ENERGETICS

5.1 Exothermic and endothermic reactions

If a reaction produces heat (increases the temperature of the surroundings) it is exothermic. If the temperature of the reaction mixture decreases (ie heat is absorbed) then the reaction is endothermic.

Exothermic -> a reaction which produces heat (ΔH has a negative value by convention, -ve)

Endothermic -> a reaction which absorbs heat (ΔH has a positive value by convention, +ve)

Enthalpy of reaction: The change in internal (chemical) energy (H) in a reaction = ΔH.

The most stable state is where all energy has been released. When going to a more stable state, energy will be released, and when going to a less stable state, energy will be gained (from the surroundings). On an enthalpy level diagram, higher positions will be less stable (with more internal energy) therefore, if the product is lower, heat is released (more stable, ΔH is -ve) but if it is higher, heat is gained (less stable, ΔH is +ve).

Formation of bonds causes an energy release (exothermic). Breaking of bonds requires energy (endothermic).

5.2 Calculation of enthalpy changes

Change in energy = mass x specific heat capacity x change in temperature --> E = m x c x ΔT

Enthalpy changes (ΔH) are related to the number of mols in the reaction...if all the coefficients are doubled, then the value of ΔH will be doubled.

When a reaction is carried out in aqueous solution, the water will gain or lose heat from (or to) the reactants. Therefore, the change in energy, and so the ΔH value, can be calculated from E = m x c x ΔT where m is the mass of water present (kilograms), and c = 4.18 kJ Kg-1 K-1. The ΔH value can then be calculated back to find the molar enthalpy change for the reaction.

Experimental

A known mass of solution should be placed in a container, as insulated as possible, to prevent as much heat as possible from escaping. The temperature is measured continuously, the value used in the equation is the maximum change in temperature from the initial reading.

The result will be a change in temperature. This can be converted into a change in heat (or energy) by using the above equation E = m x c x ΔT.

Δ-H may then be calculated for the amount of reactants present, and then this can be used to calculate for a given number of mols.

http://www.ibchem.com/IB/ibfiles/energy/ene_swf/calorimetry.swf

5.3 Hess' Law

Hess' Law states that the total enthalpy change between given reactants and products is that same regardless of any intermediate steps (or the reaction pathway).

Any equations can be mathematically manipulated using the four rules of number to construct other equations.

Calculate the enthalpy change of a reaction which is the sum of two or more reactions with known enthalpy changes. Use examples of simple two- and three- step processes. Students should be able to construct simple enthalpy cycles, but will not be required to state Hess' law.

Example 1

Monoclinic sulphur is formed in volcanic regions by reaction between sulphur dioxide and hydrogen sulphide according to the equation:

SO2 + 2H2S --> 2H2O + 3S

Draw an enthalpy diagram or cycle and calculate the standard enthalpy change for this reaction.

Here are some values.

Standard enthalpy of formation (all in kJ)

H2O(l) -286 H2S(g) - 20.2

Standard enthalpy of combustion of S (monoclinic) = -297.2kJ

equation 1: H2 + 1/2O2 --> H2O     -286

equation 2: H2 + S --> H2S     -20.2

equation 3: S + O2 --> SO2     -297.2

multiply E2 by 22H2 + 22 --> 2H2S     -40.4add E3equation 4: 2H2 + 3S + O2 --> SO2 + 2H2S     -337.6multiply E1 by 2equation 5: 2H2 + O2 --> 2H2O     -572 kJsubtract 4 from 5- 3S ---> 2H2O - SO2 - 2H2S     -234.4 kJrearrangeSO2 + 2H2S --> 2H2O + 3S     -234.4 kJ 

Example 2:

Calculate the enthalpy change in the reaction:

2KHCO3 (s) --------> K2CO3 (s) + CO2 (g) + H2O (l)

Hess' law by manipulation of the equations...

2KHCO3 --> K2CO3 + CO2 + H2O

2K + H2 + 2C + 3O2 --> 2KHCO3 ΔH = 2(-959) = -19182K + C + 3/2O2 --> K2CO3 ΔH = -1146SUBTRACTH2 + C + 3/2O2 --> 2KHCO3 - K2CO3 ΔH = -772H2 + 1/2O2 ---> H2O ΔH = -286SUBTRACTC + O2 --> 2KHCO3 - K2CO3 - H20 ΔH = -486C + O2 --> CO2 ΔH = -394SUBTRACTZERO ---> 2KHCO3 - K2CO3 - H20 -CO2 ΔH = -92REARRANGEK2CO3 + H20 + CO2 --> 2KHCO3 ΔH = -92THEREFORE2KHCO3 --> K2CO3 + H20 + CO2 ΔH = +92

As all the elements in standard states are assigned an energy value of zero, then to break apart the compound on the left hand side is the opposite of the enthalpy of formation (must be broken apart) Formation of the products molecules is the actual value of the enthalpies of formation.This makes the enthalpy change

ΔH = sum  ΔHf (Products) - sumΔHf (Reactants)note that we use the actual values of the sum ΔHf (Products) and the negative of the sumΔHf (Reactants)does that make it clearer?you can think of any reactant as having to be broken down into its elements which then must be reformed as the products...Hence sum of the ΔHf of the left hand side (reactants) but with the opposite sign (because they're being broken up)

Then they are reformed into the products --> hence sum of the ΔHf of the products but with the correct sign this time

Simply add up the two sets of enthalpy sums to give you the reaction enthalpy- sum ΔHf (Reactants) + sum ΔHf (Products) = Reaction enthalpyof course the books usually just state the rearranged formReaction enthalpy = sum ΔHf (Products) - sum ΔHf (Reactants)

Which does not particularly help in the understanding of WHY this is the case...Don't worry about the exam just remember that you have to break the reactants down to the elements in the first stage (sum the enthalpies of formation and change sign)and then build them back up to the products in the second stage (sum the enthalpies of formation)Then add it all up! Bond breaking = endothermic as you would expect

5.4 Bond enthalpy and bond dissociation enthalpy

Bond dissociation enthalpy is the enthalpy change when one mole of specific bonds are broken.

X-Y(g) -> X(g) + Y(g) : ΔH(dissociation).

Bond enthalpy is the average value of a particular type of bond which has been measured over a range of molecules.

Example:CH4 has four C-H bonds, and so will have four different bond dissociation enthalpies corresponding to the following bonds breaking:CH4 --> CH3 + HCH3 --> CH2 + HCH2 --> CH + HCH   --> C + Hthe C-H bond enthalpy is the average value of the four bond dissociation enthalpies.Bond energies (enthalpies) can be used to calculate unknown enthalpy changes in reactions where only a few bonds are being formed/broken.Bonds broken (left hand side) - bonds formed (right hand side) = enthalpy change for the reaction.(all bond values positive)

15.1 - Standard enthalpy changes of reaction

Definitions

Standard state -- Pressure 101 kPa, temperature 298 K (or 1 atm, 25 degrees celsius). The standard state of an element or compound is the form in which it exists under standard conditions (not to be confused with STP)

Standard enthalpy change of formation -- The enthalpy change when 1 mol of a substance is made from its elements in their standard states

Example: The enthalpy of formation of methane.

C(graphite) + 2H2(g) -> CH4(g)

Hess' Law

The energy difference between two states is independent of the route between them.

This allows calculation of energy changes from other data by the manipulation of chemical equations. Equations may be treated using the four rules of number (doing the same operation on the energy value).

Example: Calculate the enthalpy of formation of methane.

Data:

ΔH combustion (carbon) = -394kj ΔH combustion (hydrogen) =-242kj

ΔHcombustion (methane) =-891kj

ΔHf methane is represented by the equation:

C(s) + 2H2(g)  CH4 (g)

This equation can be constructed using the equations for combustion of the reactants (carbon and hydrogen) and the product (methane)

C(s) + O2(g)   CO2(g) -394kj enthalpy 1

   

H2(g) + 1/2O2(g)   H2O(l) -242kj enthalpy 2

multiply this equation by 2 to get 

   

2H2(g) + O2(g)   2H2O(l) -484kj enthalpy 3

Add the first and third equation together to get: 

C(s) + 2O2(g) + 2H2(g)   CO2(g) + 2H2O(l) -878kj enthalpy (1 + 3) = 4

Now take away the equation for the combustion of methane  

CH4 (g) + 2O2(g)   CO2(g) + 2H2O(l) -891kj enthalpy 5

   

And after rearrangement ( take the CH4 to the right hand side) the result is the equation for the formation of methane

 

C(s) + 2H2(g)  CH4 (g) +13kj enthalpy ( 4 - 5)

     

15.1.1: Define and apply the terms standard state, standard enthalpy change of formation (ΔHfº) and standard enthalpy change of combustion (ΔHcº)

Standard conditions are defined as a temperature of 298K and an atmospheric pressure of 101.3 kPa. The standard state of a substance is its state under standard conditions.

The standard enthalpy of formation is the energy change when 1 mole of a compound is formed from its constituent elements in their standard states. This can be represented by an equation of formation:

The standard enthalpy of formation of sulphuric acid:

H2(g) + S(s) + 2O2(g)   H2SO4(l)  ΔH = -900 kJ mol-1

 

The standard enthalpy of combustion is the energy released when 1 mole of a compound burns in excess air or oxygen. This can be represented by an equation:

The standard enthalpy of combustion of ethene:

CH2=CH2(g) + 3O2(g)   2CO2(g) + 2H2O(l)  ΔH = -900 kJ mol-1

 

15.1.2: Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion.

We can apply Hess law to reactions using the elements as an alternative route from reactants to products. Changing the reactants to their constituent elements at standard state is the reverse of the enthalpy of formation definition. Consequently the energy change is equal to the reverse of the reactants formation enthalpy changes (i.e. with the opposite sign)

Hence, we can say that the enthalpy of reaction is equal to the enthalpy of formation of the products - the enthalpy of formation of the reactants.

Reaction enthalpy = sum of the products formation enthalpies - the sum of the reactants formation enthalpies.

It is also possible when dealing with flammable substances to go from reactants to products via the combustion products:

In this case step 1 (from the reactants to the combustion products) is equal to the enthalpy of combustion of the reactants. However, step 2 is the reverse of the enthalpy of combustion of the products. Hence, to calculate reaction enthalpy we must use the actual reactant combustion enthalpies plus the reverse of the product reaction enthalpies (i.e. the sign must be changed)

Reaction enthalpy = sum of the reactants combustion enthalpies - the sum of the products combustion enthalpies.

15.2 - Born Haber Cycles

Born Haber cycles are the application of Hess' law to ionic systems. An ionic solid consists of a giant structure of ions held together in a giant lattice.

Application of Hess law tells us that the enthalpy of formation of an ionic crystal is equal to the sum of the energies of formation of the ions plus the enthalpy of the lattice. It is a several step process that is best represented by a diagram showing the individual steps as endothermic upwards and exothermic downwards.

http://www.ibchem.com/IB/ibnotes/brief/ene-hl.htm#BH2

15.2.1: Define and apply the terms lattice enthalpy and electron affinity

The enthalpy change when 1 mol of an ionic lattice is formed from its component ions at an infinite distance apart.

M+(g) + X-

(g) -> MX(s)

The value of lattice enthalpy is understood to be negative for the formation of the lattice (an exothermic - bond forming process), and positive for the breaking up of the lattice (an endothermic - bond breaking process).

Hess' law can also be applied to the formation of ionic lattices via a series of steps. In the case of ionic substances this is called a Born Haber cycle.

Lattice enthalpy increases with higher ionic charge and with smaller ionic radius (due to increased force of attraction). The smaller the ion the greater its charge density and the force of electrostatic that it can exert.

Lattice enthalpy may be seen defined in two different ways, depending on the reference literature. For the IB it is defined as the energy released when 1 mole of an ionic substance is formed from its constituent gaseous ions at infinite separation.

Na+(g) + Cl-(g)  NaCl(s)   ΔH = -ve

http://www.ibchem.com/IB/ibnotes/full/ene_htm/15.2.htm

15.2.2: Explain how the relative sizes and the charges of the ions on the lattice enthalpies affect the lattice enthalpies of different ionic compounds. The relative value of the theoretical lattice enthalpy increases with higher ionic charge and smaller ionic radius due to increased forces of attraction.

The bond between ions of opposite charge is strongest when the ions are small.

The lattice energies for the alkali metal halides is therefore largest for LiF and smallest for CsI, as shown in the table below.

Lattice Energies of Alkali Metals Halides (kJ/mol)

F- Cl- Br- I-

Li+ 1036 853 807 757

Na+ 923 787 747 704

K+ 821 715 682 649

Rb+ 785 689 660 630

Cs+ 740 659 631 604

The ionic bond should also become stronger as the charge on the ions becomes larger. The data in the table below show that the lattice energies for salts of the OH- and O2- ions increase rapidly as the charge on the ion becomes larger.

Lattice Energies of Salts of the OH- and O2- Ions (kJ/mol)

OH- O2-

Na+ 900 2481

Mg2+ 3006 3791

Al3+ 5627 15916

15.2.3: Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate an enthalpy change.http://www.ibchem.com/IB/ibnotes/full/ene_htm/15.2.htm

 15.2.4: Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character. A significant difference between the two values indicates covalent character.

It is possible to measure lattice enthalpy using thermodynamic principles in the Born Haber cycle. However, it is also possible to calculate theoretical lattice enthalpy based on the electrostatic attractions between ions.

For the theoretical calculation some assumptions must be made:

1. The ions themselves are treated as point charges.

2. The structure is purely ionic

The overall lattice enthalpy is then a function of electrostatic attraction, the lattice structure, the magnitude of the charges on the ions. There are several approaches, the best known being the Madelung equation and the Kaperinsky equation:

Equations here

For highly electropositive metals the correlation between experimentally determined values and theoretically determined values is good. However, in some ionic compounds, such as silver chloride, there is a large difference. This suggests that such compounds are not purely ionic in their structure.

The classical comparison is between the halides of the group 1 metals and silver halides.

compound theoretical lattice enthalpy/ kJ mol-1 experimental lattice enthalpy/ kJ mol-1

sodium chloride 766 771

silver chloride 770 905

The data shows that the assumptions made for the silver chloride lattice are probably not valid and that the structure has some contribution from covalent bonding. This appears to be bonding the stucture together more tightly, as in a giant covalent structure.

Curiously, the relative melting points indicate that the structure of silver chloride is actually weaker than that of sodium chloride. It seems as though the bonding changes its nature as the compound is heated and reverts back to a more ionic or simple covalent model.

15.3 - Entropy

Chemistry is concerned with the statistical likelyhood of a process occurring. This is related to the number of possible states which the particles can adopt. This can be regarded as the degree of disorder or Entropy of the system. Factors which increase disorder in a system are:

Increased number of particles (when more gas particles are produced, this is far more important than the other factors)

Mixing of particles

Change of state to greater distance between particles (solid->liquid or liquid->gas)

Increased particle movement (temperature)

Δ-S is positive when entropy increases (more disorder) and negative when entropy decreases (less disorder).

Calculations can be carried out using absolute entropy values in the same way as enthalpy values in Hess' law. The change in entropy form one state to another will always be the same regardless of the route taken.

The standard entropy change can be calculated by subtracting the absolute entropy of the reactants from that of the products.

ΔS (products) - ΔS (reactants) = standard entropy change for a reaction

15.3.1: State and explain the factors that increase the entropy of a system.

Entropy has been described as nature's arrow. It is the natural tendency of everything towards disorder.

There are two factors that contribute to the degree of disorder of a system.

1. The number of particles

2. The nature of the particles

3. The available energy of the particles

Of these three factors, by far the most important are the number of particles and the available energy (temperature)

Gases have the greatest entropy as their particles are free to move randomly

An increase in disorder can result from the mixing of different types of particles, changes of state (increased distance between particles), increased movement of particles or increased numbers of particles. An increase in the number of particles in the gaseous state usually has a greater influence than any other possible factor.

15.3.2: Predict whether the entropy change (ΔS) for a gives reaction or process is positive or negative.

The entropy of reactants can be gauged by considering the number of moles of gas on the left hand side of a reaction. The entropy of the products can be considered in the same way. To decide whether the entropy increases or decreases we look at whether the number of moles of gas increases or decreases in the course of the reaction:

Example: The Haber process

N2 + 3H2   2NH3

On the left hand side there are 4 moles of gas and on the right hand side there are 2 moles of gas. The moles of gas have decreased from 4 to 2, therefore the total entropy has decreased.

15.3.3: Calculate the standard entropy change of a reaction (ΔSº) using standard entropy values (Sº).

Calculation of standard entropy change is similar to the assessment carried out in the previous section,   only   this   time   useing   the   actual   numbers   given   for   the   absolute   entropy   ogf   each substance.

Example: The Haber process

N2 + 3H2   2NH3

The absolute entropy values are:

nitrogen: 192 J K-1 mol-1

hydrogen: 131 J K-1 mol-1

ammonia: 193 J K-1 mol-1

On the left hand side there is 1 mole of nitrogen and 3 moles of hydrogen, hence the total absolute entropy on the left hand side = 192 + 3(131) = 585 J K-1

On the right hand side there are 2 moles of ammonia, hence the total absolute entropy on the right hand side = 2(193) = 386 J K-1

Hence the entropy has decreased from 585 to 386 = -199 J K-1

As can be seen the operation that is carried out is:

Entropy change = sum of the products entropy - sum of the reactants entropy

15.4 - Spontaneity of a reaction - Gibbs Free Energy

Reactions which release heat (and so increase stability) tend to occur. Reactions which increase entropy (ΔS is positive) tend to occur, but neither can be used to accurately predict spontaneity alone.

Gibbs free energy (G) is defined as a measure of the total entropy of the universe. Hence the change in Gibbs free energy (ΔG) is the change in the total entropy of the universe. The total entropy of the universemust increase for any process to occur.

When heat is released in a reaction (exothermic change) this energy heats up the universe and effectively increases its entropy (there are a greater number of possible energy states that the particles in the universe can adopt).

The total entropy of the universe must increase and consequently exothermic reactions are favourable.

If the entropy of a reaction mixture increases then this is also favourable as the total entropy of the universe also increases.

Gibbs free energy change = ΔH - TΔS

If Gibbs free energy change is negative (convention) then the total entropy of the universe increases and the reaction is spontaneous. Why is the sign negative?

When ΔG is negative, the reaction is spontaneous, when it's positive, the reaction is not.

Gibbs free energy calculations

Enthalpy changes can be calculated indirectly by summing the enthalpy values of related equations using Hess' law. Entropy changes can be calculated in the same way. It follows then that Gibbs free energy changes can be calculated from a knowledge of Gibbs free energy values in related equations.

Spontaneity of reaction

Determined by the relationship

ΔG = ΔH - Temperature(in kelvin) x ΔS

See standard level entropy for a fuller treatment

Enthalpy change Entropy change Gibbs free energy Spontaneity

positive positive depends on T, may be + or - yes, if the temperature is high enough

negative positive always negative always spontaneous

negative negative depends on T, may be + or - yes, if the temperature is low enough

positive negative always positive never spontaneous

       

15.4.1: Predict whether a reaction or process will be spontaneous using the sign of ΔGº

Entropy has been said to be nature's arrow. Overall, everything tends towards increasing disorder, as this is overwhelmingly the most probable state for particles and associated heat energy.

Josiah Willard Gibb realised that for a process to occur the overall entropy of the universe must increase. The universe being made up of the system under study and its surroundings.

Therefore the total entropy in the universe must be equal to the entropy of the system + the surroundings, and any change in the universal entropy must be due to a change in entropy of system, surroundings or both. i.e.

ΔS(universe) = ΔS(system) + ΔS(surroundings)

The surroundings can only be affected by heat exchange with the system. For an exothermic reaction the surroundings receives energy that increases its disorder (entropy) by a factor of q/T where q (in terms of the system) must be ΔH/T

Hence universal entropy can be written:

ΔS(universe) = ΔS(system) + ΔH/T(surroundings)

But when the surroundings increases in energy so the system decreases in energy, therefore in terms of the system:

ΔS(universe) = ΔS(system) - ΔH/T(system)

Multiply through by -T to remove fractions:

-TΔS(universe) = -TΔS(system) + ΔH(system)

and rearrange:

-TΔS(universe) = ΔH(system) -TΔS(system)

Gibbs lumped together the -TΔS(universe) term and called it ΔG, Gibbs free energy.

So when Gibbs free energy is negative the entropy change of the universe must be positive. This can be measured by looking at the enthalpy and entropy change of the system under study:

ΔG = ΔH(system) -TΔS(system)

So when the universal entropy increases (a spontaneous process) then the Free energy decreases.

ΔG is a measure of the free energy of the UNIVERSE, not the system under study. You can think of free energy as that portion of the energy of the universe that is available to do work.

The universal entropy is always increasing and this decreases the available free energy.

Gibbs free energy equation relates the free energy of the universe to the systems enthalpy and entropy change:

ΔG = ΔH - TΔS

When the value of ΔG is negative this means that the universal free energy has decreased and hence the universal entropy has increased. The process is consequently favourable (thermodynamically spontaneous)

It is impossible to increase the free energy of the universe.

So what exactly is free energy?

It has been interpreted as being the energy that could be used to do work. All of this came about with the study of engines. We know that energy tends to flow from high temperature to ambient. It cannot go in reverse. So a region of high energy can be used to do work, such as boiling water to drive a steam engine. But the same energy spread out evenly throughout the universe cannot do any work. All it succeeds in doing is increasing the average heat of the universe by a miniscule fraction.

Energy does not have to exist as heat. It could also be chemical potential energy. In that state it could be transformed into heat energy and do work, before dissipating and being useless.

So the idea of 'Free energy' is energy that is still available to do work. Energy that has been released and dissipated has increased the entropy of the universe and is no longer 'useful'.

Hence an exothermic reaction releases heat energy, this increases universal entropy and decreases the free energy of the universe.

Spontaneity

A reaction is said to be thermodynamically spontaneous (i.e. the reaction can happen) if its free energy change is negative.

There are several possibilities depending on the sign of the enthalpy change and the entropy change for the reaction, remembering that:

ΔG = ΔH(system) -TΔS(system)

Sign of the enthalpy change Sign of the entropy change Spontaneity

positive (+) positive (+) The reaction is spontaneous at high temperature

positive (+) negative (-) The reaction is never spontaneous

negative (-) negative (-) The reaction is spontaneous at low temperature

negative (-) positive (+) The reaction is always spontaneous

In order to find out the temperature at which the reaction just becomes spontaneous, set ΔG = 0 and fill in the values for ΔHand ΔS to find the temperature.

 

15.4.2: Calculate Gibb's free energy change (ΔGº) for a reaction using the equation ΔGº= ΔHº - TΔ Sºand by using values of the standard free energy change of formation, ΔGfº

Gibbs free energy change is calculated using the Gibbs free energy equation:

ΔG = ΔH(system) -TΔS(system)

It is important to remember that the units of enthalpy are often given in kiloJoules whereas the units of entropy are given in Joules per Kelvin. In order to enter values into the Gibbs free energy equation all values must be in either kiloJoules or Joules.

Example: The Haber process

N2 + 3H2   2NH3   ΔH = -92 kJ

The absolute entropy values are:

nitrogen: 192 J K-1 mol-1

hydrogen: 131 J K-1 mol-1

ammonia: 193 J K-1 mol-1

On the left hand side there is 1 mole of nitrogen and 3 moles of hydrogen, hence the total absolute entropy on the left hand side = 192 + 3(131) = 585 J K-1

On the right hand side there are 2 moles of ammonia, hence the total absolute entropy on the right hand side = 2(193) = 386 J K-1

Hence the entropy has decreased from 585 to 386 = -199 J K-1

ΔGº = ΔHº - TΔSº

At 298K

ΔGº = -92 + (298 x 0.199) = -32.7 kJ

 

15.4.3: Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and enthalpy changes and the equation ΔGº= ΔHº- TΔSº

To predict the effect of changing temperature refer to the table of enthalpy and entropy signs on spontaneity.

Sign of the enthalpy change Sign of the entropy change Spontaneity

positive (+) positive (+) The reaction is spontaneous at high temperature

positive (+) negative (-) The reaction is never spontaneous

negative (-) negative (-) The reaction is spontaneous at low temperature

negative (-) positive (+) The reaction is always spontaneous

Clearly,   if   a   reaction   becomes   spontaneous   at   higher   temperatures,   then   any   increase   in temperature will make Gibbs free energy more negative (or less positive). The same logic can be applied to any reaction.

Example: The Haber process

N2 + 3H2   2NH3  ΔH = -92 kJ

The absolute entropy values are:

nitrogen: 192 J K-1 mol-1

hydrogen: 131 J K-1 mol-1

ammonia: 193 J K-1 mol-1

On the left hand side there is 1 mole of nitrogen and 3 moles of hydrogen, hence the total absolute entropy on the left hand side = 192 + 3(131) = 585 J K-1

On the right hand side there are 2 moles of ammonia, hence the total absolute entropy on the right hand side = 2(193) = 386 J K-1

Hence the entropy has decreased from 585 to 386 = -199 J K-1

ΔGº = ΔHº - TΔSº

At 298K

ΔGº = -92 + (298 x 0.199) = -32.7 kJ

As the sign of enthalpy is negative and the sign of the entropy change is negative then the value of - TΔSº becomes more positive as the temperature is increased.

A point will be reached at which the reaction is no longer spontaneous, as increasing temperature makes Gibbs Free energy change more positive and less negative.

When the value of ΔGº = 0 the reaction reaches equilibrium. Any further increase in temperature drives the reaction backwards.

Notes:

Electrostatic attraction

This is the most important force in chemistry. It depends on two factors:

1. The size of the charged particles (eg ionic radius)2. The magnitude of the charge Z1 and Z2

The electrostatic attractive force may be considered proportional to the product of the charge magnitudes divided by the distance between them. Smaller charged particles can approach closer and therefore exert a greater force of attraction.

Examples

NaCl - the charges on the ions are single plus and single minus respectively

MgCl2 - in this case the magnesium has a double positive charge and consequently exerts a much larger force than the sodium ion in sodium chloride. The MgCl2 lattice is much stronger than NaCl- higher mp and greater lattice enthalpy

LiCl - In this case the Li+ ion has the same charge as the Na+ ion but it is much smaller and can get closer to the chloride ion exerting a greater force. The lattice is stronger than NaCl and has a higher mp

Gibbs free energy equation

Why is the sign negative?

This is because of the convention adopted for enthalpy. If the enthalpy is negative it's an exothermic process and the entropy of the universe increases as explained above. The second term in the equation is a negative term (-TΔS) and if the entropy increases it makes Gibbs free energy more negative.