DUNCANRIG SECONDARY ADVANCED HIGHER CHEMISTRY

UNIT ONE

BOOKLET 6

Thermodynamic

s

Can we predict if a reaction will occur? What determines whether a reaction

will be feasible or not? This is a question that can be answered by applying the

principles of chemical thermodynamics, the study of the energy relationships

associated with chemical reactions. The term used to label a reaction that

proceeds without continual energy input is spontaneous. A spontaneous

reaction is one that will occur all by itself, once the activation energy has

been provided so that it can get started. These reactions may be fast or

extremely slow.

The burning of hydrogen, for example, is a spontaneous reaction.

Once you add a little bit of energy,

like the heat from a match,

the hydrogen continues to burn

without any outside help, until

there is no more hydrogen to burn.

It is possible to change water back into hydrogen and oxygen by passing

electricity through the water - the process is definitely NOT spontaneous.

This apparatus will stop producing hydrogen

and oxygen if the electricity supply is

switched off.

Once the external energy supply is removed

a non spontaneous reaction will stop even if

all the reactants have not been used up.

H2

H2

Just like the skier, chemical reactions tend

to be spontaneous if the natural flow of energy

is ”DOWNHILL” and so reach their lowest,

most stable, energy state.

This situation happens in exothermic reactions, where energy is

released to the surrounds. In an exothermic reaction the products have

less stored energy {ENTHALPY, H} than the reactants. The difference in

energy is the energy released to the surroundings and is labelled

Consider the combustion of methane

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

The change in enthalpy can be seen in the following reaction progress diagram

The diagram clearly shows that the products have less enthalpy than the

reactants and are therefore much more stable - a favourable condition for a

reaction to be spontaneous.

The enthalpy change is found using H = Hproducts – Hreactants

In this case : H = (-969) – (-75) = - 894kJ mol-1.

H

The activation energy, Ea, is also shown on this diagram.

In an endothermic the enthalpy of the products is greater than the reactants.

The additional energy comes from the surroundings and so an endothermic

reaction is accompanied by a fall in temperature.

Consider the cracking of ethane

C2H6(g) C2H4(g) + H2(g)

Endothermic reactions occur and so they must be feasible, even although the

products are LESS STABLE than the reactants – this is not a favourable

condition for a spontaneous process.

The enthalpy change for this reaction is given by H = Hproducts – Hreactants

H = (52) – (-85) = 137 kJ mol-1

Both these diagrams illustrate the First Law of Thermodynamics as the total

energy of the chemicals plus the energy released or absorbed from the

surroundings is constant.

H

If H for a reaction is NEGATIVE this is a good indication that a reaction will

be feasible. However, as endothermic reactions {H positive} happen, this

cannot be the only condition to determine whether a reaction will occur.

In its very simplest sense entropy is

related to how ordered something is.

Systems that are highly ordered have

LOW ENTROPY. HIGH ENTROPY

is associated with lots of DISORDER.

The drawings show examples of order and disorder – the idea of entropy.

The diagrams are illustrating that

the natural order of a system is

towards a situation where entropy

increases. It is more probable that

things become disordered rather

than more ordered.

Imagine two boxes with a gap

between them.

One box has a blue gas atom in it,

the other has a red gas atom in it.

What is the chance (probability) of

both atoms ending up in the right

hand box if they are left to diffuse?

Do the same thing with 3 different

atoms, 4 different atoms and finally

with Avogadro’s number of atoms.

Hint: there is a formula

Common sense tells us that gases are much more disordered than liquids and

liquids are more disordered than solids.

It should be fairly obvious that the molecules in steam have a far more random

nature than the molecules in water. Ice is a much more ordered substance

than water.

The change in entropy as ice is heated is shown in the following graph.

Entropy has the symbol (S)

The units for entropy are

usually quoted as

Joules per Kelvin per mole

(J K-1 mol-1)

The graph shows that, in

general, entropy increase with

increasing temperature –

disorder gets larger when

substances are heated because

they obtain more energy the

particles will move around more.

The graph also shows that

entropy increases when the

water changes state from solid

to liquid to gas.

The table shows some standard entropy values, So – the values at 25 oC.

Analysis of the table of entropy values reveals some of the trends in entropy.

1. Entropy increases from solids to liquids to gases.

2. The values for methane, ethane, propane and butane suggest that entropy

increases with increasing molecular size. Why?

The atoms joined together in a molecule move in a variety of ways. The

diagram shows some of these movements.

If there are more atoms, the there will be an increase in the number of

stretches, vibrations and rotations the atoms can make. In other words more

disorder. In general entropy will increase as molecular size increases

As already stated entropy will increase with increasing temperature. This is

due to increasing motion of the particles in the substance. As a substance

cools this motion will slow down and if the temperature is low enough it will

stop altogether.

The lowest temperature possible is known as ABSOLUTE ZERO. It has a value

of -273oc or zero Kelvin, 0 K.

At zero Kelvin the solid on the right will

have no thermal energy and the atoms will

have no motion, not even the smallest

vibration. A crystal structure like this will

be perfectly ordered and therefore have

ZERO ENTROPY.

This is summarised in the 3rd law of thermodynamics.

Use the table of entropy values on page 6 to

help with these questions.

1. Which substance in the table is most

ordered?

2. Why does the increasing trend in

entropy from methane to butane NOT

continue with pentane?

3. Calculate the entropy value for the

following.

a. 3.5 moles of sodium chloride

b. 0.25 mole of butane gas

c. 60 g of diamond

4. Calculate the increase in entropy if

2 moles of water are boiled.

5. Would you expect graphite to have a

higher or lower entropy than diamond.

Explain your answer.

In order for a chemical reaction to be feasible both enthalpy and entropy have

to be taken into consideration.

If a reaction is exothermic, this is an indication that the reaction will be

spontaneous. If the entropy change in a reaction increases when reactants

become products this is also a favourable condition to make a reaction

spontaneous.

Is the entropy change positive or negative in the following processes?

1. Mg(s) + ½O2(g) MgO(s) _______________

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

3. H2O(g) H2O(l) ________________________

4. Na+Cl-(s) Na+(aq) + Cl-(aq) _______________

5. NH4NO3(s) N2O(g) + 2H2O(g) ____________

The fact that both entropy

and enthalpy are involved in

reaction feasibility is

summarised in the Second Law

of Thermodynamics.

This means for a reaction to

be spontaneous there MUST

be an increase in the overall

entropy of the reaction

system.

Consider the burning of magnesium!

As it produces a solid product form a gas the entropy

will decrease – this appears to contradict the 2nd Law.

However, the reaction produces heat and light which

increase the entropy of the material surrounding the

burning magnesium. When considered together the

overall entropy increases – this reaction is therefore

spontaneous.

Now - three calculations.

1. Calculate the three components

of the Free Energy equation.

2. Use the Free Energy equation to

determine reaction feasibility at

a particular temperature

3. Use the Free energy equation to

find the temperature at which a

reaction becomes feasible.

The American Mathematical Physicist J.Willard Gibb’s derived an equation

which brought together the idea that enthalpy and entropy were the driving

forces in chemical reactions.

His equation, called Gibb’s Free Energy is as important to chemists as E=mc2

is to physicists.

This equation decides the feasibility of a chemical reaction. Very simply it

indicates if a reaction will proceed at a particular temperature.

We know that if a reaction is exothermic that this is a favourable sign the

reaction will be spontaneous. If the entropy value of the reaction itself also

increases this is another favourable sign.

Including these facts in the equation means that a NEGATIVE H value and a

POSITIVE S value will give a NEGATIVE value for G. It is this value which

decides if a chemical reaction is FEASIBLE .

The feasibility of a reaction can be predicted from a consideration of the

signs of H and S.

The table shows how free energy is affected by four possible conditions of

enthalpy and entropy.

Consider reaction A – the combustion of methanol. The free energy value of this reaction is

always negative no matter the temperature – the reaction is always spontaneous.

If the combustion of methanol is always spontaneous, why doesn’t it burn at room

temperature?

1. All reactions require their activation energy. At room temperature there is

insufficient energy to start the reaction.

2. Spontaneous means that once started the reaction will continue on its own.

3. Another view is that it is burning – just very slowly – thermodynamics tell us nothing

about the speed of the change just that the change is possible.

Before attempting Free Energy calculations we need to consider three important facts

regarding thermodynamics.

In data tables the values for free energy enthalpy and entropy are always quoted at what is

known as STANDARD CONDITIONS.

This is because the values can vary with temperature, pressure and concentration.

Standard conditions are Temperature of 298 K (25 oC)

Pressure of 1 atmosphere

Concentration of 1 mol l-1

The symbols denoting standard conditions are Go, Ho and So

The standard enthalpy of formation, Hfo is often used to find the overall enthalpy change

for a reaction.

The enthalpy of formation is defined as the enthalpy change when 1 MOLE of a compound is

formed from its elements under standard conditions.

Equations can be written for the enthalpy of formation

E.g Enthalpy of formation of methane is C(s) + 2H2(g) CH4(g)

Enthalpy of formation of water is H2(g) + ½ O2(g) H2O(l)

Notice that state symbols must be used to show the equations complies with the definition

and that the substance formed must be 1 mole.

The enthalpy of formation of any ELEMENT is defined as ZERO.

Given Go, Ho or So values for the chemicals in a reaction it is possible to calculate the

G, H or S for the reaction in which they are involved.

This is done by using the formula;

Where is “the sum of” and Xo is the free energy, enthalpy of entropy of the chemicals.

1. The equation for the reaction of ethyne and hydrogen chloride is

C2H2 + 2HCl CH2ClCH2Cl

The thermochemical data for the substances involved in the reaction is

shown in the table below.

Use the information to calculate both the enthalpy change and the entropy

change for this reaction.

We can use

to calculate the enthalpy change

Ho reaction = Hf° [CH2ClCH2Cl] – {2 Hf° [HCl] + Hf° [C2H2]}

= [-166] – [2(-92.3) + (227)]

= -208.4 kJ mol-1

The entropy change is calculated in exactly the same way..........

So reaction = S° [CH2ClCH2Cl] – {2 S° [HCl] + S° [C2H2]}

= [208] – [2(187) + (201)]

= -367 J K-1 mol-1

Note that as the values are given per mole of substance the stoichiometric

coefficients (balancing numbers) must be used in these calculations.

These results indicate that this reaction is exothermic (H is negative) and

that there has been a decrease in entropy (S is negative).

Compound So/J K-1 mol-1 Hfo /kJ mol-1

C2H2 201 227

HCl 187 –92.3

CH2ClCH2Cl 208 –166

2. The equation for the decomposition of zinc carbonate is

ZnCO3 ZnO + CO2

Calculate the free energy, Go, for this reaction given that Ho is

71 kJ mol-1 and So is 175.1 J K-1 mol-1

Use

Go = 71 – 298 (175.1/1000)

= 71 – 52

= 19 kJ mol-1

In the free energy equation the value for temperature must be in Kelvin.

In this example it appears that no value for the temperature has been

given. However, as the standard state symbols have been used, we can

assume a temperature of 298 K.

Notice that the entropy value has been divided by 1000. This is done to

“match up the units”. The units for enthalpy are given in kilojoules and the

units for entropy are given in Joules. Both must be the same before being

used in the free energy equation.

Not doing this is a very common mistake in exams.

As the free energy value is positive this reaction is not spontaneous at

298 K.

For this reaction to occur the temperature would need to be much higher.

This would increase the entropy value which would eventually overcome

the unfavourable positive enthalpy value.

Anyone trying to change zinc carbonate to zinc oxide at 298 K (room

temperature) would probably have to wait a very long time indeed.

3. Barium carbonate decomposes on heating.

BaCO3(s) BaO(s) + CO2(g) ΔH° = +266 kJ mol–1

a. Using the data from the table below, calculate the standard entropy

change, ΔS°, in J K–1 mol–1, for the reaction.

So reaction = {S° [CO2] + S° [BaO]} - S° [BaCO3]

= [213.8 + 72.1] – [112]

= 173.9 J K-1 mol-1

b. Calculate the temperature at which the decomposition of barium

carbonate just becomes feasible.

To answer this question we use a variation of

the free energy formula.....

This formula relies on the fact that,

at equilibrium, the value of Go is ZERO.

Substituting the values into this expression gives

This temperature

provides the activation

energy to make this

reaction proceed at a

reasonable rate.

1. Use the information in the table to calculate the free energy change in the following

reactions

a. 2Mg + CO2 2MgO + C

b. 2CuO + C CO2 + 2Cu

Explain wether both these reactions are

feasible at 298 K

2. The equation for the decomposition of magnesium carbonate is shown below

a. Use the thermodynamic data to calculate the free energy change, in kJ mol-1 at 400K

for the reaction.

b. Is the reaction feasible at this temperature?

3. Consider the thermodynamic data shown for the Haber process.

Use the data given above, along with data book values to calculate the temperature at

which the Haber process becomes feasible.

4. Chloroform was one of the first anaesthetics used in surgery.

Use the thermodynamic data to calculate a boiling point for chloroform.

CHCl3(l) CHCl3(g)

S = 94.2 J K-1 mol-1

H = 31.3 kJ mol-1

5. Consider the reactions and their thermodynamic data.

a. Why does the reaction between aluminium and oxygen have the largest difference

between Go and Ho?

b. The reaction between hydrogen and chlorine is spontaneous at 298 K.

Explain why there is no observable reaction between hydrogen and chlorine until the

mixture is exposed to ultraviolet light.

c. Why does the data confirm that ammonium chloride dissolves spontaneously in water

and that the temperature of the water will decrease as the ammonium chloride

dissolves?

6. 8.

9.

7.

10. This table contains some thermodynamic data for hydrogen, oxygen and water.

a. Calculate the temperature above which the reaction between hydrogen and oxygen to

form gaseous water is not feasible.

b. State what would happen to a sample of gaseous water that was heated to a

temperature higher than that of your answer to part a.

Give a reason for your answer.

11. The oxides nitrogen monoxide (NO) and nitrogen dioxide (NO2) both contribute to

atmospheric pollution.

The table gives some data for these oxides and for oxygen.

Nitrogen monoxide is formed in internal combustion engines. When nitrogen monoxide comes

into contact with air, it reacts with oxygen to form nitrogen dioxide.

a. Calculate the enthalpy change for this reaction.

b. Calculate the entropy change for this reaction.

c. Calculate the temperature that this reaction becomes thermodynamically feasible.