Pictorially (in 2 D):

Phase transitions between different states of matter occur at sharply defined

temperatures and pressures with taking in or giving out of heat (at constant pressurethis heat change is the enthalpy change). For example, for pure water at 1 atm

pressure:

0oC, 60.2 kJ mol

-1100

oC, 40.7 kJ mol

-1

ice water steam at 1 atm

20 cm3

18 cm3

30618 cm3

Solid

Rigid lattice

Long range order

Atoms/molecules

touching

Liquid

No rigid lattice

Short range order

Atoms/molecules

touching

Gas

No rigid lattice

No order

Atoms/molecules

very widely

separated


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NB. Water is very unusual in that the solid form (ice) is less dense than the liquid

form (water). This behaviour is due the structuring of liquid water caused by

hydrogen bonding.

Phase transition temperatures depend on pressure. Increasing the pressure causes the

lowest volume form (i.e. the highest density)of the material to be formed formed more

easily. For example, increasing the pressure to more than 1 atm will cause water toboil at a temperature higher than 100

oC. High pressures will make ice melt to water

at temperatures lower than 0oC. This last effect is why ice skating works. The high

pressure of an ice skate blade on the ice causes the ice to melt enabling the blade to

slip over a thin film of liquid water. The thin film of liquid water re-freezes when the

ice skate blade is removed.

Note that skating would not work on the frozen forms of materials other than water.

Because these solids are more dense than the corresponding liquid forms, high

pressure has the opposite effect on the melting points of such solids i.e. it causes

increased freezing rather than melting.

The graph below shows how the temperature varies as heat is supplied at a constantrate into a fixed mass of H2O at a pressure of 1 atm which is initially frozen ice

at -50oC.

-50

0

50

100

150

200

0 2 4 6

time/arbitrary units

temperature/oC

For a particular substance, the state of matter is determined by the balance between

inter-molecular force energy (a potential energy which depends on the

atomic/molecular positions and can be changed by pressure) and the kinetic energy of

The heat

supplied raises

the temperature

of the ice.

The heat supplied

converts the water

into steam at

100oC

The heat

supplied raises

the temperature

of the steam.

The heat

supplied raises

the temperature

of the water.The heat

supplied coverts

the ice to water

at 0oC.


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the atoms/molecules (kinetic energy is determined by atom/molecule velocities) and is

controlled by temperature.

Gases favoured by high temperatures, low pressures (corresponds to large atom or

molecule separations) and intermolecular forces in which repulsion is stronger than

attraction.

Solids favoured by low temperatures, high pressures (corresponds to small atom or

molecule separations) and intermolecular forces in which attraction is stronger than

repulsion.

Inter-molecular Forces & Energies

Some key points to understand first.

Energy = force x distance moved (when the force is constant).In terms of units, 1J = 1Nm

Inter-molecular forces are weaker than intra-molecular forces (covalent,ionic or metallic bonds. In terms of energies, this means that the energy

needed to break one mole of inter-molecular bonds is much less than the

energy needed to break one mole of intra-molecular bonds, e.g. C-C bond

energy is approximately 440 kJ mol-1

.

Inter-molecular forces become weaker when the separation between atoms ormolecules is increased. They are zero when the separation is infinite

(approximately true for a gas at low concentration).

Because the energy associated with inter-molecular forces depends on theatom/moleculepositions, the energy is apotential energy.

The energy required to pull atoms or molecules apart from a certain separationto infinite separation ispositive (requires energy input) when the net force is

attractive and negative (gives energy out) when the net force is repulsive. The energy needed to break all the inter-molecular forces in a material is the

energy required to turn it into a gas where all the molecules are widely

separated and inter-molecular forces are negligibly small.

The basis for all intermolecular forces can be understood in terms of the rules of

electrostatics

Like charges repel

Unlike charges attract

Inter-molecular forces depend on the net charge and charge distribution of the

interacting atoms or molecules and can be summarised under the following headings.

Ion-Ion

Example: Na+

with Cl-

Energy of interaction = +40-400 kJ mol-1 (depends on solvent)

The + sign indicates that can be attractive or repulsive.


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Digression about permanent dipole moments (dipoles).

Uncharged molecules can be polar or apolar. In a molecule such as ethane where all

the atoms have similar electronegativities, then the charge distribution is even over

the whole molecule which is then described as apolar. In a molecule such as

chloroethane, the chlorine atom is more electronegative than the other atoms and

tends to attract more than its fair share of electrons to itself. This uneven electrondistribution causes the chlorine end of the molecule to be slightly negative and the

other end to therefore be slightly positive. This type of molecule is polar and has a

permanent dipole moment. The word permanent here means that the dipole

moment remains in the absence of an applied electric field. The chlorine end of the

chloroethane molecule has a small negative charge (units = Coulomb C) and the other

end has a slight positive charge.

CH3 CH3 CH3 CH2Cl

Apolar Polar

The dipole moment has the following definition. For a particle which has one end

with a charge of +q and the other end with a chage of q with a separation of xbetween them, then the dipole moment is qx and has units of Cm.

How big is a dipole moment? We can estimate the order of magnitude very roughly

as follows. Intuitively we might expect that the charge on one end of a polar

molecule might be (say) 10% of one electron (electronic charge = -1.602 x 10-19 C).

The separation will be roughly one bond length which is about 0.1 nm. Assuming

these values, we get that a typical dipole moment will be about 1.6 x 10-30

Cm. It can

be more convenient to use Debye units for dipole moments where 1 D is equal to

3.336 x 10-30

Cm. Our estimated dipole moment is therefore of the order of 0.5 D.

Compare this order-of-magnitude estimate with actual values for molecules in the gas

phase as shown in the table below. As you can see, the order of magnitude of the

actual dipole moment values observed do indeed correspond to charges equal to a

fraction of an electron separated over a distance comparable to a typical bond length.

molecule Dipole moment/D

water 1.85

chloroethane 2.05

Carbon dioxide 0

Cyanoethane (= propionitrile) 4.02

Question: Can you work out why water has a non-zero dipole moment whereas

carbon dioxide does not?

Ion-Dipole

Example: Na+

with H2O

Energy of interaction = 4-40 kJ mol-1

(always attractive)

In this example, the positive sodium ion is attracted to the negative end of the water

molecule (i.e. the oxygen atom). A negative ion is attracted to the positive end of a

dipole. Attractive ion-dipole forces between ions and polar solvents such as water are

the reason why salts are soluble only in polar solvents.


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Dipole-Dipole

Example: Chlorobenzene with Chlorobenzene

Energy of interaction = 0.4-4 kJ mol-1

(always attractive)

The molecules rotate until the negative end of one (the Cl atom in this example) is

aligned with the positive end of another which has the result that the force is always

attractive.

Hydrogen bond

The hydrogen bonds is an especially strong case of a dipole-dipole interaction which

occurs between H and electronegative atoms such as O, N, F, Cl, etc. We can

represent it as follows.

R-E-H..E-R

where E and E are electronegative atoms and R and R represent the rest of each of

the molecules. The H bond is represented as the dotted line.

Example: water with water

Energy of interaction = 10-40 kJ mol-1 (always attractive)The H bond is particularly strong because the H atom has only one electron

surrounding its nucleus consisting of a single proton. When the single electron gets

pulled towards the electronegative atom, it leaves the proton of the H atom

unshielded where its positive charge can interact strongly with the negative charged

end of the second molecule. When outer-shell electrons are pulled away from atoms

other than hydrogen, there are always inner-shell electrons shielding the nucleus.

Dipole-Induced dipole

When a molecule which is uncharged and has no permanent dipole moment is placed

in an electric field, the electron distribution is distorted by the electric field to give an

induced dipole which goes to zero when the electric field is removed. The electric

field can be applied by electrodes (electric field = voltage difference divided by

electrode separation, units V m-1

), by being close to a charge or another dipole or by

electromagnetic radiation (= an oscillating electric and magnetic field). The size of an

induced dipole depends on the magnitude of the applied electric field and the

polarisability of the molecule. Molecules such as benzene which contain their

outermost electrons in big, delocalised orbitals (i.e. electrons are far from the positive

nuclei and therefore weakly held) are relatively highly polarisable.

Example: Chlorobenzene (permanent dipole) with benzene (no perm. dipole)


(always attractive)

When close together the permanent dipole induces the opposite sign dipole in theother molecule. Hence the force is always attractive.

Dispersion forces (also called London forces after their discoverer).

Even molecules with no permanent charges or dipoles attract each other. For

example, we know that He attracts He because it is possible to liquefy the gas at low

enough temperatures. How does this attraction work when we know that the average

electron distribution in the 1s orbitals is spherically symmetrical? The answer is that


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the electrons are moving from one end of the atoms the other on a timescale of

approximately 10-15 seconds. On this very fast timescale the atoms will have

instantaneous transient dipoles which are continuously changing direction. When two

He atoms are close you might think that their transient dipoles will spend as much

time repelling (when dipoles are aligned) as attracting (when dipoles are opposite) so

that the time-average net force is zero. However, when they are close the transient

dipoles interact and tend to change direction synchronously such that there is a nettime-average attraction. This should not be too surprising as you would expect the

system to do whatever gives the lowest energy. The net result is that there is always

an attractive force. Because all molecules have electrons which are moving, all

molecules have attractive dispersion forces.

Example: Helium with Helium


(always attractive, bigger for bigger

molecules)

Short-range repulsion

When molecules or atoms are very close together (i.e. the separation is approximatelyequal to the size of the outermost electron orbitals, typically less than 0.15 nm) then

the electron clouds repel each other very strongly. This short range repulsion

increases very strongly if you try to push the atoms closer.

This short range repulsion is why things feel solid. When you look at atomic

diagrams it is obvious that the actual nuclei and electrons occupy only a small fraction

of the actual space taken up by atoms and molecules. Most of this space occupied by

molecules is empty! The reason you cannot push your hand through a wall is that the

electron clouds of the atoms of your hand feel a strong short range repulsion with the

electron clouds of the atoms of the wall.

Inter-molecular interaction energy diagram

As we have seen, all molecules interact with a mixture of attractive and repulsive

forces. We can represent this as a plot of the energy required to bring the molecules

to a certain separation from infinite separation where the interactions are zero. In

such a plot (shown below), negative energies correspond to attractive forces and

positive energies correspond to repulsive forces.

The force between the molecules at a particular separation is given by the slope of the

plot, i.e. it is equal to the differential d(energy)/d(separation). The negative slope

region corresponds to a repulsive force and the positive slope region to an attractive

force. At the separation corresponding to minimum energy, the force (= slope) is zeroand hence this is the equilibrium separation of the two atoms or molecules.

At an absolute temperature T, the thermal or kinetic energy of a molecule is kT where

k is Boltzmanns constant. For a mole of molecules, the thermal energy is NkT where

N is Avogadros number. The product Nk is equal to the gas constant R (equal to

8.314 J mol-1

K-1

) and hence the thermal energy of one mole of molecules is RT. At

room temperature, this thermal energy of a mole of molecules is roughly 2.5 kJ mol-1

.


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How does the inter-molecular interaction energy diagram determine whether a

particular material is a solid, liquid or gas at some particular absolute temperature? It

depends how deep the interaction energy minimum is relative to the thermal energy.

If the interaction energy minimum is large compared with thermal energy then no

molecules have enough thermal energy to escape from the energy minimum.

Hence, all molecules will have a separation corresponding to the equilibrium

separation and the material will be a liquid or solid. If the minimum is smallcompared with the thermal energy, then the thermal energy is enough to enable the

molecules to escape from the minimum to infinite separations, i.e. to form a gas.

Gases will be formed when either (a) the attractive interaction energy is weak or (b)

when the thermal energy is high by increasing the temperature. Obviously all

molecules fall to the bottom of the energy minimum when the absolute temperature is

zero.

-0.6

-0.4

-0.2

0

0.2

0 0.5 1 1.5 2 2.5

typical separation/nm

energyrequiredtobringthemoleculesfr

infiniteseparation/arbitraryunits

Ideal gas behaviour

All gases behave ideally under conditions when all inter-molecular forces are

negligible. These forces include long-range attraction and short-range repulsion. All

inter-molecular forces become negligibly small when all the atoms or molecules are a

long way apart and the actual space occupied by the gas molecules is negligible

compared with the overall volume of the gas. This happens when the concentration

(=n/V) of the gas is low which, at fixed moles, temperature and volume, can beachieved by decreasing the pressure.

How weak do inter-molecular forces have to be to be negligible? Intermolecular

forces are insignificant if they are not strong enough to attract or repel the gas

molecules away from completely random flight paths. This is true when the potential

energy of the inter-molecular forces (i.e. the energy needed to pull the molecules

against their inter-molecular forces to infinite separation) is small relative to the


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thermal or kinetic energy (= RT for one mole) making the gas molecules fly about.

Hence, the inter-molecular forces can be made negligibly small in tow ways. Firstly,

you can weaken the forces by separating the gas molecules by lowering the

concentration as described above. Secondly, you can raise the thermal energy by

increasing the temperature. In summary, gases behave ideally in the limits of low

pressure and high temperature. They deviate from ideal behaviour more and more as

you increase the pressure and/or lower the temperature, i.e. as you move towardscondensing the gas into a liquid. Whether or not the assumption of ideal gas

behaviour is OK for a particular calculation depends on how accurately you need the

answer. Gases are likely to behave reasonably ideally when the pressure is lower than

one atmosphere and the temperature is more than 100oC above the normal boiling

point of the liquid.

When gases are ideal, they obey the following equation

PV = nRT

where P is pressure, unit Pa (1 Pa = 1 N m-2

)

V is the total volume of the gas, unit m3n is the number of moles of the gas, unit mol

R is the Gas constant, equal to 8.314 J mol-1

K-1

T is the absolute temperature, unit K (equal tooC + 273.15)

Hint. As for all equations, you can check this equation is correct by testing that the

units of the LHS and RHS are equal.

LHS of equation PV units are N m-2

m3

= N m = J

RHS of equation nRT units are mol J mol-1

K-1

K = J

Units match and so the equation is OK.

Sample calculation

An ideal gas has a pressure of 400 mm Hg in a container of volume 300 cm3 at a

temperature of 25oC. What will the pressure be if the temperature is raised to 100oC?

[The gas constant R = 8.314 J mol-1

K-1

]

Long method answer

Since PV = nRT, we can re-arrange to get P = nRT/V. In order to calculate the new

pressure P at 100oC, we have to first calculate the number of moles n using the known

pressure at 25oC. By re-arranging the equation we get n = PV/RT.

Converting all units to SI units will give n in units of mol.

P = 400 mm Hg = 400/760 atmosphere = (400/760) x 101325 Pa = 53329 Pa

T = 25oC = 298.15 K

V = 300 cm

3

= 300 x 10

-6

m

3

Hence, n = (53329 x 300 x 10-6

)/(8.314 x 298.15) = 6.454 x 10-3

mol

We now use this value of n to calculate P at 100oC (= 373.15 K)

P = nRT/V = (6.454 x 10-3

x 8.314 x 373.15)/(300 x 10-6

) = 66742 Pa

We can convert this pressure back into units of mm Hg.

P = 66742 Pa = (66742/101325) atmosphere = (66742/101325) x 760 mm Hg

P = 500.6 mm Hg


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Short method answer

Re-arranging PV = nRT gives P/T = nR/V which is a constant if the number of moles

and volume is kept constant as they are in this problem. If we have two values (P1 at

T1 and P2 at T2), then P1/T1 = constant = P2/T2. Putting in numbers gives:

400/298.15 = P2/373.15

Re-arranging gives P2 = (400/298.15) x 373.15 = 500.6 mm Hg. Note here that P2

comes out in the same units as those used for P1 (i.e. mm Hg used here). The finalanswer by this much quicker method is, of course, the same as above.

P = 500.6 mm Hg

Equilibrium constants

For any chemical reaction when equilibrium is reached and the concentrations stop

changing, the ratio of the equilibrium activities or concentrations of the products

divided by the equilibrium activities or concentrations of the reactants is equal to a

constant called the equilibrium constant K.

For any reaction, the true, thermodynamic equilibrium constant K is equal to theequilibrium activities of all the product species divided by the equilibrium activities of

all the reactant species. For the particular example reaction below:

A + B + C + D X + Y + Z

K is equal to

DCBA

ZYX

aaaa

aaaK=

where aX is the activity of the species X. The activity of a species is its actual activitydivided by its activity at a standard concentration which we choose. The standard

concentration is called the standard state and is usually chosen to be 1 M for solutes in

solution. Note that activity defined in this way is actually dimensionless but requires

that the standard state should be specified. However, you are likely to see statements

like activity = 0.1 M. This is a shorthand way of saying that the activity is 0.1 of

the activity of the same species when in its standard state of 1 M. Because activity is

unitless, all true thermodynamic equilibrium constants are also unitless. When

apparent units are shown, they actually indicate the standard state which has been

assumed.

When we choose 1 M as the standard state and the solute behaves ideally, then the

activity is numerically equal to the concentration in M units. For real solutions which

do not behave completely ideally, then concentration is only approximately

numerically equal to activity. For normal approximate work, we can ignore non-

ideality effects and equate activity and concentration. With this approximation, the

equilibrium constant for the reaction above is

[ ][ ][ ][ ][ ][ ][ ]DCBA

ZYXK=


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As written in terms of concentrations, K for this reaction formally has units of M-1.

However, these units are better regarded as a label telling you that the standard

states which have been chosen for the solutes A, B, C, D, X, Y and Z are all 1 M.

Equilibrium constants tell you about the extent to which a reaction will proceed if

given enough time to reach equilibrium. How far a reaction will go can also beexpressed in terms of the reaction yield. Hence, it should be clear that K and yield are

related as illustrated below.

Let us consider the simple reaction A B for which[ ][ ]AB

K= . We can then calculate

the yields we would get for different values of K assuming that we start the reaction

with an initial concentration of A equal to [A]0. The equations we use are shown

below.

Initial concentration of A is [A]0 and the initial concentration of B is zero

Therefore, [A] = [A]0 [B] and hence

[ ][ ]

[ ][ ] [ ]{ }BA

B

A

BK

==

0

and re-arranging gives [ ][ ]

{ }KAK

B+

=1

0

The equilibrium yield is equal to [B]/[A]0

The table below shows a set of values calculated for [A]0 = 1 M. It can be seen that a

high value of[ ][ ]AB

K= corresponds to a high reaction yield at equilibrium.

K Equilibrium [A] Equilibrium [B] Equilibrium yield/%

0.01 0.9901 0.0099 0.99

0.1 0.909 0.091 9.1

1 0.5 0.5 50.0

10 0.091 0.909 90.9

100 0.0099 0.9901 99.0

1000 0.000999 0.9990 99.9

Equilibrium constants provide a measure of the extent to which a reaction will go

forward if left long enough to reach equilibrium. Whether or not a reaction will go

forward depends on the difference in energies between the reactants and products and

so equilibrium constant must be related to these energy differences. Consideration ofthese energies is the realm of thermodynamics.

Enthalpy H

Enthalpy change H (delta H) is the heat change of a process or reaction measuredat constant pressure. Heat given out (the beaker gets hot) means the reaction is


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exothermic and, by convention, H is negative. Heat taken in (the beaker gets cold)means the reaction is endothermic and H is positive.

For reactions H is normally quoted per mole of reaction as written proceeding left toright. The units ofH are J mol-1 or kJ mol-1. Consider a reaction:

A B for which H = -100 kJ mol-1

If you allow 0.1 mol of A to react to give 0.1 mol of B, then the actual measured

enthalpy change will 0.1 x (-100) = -10 kJ (not kJ mol-1

). This is negative and so

corresponds to heat given out. The enthalpy changeH is simply (Hproducts Hreactants)and equals (HB HA) for the reaction shown above. The H value refers to thechemical reaction as written and hence, the H for the reverse reaction B A must

be +100 kJ mol-1

.

Enthalpy change in a chemical reaction comes from bond making and bond breaking

and so is related to the relative strengths of the chemical bonds in the products and

reactants of the reaction. Making bonds releases heat and gives an exothermicreaction whereas breaking bonds requires heat input (endothermic).

Note that only enthalpy changesH can be measured directly. Individual H valuesfor separate chemicals cannot be measured directly. However, we can define a zero

point that the enthalpy of a pure element in its most stable form at 298.15 K and 1 atm

= 0. Relative to this zero point that we have defined, we can then measure the

enthalpy of a compound. For example, the enthalpy of CO2 at 298.15 K and 1 atm

will be equal to the measured H for the reaction

C(diamond) + O2 (gas) CO2(gas)

since the measured H = (Hproducts Hreactants) = (HCO2 - (Hdiamond + Hoxygen)) = HCO2since both Hdiamond are Hoxygen are zero by definition. Although this can seem subtle

and hard to understand, in fact it is the same process that we use to define altitude.

We can only directly measure altitude differences and so we define the altitude of sea

level as being zero. The altitudes that are listed on maps are equal to the altitude

differences between the relevant point on the map and sea level. Similarly, individual

enthalpies of chemical compounds correspond to the enthalpy differences for the

reactions in which the compounds are formed from the most stable forms of their

elements.

Hesss Law

This can be expressed in several different ways.

The overall enthalpy (or heat) change for a process or reaction is equal to thesum of the enthalpy (or heat) changes for all the individual steps.

Enthalpy is a state function, i.e. the enthalpy change depends only on theinitial and final states and not on the path taken.

Hesss Law can be used to calculate enthalpy changes for reactions from the enthalpy

changes of the individual steps.


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Example. The heats of combustion of ethane(g), carbon(s) and hydrogen(g) are -1558,

-393 and -285 KJ mol-1 respectively. Calculate the enthalpy of formation of ethane

from its elements.

Answer. From the question, we know:

1. C2H6 + 3.5O2 2CO2 + 3H2O H = -1558 kJ mol-12. C + O2 CO2 + H = -393 kJ mol

-1

3. H2 + 0.5O2 H2O H = -285 kJ mol-1

We wish to know the enthalpy change for the formation reaction:

2C + 3H2 C2 H6 H = ? kJ mol-1

We can obtain the target formation reaction by taking the following combination of

reactions 1 3.

(2 x reaction 2) + (3 x reaction 3) (1 x reaction 1) which gives:

2C + 2O2 + 3H2 + 1.5O2 + 2CO2 + 3H2O 2CO2 + 3H2O + C2H6 + 3.5O2

We then cancel all species which appear on both the LHS and RHS to give the final

reaction which we can see agrees with the target reaction.

2C + 3H2 C2H6

We then combine the H values for the individual reactions in precisely the same wayas we did for the reactions, i.e.

H = (2 x -393) + (3 x -285) (1 x -1558)H = -83 kJ mol-1

Does enthalpy change alone decide whether or not a reaction will proceed

forwards?

This is the same as asking whether reactions or processes will only proceed if

they are exothermic more or stronger bonds are formed relative to the ones that are broken

The answer is NO. Enthalpy change alone does not predict whether a reaction will

proceed. The following are all examples of reactions or processes which proceed in

spite of being enthalpically unfavourable (i.e. endothermic with H positive).

1. Hg2Cl2(s) + H2(g) 2Hg(l) + 2HCl(g)

2. SOCl2(l) + H2O(l) SO2(g) + 2HCl(g)

3. Liquid gas

4. some salts + water aqueous solution

Why do these reactions or processes all proceed even though they cost enthalpy?

We can see a clue by looking at the states of matter in all the examples.


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1. (s) + (g) (l) + (g)

2. (l) + (l) (g) + (g)

3. (l) (g)

4. (s) + (l) (l)

In each of these examples there is a net change from (s) (l) or(l) (g). In all casesthere is a loss of order which corresponds to an increase in randomness. Clearly, in

addition to H, there must be a function related to this randomness or disorder which

acts to determine whether reactions proceed. This function is called entropy S and

S = Sproducts Sreactants

When S ispositive, it means the products are more disordered than the reactants andthisfavours the reaction proceeding forward. When S is negative, it means the

products are less disordered than the reactants and this disfavours the reaction

proceeding forward. (N.B. This is the opposite way round to H for which a negative

value favours reaction.)

As we shall see, the equilibrium position of reactions is determined by a combination

of H and S which is called Gibbs Free Energy G

G = H TS (where T is the absolute temperature)

G = (Gproducts Greactants) = H - TS

The product TS has units of energy per mol (like H since quantities can only beadded together if they have the same units). Hence, S and S have units of J mol-1K-1.G is the Gibbs free energy change for one mole of reaction proceeding in theforward direction as written.

Gibbs free energy and equilibrium constant

All reactions are reversible, albeit to an infinitesimally small extent for some

examples. In principle, every reaction can be made to go eitherforwards or

backwards to reach its equilibrium ratio of concentrations by starting at different

concentrations of reactants or products. For example, consider the reaction

A B with equilibrium constant K = 1

If we start the reaction by taking a solution initially containing 1 M of A with zero B.

The reaction will then proceed forward (i.e. A will form B) with the concentration of

A dropping and the concentration of B increasing. The reaction will stop when both

concentrations are equal to 0.5 M and the ratio [B]/[A] = 1 = K. Alternatively, we can

start the same reaction by taking a solution initially containing 1 M of B with zero A.

In this case, the reaction will proceed backwards until the same equilibrium

concentrations are reached. It is important to understand that reactions can be driven

to go either forwards or backwards by changing concentrations.


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G determines whether a reaction will proceed forwards or backwards. It can beshown (see Dr. Paunovs course on thermodynamics) that:

IfG = +ve, the reaction will proceed right to left (i.e. backwards) IfG = 0, the reaction will be at equilibrium IfG = -ve, the reaction will proceed left to right (i.e. forwards)

Accepting this, then it is clear from our previous discussion that G must depend on the

concentration of a species. If changing concentrations can drive reactions either

forwards or backwards, then it must also change the sign ofG. G is a measure of thefree energy per mol of a species which it has available to push chemical reactions.

Because G is expressedper mole, its value does notdepend on how much solution

you have. For example, the G of a species A in 1 ml of a 1 M solution of A is

identical to its G value in 100 litres of 1 M solution. However, G does depend on the

concentration of A. We know that high concentrations are better at making reactions

proceed and hence we expect G to be higher for higher concentrations. The

relationship is as follows.

G = Go

+ RTln(C/Cstandard state)

where G is the Gibbs free energy of one mole of a species at concentration C

R is the gas constant (= 8.314 J mol-1

K-1

)

T is the absolute temperature

Cstandard state is the standard state concentration (which we can choose

and which we normally take to be 1 M for species in solution).

Go

is the standard Gibbs free energy and is equal to the Gibbs free

energy of one mole of a species at concentration equal to Cstandard state.

We can now work out the relationship between G and equilibrium constant K.

Consider the reaction

A + B C with equilibrium constant K

Let GA, GB and GC be the Gibbs free energies per mole for species A, B and C. The

G for 1 mole of the reaction in the forward direction as written is

G = GC (GA + GB)

We can write equations for each of the G values in terms of the concentrations of

each. Note that we have not included Cstandard state in these equations because we are

taking it to be 1 M. These equations are then substituted into the equation for

G.

A

o

AA CRTGG ln+=

B

o

BB CRTGG ln+=

C

o

CC CRTGG ln+=

( ){ } ( ){ }BACoBoAoC CRTCRTCRTGGGG lnlnln +++=


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( ){ }BACo CCCRTGG lnlnln ++=

+=BA

Co

CC

CRTGG ln

QRTGG o ln+=

This last equation tells you what G would be for any ratio of product and reactantconcentrations (equal to Q). Inspection of this equation shows that if we make the

product concentrations high and reactant concentrations low, then Q is large, ln Q is

large and positive and so G will be positive. In this situation, the reaction will gobackwards. If we make the product concentrations low and the reactant

concentrations high, then Q will be less than 1 and lnQ will be large and negative.

This will make G negative and hence the reaction will go forward. We can see thatthe equation correctly predicts what we already know.

We have a special situation when the reaction reaches equilibrium. At equilibrium,

we know the following:

G = 0The general concentration ratio Q = the equilibrium constant K

KRTGo ln0 += and hence

KRTGo ln= vant Hoff equation

This important equation tells us that we can relate K to Go. Go is what G wouldbe if all reactant and product concentrations were all equal to the standard state

concentration. The sign ofGo tells whether the reaction will go forwards orbackwards if all reactant and product concentrations were all equal to the standard

state concentration. When the concentrations are equal to the equilibrium

concentrations G is always zero. Obviously, you need to take care not to write Gwhen you mean Go and vice versa.

Why do we bother to compile tables ofGo values for different reactions? If we wantto compare the extent to which different reactions will proceed (which is what

chemistry is all about!) why dont we compare easier quantities such as reaction

yield? The answer is the following. If you are told that reaction 1 has an equilibrium

yield of 40% and reaction 2 has an equilibrium yield of 60%, which reaction has the

greater tendency to proceed? We cannot tell since we have not been told the all the

concentrations and we know that any reaction can be made to go forwards or

backwards by changing the concentrations. To compare the reactivities of different

reactions fairly, we need to compare reactions at thesame concentrations (lets choose

standard state concentrations). This is exactly what Go values do. A more negativevalue ofGo corresponds to a more reactive reaction.

Uses of the vant Hoff equation

We can derive how equilibrium constants vary with absolute temperature T as

follows. We know:


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KRTGo ln=

andooo

STHG =

where Ho and So are the values ofH and S when all species are in their standardstates.

Hence oo STHKRT = lnoo STHKRT = ln

ooSTHKRT +=ln

FinallyR

S

RT

HK

oo +

=ln

If we measure a series of K values at different absolute temperatures, then a plot of

lnK as the Y axis (called the ordinate) versus (1/T) as the X axis (called the abscissa)

will give a straight line graph with

Graph slope = -(Ho/R)Intercept on the Y axis = (So/R)

temperature/o

C K T/K (1/T)/K-1

lnK

0 0.5 273.15 0.003661 -0.693147

10 1.1 283.15 0.003532 0.09531

20 2 293.15 0.003411 0.693147

30 3.9 303.15 0.003299 1.360977

40 8.4 313.15 0.003193 2.128232

50 17 323.15 0.003095 2.833213

y = -6144x + 21.745

-1

0

1

2

3

4

0.003 0.0032 0.0034 0.0036 0.0038

(1/T)/K-1

lnK

From the graph, the slope = -6144 K (N.B. remember to include both the sign and the

units!) Ho = -slope x R = -(-6144 x 8.314) = +51081 J mol-1Ho = +51.1 kJ mol-1

From the graph, we get that the intercept on the Y axis = 21.745 = (So/R).Hence So = intercept x R = 21.745 x 8.314 = 180.8 J mol-1 K-1

So = 180.8 J mol-1 K-1


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This example shows how we can separate the total Gibbs free energy driving force for

this reaction into the contributions coming from bond making/bond breaking (i.e. the

enthalpy change) and from order/disorder (i.e. TSo). For this example, K increaseswith increasing temperature which means the reaction is endothermic, i.e. Ho is

positive and disfavours the reaction. The standard entropy change is positive which

means the products are more disordered than the reactants. The entropy change isdriving the reaction forward against an unfavourable enthalpy change.

In many exam questions, you will not be asked to plot a graph. Instead you will be

given only two values of K (lets call them K1 and K2) at two values of T (T1 and T2).

In the following example, we will use two values from the data table and graph shown

above. Obviously, we should get roughly the same final answers (not exactly since

the example graph data has some scatter due to uncertainties).

We choose K1 = 0.5 at T1 = 273.15

and K2 = 17 at T1 = 323.15

We apply our basic equation to both sets of values as follows

R

S

RT

HK

oo +

=

1

1ln

R

S

RT

HK

oo +

=

2

2ln

Looking at these two equations, we can see that we can get rid ofSo/R (a constant inboth equations) by subtracting one equation from the other.

21

21 lnlnRTH

RTHKK

oo+=

Some re-arranging gives

=

212

1 11lnTTR

H

K

K o

Put in the numbers

= 15.3231

15.2731

314.8175.0ln

o

H

Hence, Ho = 51.8 kJ mol-1(as expected, this agrees roughly with our previous answer)

To calculate the So, we can take one value of K and work out Go. We will take thevalue at T = 273.15 K (although we could take either value and still get the same

answer).


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At T = 273.15, K = 0.5 and Go = -RTlnK = -(8.314 x 273.15) x ln(0.5)Go = +1574 J mol-1 = Ho - TSo = 51800 273.15 x So

Hence So = -(1573 51800)/273.15 = +184 J mol-1 K-1So = +184 J mol-1 K-1

(again, this roughly agrees with our previous answer)

More general exam question tips.

If asked to quote an equation, take care to properly define all the terms. If asked to plot a graph, take care to label all axes properly with their units When starting a calculation, try to think in advance on any calculation:

will the answer will positive or negative?

roughly how big is the answer?

what are the units of the answer?

Tabulated values of So

and Ho

can be used to calculate equilibrium constants

We have seen that define H

o

values for stable elements as zero enables us to obtain H

o

values for compounds. It is also true that So

for perfect crystals at 0 K will be zero

because they will be perfectly ordered under these conditions and hence have zero

disorder or entropy. Using this fact with additional equations from thermodynamics

and measured quantities, So

values can be derived for any compound at any

temperature and pressure. Tabulated values of So

and Ho

can then be used to calculate

equilibrium constants as shown in the example below.

Question. Find the equilibrium constant for the reaction, at 25oC

0.5N2(g) + 0.5(O2(g) NO(g)given that Ho at 298.15 K is 89.9 kJ mol-1 for the reaction and the standard molarentropies at 298.15 K of NO, O2 and N2 are 210, 205 and 192 J mol

-1 K-1 respectively.

Answer. We first calculate So

So = So for NO (0.5So for O2 + 0.5So for N2)

So = 210 102.5 96So = 11.5 J mol-1 K-1

Hence Go = Ho TSo = 89900 (289.15 x 11.5) = 86471 J mol-1

From KRTG

o

ln=

we get that

=

RT

G

K

o

exp

Putting in numbers gives K = 6.9 x 10-16

This is a very small number which tells us that the equilibrium product concentrations

are very much less that the equilibrium reactant concentrations. In fact, this

equilibrium constant is so low that it is very hard to measure. Thermodynamics is

enabling us to use other measurements to calculate equilibrium constants which we

cannot measure.


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Some final points about standard states.

We have focussed on species in solution for which the usual standard state is 1 M.

Other concentration scales can be used. For example, the standard state can be taken

to be unit mole fraction, i.e. the pure material. For gases where the concentration is

proportional to pressure, common standard states are either unit pressure or unit

concentration. For solids or immiscible liquids (where the concentration is fixed by

the density) the standard state is usually taken to be the pure solid.

Chemical Kinetics

Equilibrium thermodynamics tells us about the extent to which reactions will proceed

if allowed infinite time. Chemical kinetics tells us about how fast reactions proceed

towards their equilibrium state. We first need to define some key terms.

Reaction rate is the rate of change of concentration with respect to time. It is helpful

to consider plots of concentration versus time for a simple reaction. We will consider

the reaction

A B

Mathematically, the reaction rate =[ ] [ ]

dt

Bd

dt

Ad+= The rate is a differential quantity

which is equal to the negative slope of the tangent of a plot of concentration of A

versus time or to the positive slope of a plot of concentration of B versus t.

Obviously, the rate cannot be negative which is why the minus sign must be in front

of the negative slope d[A]/dt but not in front of the positive slope d[B]/dt. Because

one molecule of B is formed for every molecule of A lost, it is clear that the plot of

[B] versus time must be an exact mirror image of the plot of [A] versus time.

For a more complicated reaction stoichiometry, such as

A + 2B + 3C 2X + Y + Z

[ ] [ ] [ ] [ ] [ ] [ ]dt

Zd

dt

Yd

dt

Xd

dt

Cd

dt

Bd

dt

Adrate +=+=+====

2

1

3

1

2

1

The normal units of rate are M s-1

.

The Rate Law

Reaction rates generally depend on the concentrations of one or more of the different

reactants of the reaction in a way that depends on the mechanism of the reaction. Therelationship between rate and all the reactant concentrations is called the rate law forthat particular reaction and must be measured experimentally. Consider a generalreaction

A + B + C + products


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The rate law always has the general form

Rate = k [A]a[B]

b[C]

c. and so on

where k is the rate constant. This is what the rate would be if theconcentrations of all the reactants were all equal to unity.

[species] is the time-dependent concentration of a speciesThe superscript a, b, and c are the ORDERS with respect to the speciesA, B and C. Chemical reaction orders are generally either 0, 1 or 2 butthere are a few exceptions.The overall order of a chemical reaction is the sum of all the orderswith respect to the individual species.

The units of a rate constant depend on the overall order of the reaction rate law. Asusual, the units of k for rate laws with different overall orders can be worked out byrecognising that the units of the LHS of an equation must equal the units of the RHS.This is shown in the table below for a general rate law of the form

Rate = k [X]n

Overall order n Units of rate Units of k Units of [X]n

0 M s-1

M s-1

No units1 M s

-1s

-1M

2 M s-1

M-1

s-1

M2

3 M s-1 M-2 s-1 M3n M s

-1M

-(n-1)s

-1M

n

Note that you can only compare rate constants with the same overall order and hencethe same units. It is obviously nonsense to say something like the first-order rate

constant of 10 s

-1

is larger than the second-order rate constant of 3 M

-1

s

-1

. This islike saying that 10 kg is larger than 2 m.

How to derive the equation for concentration versus time from the rate law

A rate law is a differential equation since it involves the rate which is a differentialquantity. We can obtain the equation for how the reactant concentration depends ontime (which is normally what is measured in a kinetics experiment) by integration ofthe rate law. We shall show this for zero, first and second-order rate laws.

Zero order rate law

For a reaction A products

For which the rate law is measured to be

-d[A]/dt = k [A]0

= k

since any quantity raised to the power zero is simply equal to 1. We proceed by re-arranging the equation so that all concentration terms are on the LHS and everythingelse is on the RHS. Both sides are then integrated (see sheet list all the simple


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integrals which you need to know) and the constant of integration is added to theRHS.

[ ] = kdtAd [A] = -kt + constant of integration

We find the constant by noting that [A] = [A]o (the initial concentration of A) when t= 0. Hence the constant of integration is equal to [A]o. The final integrated rateequation is then:

[A] = [A]o kt

We can see from this result that, if the kinetics are zero order, a plot of [A] (asordinate versus t as abscissa will be linear with slope equal to k and intercept on theordinate equal to [A]o.

k/M s-1

[A]o/M

0.002 0.3

ZERO ORDER KINETICS

time/s [A]/M

0 0.3

10 0.28

20 0.26

30 0.24

40 0.22

50 0.2

60 0.18

70 0.1680 0.14

90 0.12

100 0.1

110 0.08

120 0.06

130 0.04

140 0.02

150 0

160 -0.02

0

0.1

0.2

0.3

0 50 100 150time/s

concentration/M

First order rate law

For a reaction A products


-d[A]/dt = k [A]1

= k [A]


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We again proceed by re-arranging the equation so that all concentration terms are onthe LHS and everything else is on the RHS. Both sides are then integrated (see sheetlist all the simple integrals which you need to know) and the constant of integration isadded to the RHS.

[ ]

[ ] =

kdtA

Ad

ln[A] = -kt + constant of integration

We find the constant by noting that [A] = [A]o (the initial concentration of A) when t= 0. Hence the constant of integration is equal to ln[A]o. The final integrated rateequation is then:

ln[A] = ln[A]o kt

which can also be written as [A] = [A]oexp(-kt)

We can see from this result that if the kinetics are first order, a plot of ln[A] (asordinate versus t as abscissa will be linear with slope equal to k and intercept on theordinate equal to ln[A]o.

k/s-1

[A]o/M

0.03 0.3

FIRST ORDER KINETICS

time/s [A]/M [B]/M ln([A]/M)

0 0.3 0 -1.203972804

10 0.22225 0.077754534 -1.503972804

20 0.16464 0.135356509 -1.803972804

30 0.12197 0.178029102 -2.103972804

40 0.09036 0.209641736 -2.403972804

50 0.06694 0.233060952 -2.703972804

60 0.04959 0.250410334 -3.003972804

70 0.03674 0.263263072 -3.303972804

80 0.02722 0.272784614 -3.603972804

90 0.02016 0.279838346 -3.903972804

100 0.01494 0.285063879 -4.203972804

110 0.01106 0.28893505 -4.503972804

120 0.0082 0.291802883 -4.803972804

130 0.00607 0.293927427 -5.103972804

140 0.0045 0.295501327 -5.403972804

150 0.00333 0.296667301 -5.703972804

160 0.00247 0.297531076 -6.003972804

0

0.1

0.2

0.3

0 50 100 150time/s

concentration/M

[A]

[B]

y = -0.03000x - 1.20397

-7

-6

-5

-4

-3

-2

-1

0

0 50 100 150

time/s

ln(concentration/M)

It is a particular feature of first-order reactions that the half-life (equal to the timetaken for the concentration to halve) is the same which ever concentration you startfrom (see graph below). It takes exactly the same time to go from 1 M to 0.5 M as itdoes to go from 0.0001 M to 0.00005 M. For a first-order reaction the half-life isequal to (ln2/k) which you can see does not depend on the concentration. For allother order reactions, the half-life does depend on the initial concentration. Try


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drawing in the 1st, 2

nd, 3

rd, etc. half-lives on the zero- and second-order concentration

versus time plots to demonstrate that this is the case.

0

0.1

0.2

0.3

0 50 100 150time/s

concentration/

You can see from this graph showing the 1st

, 2nd

, 3rd

, etc. half-lives for a first-orderreaction that there is no time when the reaction is complete. Each new half -lifeonly halves the concentration remaining. Zero concentration is never reached.

There are two types of second order rate law

Type 1. For a reaction A products


-d[A]/dt = k [A]2

We again proceed by re-arranging the equation so that all concentration terms are onthe LHS and everything else is on the RHS. Both sides are then integrated (see sheetlist all the simple integrals which you need to know) and the constant of integration isadded to the RHS.

[ ][ ]

= kdtA

Ad2

1/[A] = kt + constant of integration

We find the constant by noting that [A] = [A]o (the initial concentration of A) when t

= 0. Hence the constant of integration is equal to 1/[A]o. The final integrated rateequation is then:

1/[A] = 1/[A]o + kt

We can see from this result that if the kinetics are first order, a plot of 1/[A] (asordinate versus t as abscissa will be linear with slope equal to k and intercept on theordinate equal to 1/[A]o.


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k/M-1

s-1

[A]o/M

0.3 0.3

SECOND ORDER KINETICS

time/s [A]/M (1/[A])/M-1

0 0.3 3.333333333

10 0.15789 6.33333333320 0.10714 9.333333333

30 0.08108 12.33333333

40 0.06522 15.33333333

50 0.05455 18.33333333

60 0.04688 21.33333333

70 0.0411 24.33333333

80 0.03659 27.33333333

90 0.03297 30.33333333

100 0.03 33.33333333

110 0.02752 36.33333333

120 0.02542 39.33333333

130 0.02362 42.33333333

140 0.02206 45.33333333

150 0.02069 48.33333333

160 0.01948 51.33333333

0

0.1

0.2

0.3

0 50 100 150time/s

co

ncentration/M

y = 0.30000x + 3.33333

0

10

20

30

40

50

60

0 50 100 150

time/s

(1/conce

ntration)/M-1

Type 2. For a reaction A +B products


-d[A]/dt = k [A]1

[B]1

This equation is harder to integrate and we will simply quote the final answer. (Seefor example P.W. Atkins, 7th edition, page 875 for the maths)

[ ] [ ][ ][ ][ ][ ]

=

0

0

00

ln1

BA

AB

ABkt

This type of second-order reaction (overall second order but first order with respect totwo different reactants) is the most common type.

What does a 2nd

order rate constant tell us? For a 2nd

order reaction

A + A products

which is second order in A, then the half-life t1/2 (which does depend on the initialconcentration) is

[ ]02/1

1

Akt =


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The first half-life (i.e. the time to go from [A]o to [A]o/2) depends on both the rate

constant k and the initial concentration [A]o. The table below shows some values

which help to relate k, concentration and t1/2 which provides a measure of the time

taken by a reaction to reach a know amount of conversion to products.

k/M-1 s-1 [A]o/M Half-life/s Comments1 1 1

1 10-3

1000 About quarter of an hour

10-6 1000000 About a fortnight

10-6

10-3

1000000000 About 30 years, too slow to measure kinetics as

research projects usually only last 3 years!

103

10-3

1

106

10-3

10-3

1010 10-3 10-7 Fastest possible (diffusion controlled) value of k

in water at room temperature. The half-life is

100 ns. For comparison, light travels

approximately 100 ft in 100 ns

Reactions cannot go infinitely fast, i.e there is a maximum value of the rate constant k.

For 2nd

order reactions involving two solute molecules reacting together in a solvent,

the maximum rate is controlled by the speed at which the molecules can diffuse

together. For water as solvent at room temperature, the maximum, diffusion

controlled rate constant is approximately 1010

M-1

s-1

. There are also maximum

values for different types of reactions. As you can see from the table, you need to be

able to measure reaction kinetics in ns in order to measure the fastest possible 2nd

order rate constants with mM concentrations. See your Physical Chemistry text for a

description of some of the special fast techniques used to do this.

Reactions with a significant backwards rate which approach equilibrium

So far, we have considered reactions which proceed virtually to completion. What

happens if a reaction proceeds to an equilibrium position corresponding to only partial

conversion to products? We have to consider both the forward and reverse rates as

seen for the example reaction below.

A + B C

with forward rate constant kf(2nd

order) and reverse rate constant kr(1st

order). The

rate law now has to consider both the forward and reverse rates which depend on both

the reactant and product concentrations as show below.

rate = -d[A]/dt = -d[B]/dt = +d[C]/dt = {kf[A] [B] kr[C]}

This rate law can be integrated in the usual way but is more involved (consult a more

text for more details). As equilibrium is reached, the forward rate (= kf[A] [B])

decreases as the concentrations of A and B decrease whereas the reverse rate (= kr

[C]) increases as the product concentration increases. Equilibrium is reached when


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the overall rate is zero which happens when the forward rate is balanced by the

reverse rate. Reactions do not stop at equilibrium; they simply proceed equally fast in

both the forward and reverse directions.

At equilibrium rate = 0 ={kf[A] [B] kr[C]}

Hence kf[A] [B] = kr[C]

Re-arranging gives KBA

C

k

k

r

f ==]][[

][(K is the equilibrium constant)

Show for yourself that this analysis works for reactions with any numbers of reactant

and product species. For any reaction, the equilibrium constant is always equal to the

forward rate constant divided by the reverse rate constant.

Effect of temperature on reaction rate constants

Svante Arrhenius first described the relationship between k (any order) and absolutetemperature T. It can be written in two equivalent ways as shown below.

lnk = lnA Ea/RT

or k = A exp(-Ea/RT)

where A is a constant called the pre-exponential factor

Ea is the activation energy equal to the difference in enthalpy between

the reactants and the highest enthalpy point of the reaction profile.

R is the gas constant

Note that the Arrhenius equation has the same form as the vant Hoff equationdescribing how an equilibrium constant varies with temperature. This can be seen by

comparing your notes for K with those below. If we measure a series of k values at

different absolute temperatures, then a plot of lnk as the Y axis (called the ordinate)

versus (1/T) as the X axis (called the abscissa) will give a straight line graph with

Graph slope = -(a/R)Intercept on the Y axis = lnA


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temperature/o

C k/s-1

T/K (1/T)/K-1

ln(k/s-1

)

0 0.5 273.15 0.003661 -0.693147

10 1.1 283.15 0.003532 0.09531

20 2 293.15 0.003411 0.693147

30 3.9 303.15 0.003299 1.360977

40 8.4 313.15 0.003193 2.128232

50 17 323.15 0.003095 2.833213

y = -6144x + 21.745

-1

0

1

2

3

4

0.003 0.0032 0.0034 0.0036 0.0038

(1/T)/K-1

ln(k/s-1)

From the graph, the slope = -6144 K (N.B. remember to include both the sign and the

units!) a = -slope x R = -(-6144 x 8.314) = +51081 J mol-1

Ea = +51.1 kJ mol-1

As we shall see, the activation energy Ea is always positive. This means that rate

constants (and hence rates at fixed concentrations) always increase with increasing

temperature. As you can see from the table above, an activation energy of about 50 kJ

mol-1

corresponds to the rate constant approximately doubling every 10oC or so. An

activation energy of about 100 kJ mol-1 corresponds to the rate constantapproximately doubling every 5oC or so. An activation energy of zero means that the

rate constant will not vary with temperature.

In many exam questions, you will not be asked to plot a graph. Instead you will be

given only two values of k (lets call them k1 and k2) at two values of T (T1 and T2).

In the following example, we will use two values from the data table and graph shown

above. Obviously, we should get roughly the same final answers (not exactly since

the example graph data has some scatter due to uncertainties and taking the best

slope gives an averaging effect).

We choose k1 = 0.5 at T1 = 273.15

and k2 = 17 at T1 = 323.15

We apply our basic equation to both sets of values as follows

ART

Ek a lnln

1

1 +=


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ART

Ek a lnln

2

2 +=

Looking at these two equations, we can see that we can get rid of lnA (a constant in

both equations) by subtracting one equation from the other.

21

21 lnlnRT

E

RT

Ekk aa +=

Some re-arranging gives

=

212

1 11lnTTR

E

k

k a

Put in the numbers

=

15.3231

15.2731

314.8175.0ln aE

Hence, Ea = 51.8 kJ mol-1

(as expected, this agrees roughly with our previous answer)

As you can see, the procedure to use the Arrhenius equation is obviously the same asfor the vant Hoff equation describing the variation of K with T.

Meaning of Ea and A in the Arrhenius equation

For almost all chemical reactions, some of the bonds of the reactants must be fully orpartially broken before they can form the new bonds of the products. This bondbreaking costs enthalpy which therefore goes up. The subsequent formation of newbonds releases enthalpy. Hence, as seen below, the enthalpy first increases and thendecreases. The difference in enthalpy between the maximum point (this is called thetransition state of the reaction) and the reactants is the activation energy.

0

0reaction co-ordinate

reaction

enthalp

reactants

products

Transition state

Activationenergy


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How big do we expect activation energies to be? From the diagram above, we can seethat Ea is the energy needed topartially break the chemical bonds which must bebroken to form the reaction products. We know that the energy need to fully break amole of chemical bonds (the usual bond energy) is generally in the range 200-500 kJmol

-1(The C-C bond energy is 440 kJ mol

-1). Because reactive bonds will be

relatively weak and we only need to partially break them before the new bonds start to

form, activation energies of 0-200 kJ mol-1

seem intuitively reasonable.

Molecules have a distribution of thermal energies around the average energy equal toRT for one mole of molecules at absolute temperature T. At any time during areaction, a particular molecule (or set of molecules) will only react if their thermalenergy from a collision with neighbours is equal to or greater than the energy barrierto the reaction which is the activation energy. As derived by Boltzmann, the fractionof molecules having an energy grater than or equal to E a is

fraction = exp(-Ea/RT)

We can recognise this equation as the exponential part of the Arrhenius equation. It

explains why reactions are slow when either the temperature is low or the activationenergy is high. Under these conditions, only a very small fraction of the molecularcollisions going on all continuously have enough energy to react. Hence we have towait a long time before a rare collision with enough energy for reaction occurs. Thetable below shows how this fraction changes with Ea and T.

Ea/kJ mol-

1

T/K Fraction of molecules with

energy Ea or greater

Successful collisions

resulting in reaction

50 300 (room temp) 2.0 x 10-9

1 in 500,000,00050 1000 2.4 x 10-3 1 in 40050 10000 5.5 x 10

-11 in 2

0 300 1 All of them20 300 3.3 x 10-4

1 in 300050 300 2.0 x 10

-91 in 500,000,000

100 300 3.9 x 10-18

1 in 2.6 x 1017

200 300 21.5 x 10

-351 in 6.7 x 10

34

For typical temperatures and activation energies, the fraction of collisions which issuccessful is rather small. Obviously, when the activation energy is zero, allcollisions are successful. This can be the case for reactions of free radicals for whichno bonds have to be broken before news bonds can form. The fraction of collisionswhich are successful can also reach unity when the absolute temperature is infinitelylarge. Inspection of the Arrhenius equation shows that, when eitherEa = 0 or T = ,

then the exponential part becomes equal to 1 and the rate constant k becomes equal tothe pre-exponential factor A. This then tells us the meaning of A in the Arrheniusequation. A is what the rate constant would be if all collisions had enough energy toreact, i.e. ifeitherEa = 0 or T = .

Activation energies for reversible reactions

We saw earlier that measuring the forward and reverse rate constants for a reversiblereaction enables us to work out the equilibrium constant. In a similar way, measuring


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the activation energies for both the forward and reverse reactions of a reversible

reaction enables us to calculate the equilibrium value ofH for the reaction. This canbe seen in the diagram below.

0


reaction

enthalp

Catalysis

Catalysis is the process whereby a reaction rate is increased by a catalyst. Catalysts :

are not consumed in the reaction increase the rates of both the forward and reverse reactions do not change equilibrium constants

Catalysts work by binding with (and hence lowering the energy of) reactionintermediates including the transition state. Because the energy of the transition stateis this reduced relative to the original (unbound) reactants or products, the activationenergy is reduced and the reaction rate is increased for both the forward and reverse.

The catalyst does alter the energies of the unbound reactant and products and hencethe overall equilibrium constant of the reaction is not affected by the catalyst. This isshown schematically in the energy diagram below.

0


reaction

energy

uncatalysed

catalysed

Ea (forward)

Ea (reverse)

E uilibrium H


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Spectroscopy

We will cover here only the basic concepts of spectroscopy. Fundamentally,spectroscopy tells us about the amount of light absorbed and the energy leveldifferences in atoms or molecules. This information can be used to identify unknownmolecules, estimate concentrations and determine molecular properties such as bond

lengths and bond strengths with very good precision.

There are many different forms of spectroscopy but most spectrometers have thefollowing common features.

The light source produces a range of wavelengths, e.g. a normal light bulb emitting inthe visible region emits wavelengths from about 400-700 nm. The monochromator

enables the selection of light of a single wavelength which is then incident onto thesample. Within the sample, some fraction of the light might be absorbed by thesample whereas the remainder is transmitted through the sample and detected by thedetector. In general, light is absorbed by the sample if the difference between twoenergy levels in the sample matches the energy of the light (which corresponds to thewavelength, frequency or colour of the light). The transmitted light intensity ismeasured as a function of the light energy (expressed appropriately) which produces a

spectrum of the sample. Spectra can expressed in various ways for both the ordinate(which might be expressed as transmitted light intensity or absorbance) and theabscissa (which might be wavelength, frequency, energy or wavenumbers). Thediagram below shows a schematic spectrum as a plot of transmitted light intensityversus light energy. The light energy values when the transmitted intensity dipsbelow the light intensity which is incident on the sample (= I0, shown as thehorizontal dashed line) correspond to the absorption bands of the sample.

light energy

transmitted

lightintensit

Light sourceemitting a rangeof wavelengths

Monochromatorto select a singlewavelength

Sample Lightdetector


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The different forms of spectroscopy differ mainly in the type of light used which, inturn, controls the energy range covered in the spectrum and the type of molecularprocesses which are induced by adsorption of light of that energy.

Electromagnetic radiation consists of a wave of oscillating electric and magnetic

fields. The wavelength (greek letter lambda) and the frequency (greek letter nu)

are related by

= c/

where = wavelength (units m)c = speed of light (units m s

-1)

= frequency (units s-1 which are the same as Hz)

Other properties of electromagnetic radiation which it is useful to consider include the

following. Theperiodof the wave is equal to 1/ and is the time for one completeoscillation. The wavenumber of a wave is simply the number of waves within a unit

length and is equal to 1/

. Some spectroscopists use wavenumber as the light energyscale in spectroscopy because the photon energy is proportional to wavenumber (havea go at proving this for yourself).

The energy of a single photon = h where h is Plancks constant (h = 6.6 x 10-34 J s).In the context of chemistry, it is useful to consider the energy of one mole of photons

which is (obviously) equal to Nh where N is Avogadros number. We can thencompare this energy with other chemical energies that we know such as bond energies(e.g. 440 kJ mol

-1for C-C bonds) and thermal kinetic energy (approximately equal to

2.5 kJ mol-1 at room temperature. The table below shows the key properties ofdifferent forms of light.

c/ m s-1

h/J s N/mol-1

3.00E+08 6.63E-34 6.02E+23

type of light wavelength/m frequency/s-1

period/s wavenumber/m-1

energy of 1 photon/J energy of 1 mol of photons/kJ mol-1

molecular process

X ray 1.00E-12 3.00E+20 3.34E-21 1.00E+12 1.99E-13 1.20E+08 nuclear excitation

X ray 1.00E-11 3.00E+19 3.34E-20 1.00E+11 1.99E-14 1.20E+07 excitation of inner shell electrons



vac uum UV 1. 00 E-08 3. 00 E+ 16 3. 34 E-17 1 .0 0E +0 8 1. 99E -17 1. 20 E+ 04 e xci tat ion of o ut er she ll ( bond ing) el ec tron s

vac uum UV 1. 00 E-07 3. 00 E+ 15 3. 34 E-16 1 .0 0E +0 7 1. 99E -18 1. 20 E+ 03 e xci tat ion of o ut er she ll ( bond ing) el ec tron s

UV/vi s 1.00E-06 3.00E+14 3.34E-15 1.00E+06 1.99E-19 1.20E+02 exci tat ion of outer shell (bondi ng) electrons

infrared 1.00E-05 3 .00E+13 3.34E-14 1.00E+05 1.99E-20 1.20E+01 molecular vibration

far infrared 1.00E-04 3.00E+12 3.34E-13 1.00E+04 1.99E-21 1.20E+00 molecular vibration

microwave 1.00E-03 3.00E+11 3.34E-12 1.00E+03 1.99E-22 1.20E-01 molecular rotation



radio 1.00E+00 3.00E+08 3.34E-09 1.00E+00 1.99E-25 1.20E-04 nuclear spi n exci tat ion in a magnet ic fi eld

radio 1.00E+01 3.00E+07 3.34E-08 1. 00E- 01 1.99E-26 1.20E-05 nuclear spi n exci tat ion in a magnet ic fi eld



Considering this table, we first discuss the energies of 1 mole of the differentphotons. We can see that chemical bond energies (100s of kJ mol-1

)correspond to UV radiation. It then follows that UV light is used to do

photochemistry in which light is used to break chemical bonds.

Higher energy X ray photons excite inner orbital electrons (inner electrons arecloser to the nucleus and so require a higher energy photon to kick then out)and are therefore used to identify atoms. X ray spectroscopy does not tell usabout outermost filled (bonding) orbitals.


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The thermal energy of molecules at room temperature (approx. 2.5 kJ mol-1)corresponds to the far infrared. We can then see that thermal energy is highenough to enable molecular processes of moving, rotation and vibration. It isnot enough to excite electrons.

Radiowaves are very low energy photons. They have enough energy to flipspinning nuclei (which behave like tiny magnets) in an external magnetic

field. This is the basis for nuclear magnetic resonance (NMR). The relativephoton energies of X Rays and radiowaves are one reason which NMRscanners are gaining in popular use in hospitals. When you are X rayed inhospital, high energy X ray photons smash through your molecules creatingreactive free radicals which can attack your DNA and cause a variety ofdiseases. In contrast, having an NMR scan exposes you to very low energyphotons which cause no permanent molecular changes and is safe.

In the table, it is also interesting to look down the column showing the periodof the wave. When absorption occurs, the light period corresponds to thetimescale of the molecular process occurring as a result of light absorption.For example, microwave absorption in liquid water tells us that a watermolecule takes about 8 ps to rotate in liquid water at room temperature.Electronic transitions of outer shell electrons take some fs.

Consult your physical chemistry textbook to learn more details of spectroscopy.Check that you can use the equations above correctly by calculating a few of thedifferent entries in the table for yourself.

Absorbance and concentration the Beer Lambert Law

We know from everyday experience that the amount of light absorbed by a solution ofan absorbing solute depends on its concentration. Using measurements of the lightabsorbed to determine concentration is a very useful aspect of spectroscopy,particularly UV/visible spectroscopy.

We consider the absorption of monochromatic (= single wavelength) light by a solute

of concentration c contained within a cell of path length l.

The concentration of the absorbing species is proportional to the absorbance A (NOTto the transmitted light intensity). The absorbance A is defined in terms of themeasured incident and transmitted light intensities as

Incident lightintensity = I0

Transmitted lightintensity = I

Sample path length = l

The light intensity decreases exponentiallyfrom I0 to I as the light beam passes

through the sample


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A = log10(I0/I)

Absorbance has no units. Use the table below to understand what the differentabsorbance values mean.

absorbance % of incident light absorbed % of incident light transmitted

0 0 1001 90 102 99 13 99.9 0.14 99.99 0.01

For particular wavelength of light and a particular absorbing solute, the absorbance isproportional to both the sample path length and the solute concentration. This isshown in the Beer-Lambert Law.

A = log10(I0/I) = c l

where A is the absorbance measured in a spectrometer (unitless)

is the extinction coefficient for the particular solute at thatwavelength. Units = M-1 cm-1c is the solute concentration in M unitslis the sample path length in cm units.

Question. The extinction coefficient of a protein is 2.53 x 106

M-1

cm-1

at awavelength of 280 nm. The absorbance at 280 nm of an unknown solution of theprotein is 1.342 in a 5 mm path length cell. What is the protein solutionconcentration?

Answer. We know that A = log10(I0/I) = c l

Hence c = A/ l = 1.342/(2.53 x 106 x 0.5)Note that we have converted the path length units from mm to cm so that they cancelwith the cm units in the extinction coefficient. To avoid mistakes in yourcalculations, always check when you enter numbers in an equation that the units ofthe LHS and RHS match as they should.

Hence Protein concentration = 1.061 x 10-6 M


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Basic Physical Chemistry. (Part of module 06512 Foundations of Chemistry 2)

Practice questions

General

1. Calculate the following unit conversions

volume = 1 cm3

= ? m3

volume = 1 cm3 = ? nm3density = 1 g cm-3 = ? kg m-3(pressure x volume) is an energy. 1 dm

3atm = ? J

Inter-molecular forces

2. Estimate the average separation between water molecules in liquid water.Hint. Imagine each water molecule exactly fills a cube. Take the side lengthof the cube to be approximately equal to the average separation.[Avogadros number = 6.02 x 10

23mol

-1, RMM of water = 18 g, density of

water = 1.00 g cm-3]

3. For each of the molecules below, state whether they have a non-zeropermanent dipole moment and whether their polarisability will be low or high

benzenenitrobenzenecyclohexanechlorocyclohexane

para-dichlorobenzeneortho-dichlorobenzene

Ideal gases

4. Re-arrange the ideal gas equation to obtain an equation for the concentrationof an ideal gas. Use your equation to calculate the concentration in units of Mof an ideal gas at 298 K and pressure of 560 Pa.[Gas constant R = 8.314 J mol

-1K

-1]

5. Calculate the pressure when 0.5 g of nitrogen and 1 g of oxygen are sealed in acontainer of volume equal to 10 litres at 25

oC.

[Gas constant R = 8.314 J mol-1

K-1]

Equilibrium constants and thermodynamics

6. A reaction A B is found to have an equilibrium yield of 50% at 25oC and75% at 42oC. Calculate Ho for the reaction.[Gas constant R = 8.314 J mol

-1K

-1] Hint. First calculate K at each

temperature and then use the vant Hoff equation. Before you start

calculating, estimate first whether you expectHo

to be positive or negative.


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7. A reaction has an equilibrium constant equal to 1.7 x 104

at 30oC and 1.3 x 10

3

at 70oC. Calculate Ho and So for the reaction. Is the reaction driven by

enthalpy or entropy?[Gas constant R = 8.314 J mol-1 K-1]

Reaction kinetics

8. The rate of a reaction is measured at different concentrations of one of thereactants (species A) whilst keeping all other concentrations and thetemperature constant. The rate is 0.0012 M s

-1when [A] = 0.15 M and

0.00037 when [A] = 0.083 M. What is the order of the reaction with respect tothe species A?

9. A first order reaction has a rate constant of 0.23 s-1

. Calculate the half-life.The reaction is started with and initial reactant concentration of 0.1 M.Calculate the concentration remaining after 25 s.

10. A reaction has a rate constant of 7 M-1 s-1

at 25oC and an activation energy of

72 kJ mol-1. What is the overall order of the reaction? Calculate thetemperature required to increase the rate constant to 43 M

-1s

-1.

[Gas constant R = 8.314 J mol-1

Kamalkant Chem for All Bbsr 1 (14)

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