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ENTHALPY CHANGE H

If the system gives out heat energy during a reaction, enthalpy is lost to the surroundings, therefore the enthalpy change, H, has a negative value. This is an exothermic reaction.

If the system takes in heat energy during a reaction, enthalpy is gained from the surroundings, therefore the enthalpy change, H, has a positive value. This is an endothermic reaction.

The size of an enthalpy change depends on the amount of substance used and on the conditions of measurement. In order to make useful comparisons between different measurements, standard amounts and standard conditions must be defined.

AMOUNT OF SUBSTANCEThe standard amount of substance is the mole. Thus an enthalpy change in a reaction will be presented as, for example:

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

This means that when the molar quantities represented by the balanced equation react, the heat given out by the reaction, at constant pressure, is 92 kJ.

H = - 92 kJ. mol-1Heat change at constant pressure exothermic size of heat change for the molar quantities specified in the eqn.

STANDARD CONDITIONSThe size of the measured enthalpy change for a given quantity of reactant(s) in a given reaction will depend also on the temperature and pressure at which it is measured. Therefore, it is usual to quote enthalpy changes which have been measured under agreed standard conditions.

Changes occurring under standard conditions are referred to as standard enthalpy changes, indicated by the symbol Ho.

STANDARD STATESElements and compounds in their normal, stable state at 298K and 100kPa are said to be in their standard state. The physical state is normally clarified further by the inclusion of a state symbol. However, some elements can exist in allotropic forms.

TOPIC 13.17: THERMODYNAMICS 1

Enthalpy change is the heat energy change in a reaction measured under conditions of constant pressure.

Standard Pressure: The standard pressure chosen is 105 Pa (also called 1 bar).Standard Temperature: The most common reference temperature is 298K.

Allotropes are different forms of the same element capable of existing in the same physical state, e.g. graphite and diamond.

Where allotropes exist, the particular allotrope should be identified in the balanced equation; e.g. C (s, graphite) or C (s, diamond). If the allotrope is not specified, it is assumed to be the more stable form under standard conditions: graphite in the case of carbon.

STANDARD ENTHALPY CHANGESThe definitions of some standard enthalpy changes need to be learned verbatim.

Many enthalpies of formation cannot be determined directly, because the reaction concerned does not go, and they must instead be determined indirectly using calculations based on Hess’s Law.


Standard Enthalpy of Formation, Hof

The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, all reactants and products being in their standard states.

e.g. Na(s) + 1/2Cl2 (g) NaCl(s)

Ca(s) + C(s,graphite) + 3/2O2(g) CaCO3 (s)

By definition, the standard enthalpy of formation of an element must be zero.

Standard Enthalpy of Combustion, Hoc

The standard enthalpy of combustion is the enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, all reactants and products being in their standard states.

e.g. C (s) + O2 (g) CO2 (g) H = -394 kJ.mol-1

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890 kJ.mol-1

Ionisation Enthalpy, HoI

The molar first ionisation energy is the enthalpy change required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous positive ions.

e.g. K(g) K+(g) + e- HI = +418kJ.mol-1

Ionisation Enthalpy, HoI

The molar second ionisation energy is the enthalpy change required to remove one mole of electrons from one mole of gaseous unipositive ions to form one mole of gaseous dipositive ions.

e.g. K+(g) K2+(g) + e- HI = +3070kJ.mol-1

The second ionisation enthalpy is always greater than the first, because the second electron is being removed from a positive ion which is considerably smaller than the atom. This requires more energy than removing an electron from a larger, neutral species.

The first electron affinity is usually exothermic.

The second electron affinity involves adding an electron to a negative ion. Owing to the mutual repulsion of the ion and the electron, the enthalpy change is highly endothermic.

The overall process of adding two electrons to a neutral atom is highly endothermic.O(g) + 2e- O2-(g) H = + 702 kJ.mol-1

The species produced are free radicals and contain an unpaired electron. This can be indicated by placing a dot after the formula, e.g. CH3

The dot is usually omitted in thermodynamic equations.


Electron Affinity, Hoea

The electron affinity is the enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous negative ions.

e.g. Cl(g) + e- Cl-(g) Hea = -364kJ.mol-1

Hea /kJ.mol-1

H(g) - 72F(g) -348Cl(g) -364Br(g) -342I(g) -314O(g) -142O-(g) +844

Bond Dissociation Enthalpy, Hodiss

The bond dissociation enthalpy is the standard molar enthalpy change which accompanies the breaking of a covalent bond in a gaseous molecule to form two gaseous free radicals.

e.g. H2(g) 2H(g) Hdiss = +436kJ.mol-1

CH4(g) CH3(g) + H(g)Hdiss = +435kJ.mol-1

.

Bond Dissociation Enthalpies (kJ.mol-1)

H-H 436 C-C 348N-H 391 O-H 463C-Cl 338 H-Cl 431Cl-Cl 242 H-Br 366C-H 412 O=O 496C=O 805 N N 945C-O 358

The relationship between the enthalpy of atomisation and other enthalpy changes depends on the nature of the element:

C(s) C(g) Hat = +715 kJ.mol-1Hat = Hsub

Hg(l) Hg(g) Hat = +61 kJ.mol-1 Hat = Hvap

1/2Cl2(g) Cl(g) Hat = +121 kJ.mol-1Hat = 1/2 Hdiss

1/2Br2(l) Br(g) Hat = +112 kJ.mol-1Hat = 1/2 Hvap + 1/2 Hdiss

1/2I2(s) I(g) Hat = +107 kJ.mol-1Hat = 1/2 Hsub + 1/2 Hdiss

Sometimes a lattice enthalpy is quoted as the enthalpy of lattice formation. This involves the formation of a solid ionic lattice from its constituent gaseous ions and is the reverse of the above process. Therefore the enthalpy change is exothermic rather than endothermic but is equal in magnitude.

e.g. Na+(g) + Cl-(g) NaCl(s) Hlatt = -771 kJ.mol-1

Born-Haber CyclesTOPIC 13.17: THERMODYNAMICS 4

Enthalpy of Atomisation, Hoat

The enthalpy of atomisation is the standard enthalpy change which accompanies the formation of one mole of gaseous atoms from an element in its standard state.

e.g. 1/2H2(g) H(g) Hat = +218 kJ.mol-1

Mg(s) Mg(g) Hat = +150 kJ.mol-1

Enthalpy of Lattice Dissociation, Holatt

The enthalpy of lattice dissociation is the standard enthalpy change which accompanies the separation of one mole of a solid ionic lattice into its gaseous ions.

e.g. NaCl(s) Na+(g) + Cl-(g) Hlatt = +771 kJ.mol-1

The enthalpy of formation of an ionic solid can be broken down into a number of other enthalpy changes, which can be arranged in a cycle known as a Born-Haber cycle. One of the steps shown in the cycle is the enthalpy of lattice dissociation; this change cannot be measured directly but can be deduced from the cycle by the application of Hess’s Law. The cycle for sodium chloride is shown below.

Applying Hess’s Law:Hf = Hat + HI + Hat + Hea - Hlatt

hence Hlatt = Hat + HI + Hat + Hea - Hf

Hlatt = + 109 + 494 + 121 + (-364) - (-411)

Hlatt = + 771 kJ.mol-1

Deviation from the ionic model


EnergykJ.mol-1 Hlatt = +771

Hea = -364

Hat = +121

HI = +494

Hat = +109

Hf = -411

enthalpy of lattice dissociation of NaCl

Electron affinity of Cl

enthalpy of atomisation of Cl

1st ionisation enthalpy of Na

enthalpy of atomisation of Na

enthalpy of formation of NaCl

NaCl(s)

Na+(g) + e- + Cl(g)

Na+(g) + e- + 1/2Cl2(g)

Na(g) + 1/2Cl2(g)

Na(s) + 1/2Cl2(g)

Na+(g) + Cl-(g)

The IONIC MODEL can be used to calculate theoretical lattice enthalpies. This model assumes that ions are spherical in shape and that the charge on the ion is evenly distributed.

The values of lattice enthalpy found using the Ionic Model and those found experimentally using a Born-Haber cycle often differ significantly.

Compound Hlatt (experimental) Hlatt (theory)NaCl -787 -766NaBr -742 -731NaI -698 -686

In the table above the theoretical and experimental values agree very well suggesting that the bonding present in the sodium halides is essentially ionic.

However, the agreement between the magnesium halides is not so good.

Compound Hlatt (experimental) Hlatt (theory)MgCl2 -2526 -2326MgBr2 -2440 -2097MgI2 -2327 -1944

The experimental (real) values are some 10-15% higher than the theoretical values. This suggests that the bonding involved is stronger than the ionic model predicts. The difference is that, due to the small and more highly charged nature of the magnesium ion, the halide ions have been polarised (and hence their shape distorted) and there is additional degree of covalent character within bonding. The smaller and more highly charged the metal ion the greater the deviation from the ionic model.

ENTHALPY OF SOLUTION


Enthalpy of Solution, Hosol

The enthalpy of solution is the standard enthalpy change for the process in which one mole of an ionic solid dissolves in an amount of water large enough to ensure that the dissolved ions are well separated and do not interact with one another.

e.g. NaCl(s) + aq. Na+(aq) + Cl-(aq) Hsol = +2 kJ.mol-1

When an ionic solid dissolves in water, the process can be broken down into two separate processes:

the breaking down of the solid lattice into gaseous ions, which will require the enthalpy of lattice dissociation

e.g. NaCl(s) Na+(g) + Cl-(g)

the hydration of the gaseous ions, which will release energy equivalent to the enthalpy of hydration

e.g. Na+(g) + Cl-(g) + aq. Na+(aq) + Cl-(aq)

The symbol (+/-) is used to indicate either a cation or an anion.

The energy diagram below illustrates the enthalpy of solution of sodium chloride.


Enthalpy of Solution, Hosol

The enthalpy of solution is the standard enthalpy change for the process in which one mole of an ionic solid dissolves in an amount of water large enough to ensure that the dissolved ions are well separated and do not interact with one another.

e.g. NaCl(s) + aq. Na+(aq) + Cl-(aq) Hsol = +2 kJ.mol-1

-+ -+

Enthalpy of Hydration, Hohyd

The enthalpy of hydration is the standard molar enthalpy change for the process:

e.g. X (g) X (aq)

Hhyd = -364

enthalpy of hydration of Cl-

EnergykJ.mol-1

Hlatt = +771

Hhyd = -405

Hsol = +2

enthalpy of lattice dissociation of NaCl

enthalpy of hydration of Na+

NaCl(s) + aq.

Na+(aq) + Cl-(g) + aq.

Na+(aq) + Cl-(aq)

Na+(g) + Cl-(g) + aq.

Calculation of the standard enthalpy change for a reaction from standard enthalpies of formation Ho

f

In general, for any reaction:


Ho = Hof(products) - Hof(reactants)Example: Calculate the standard enthalpy change for the reaction:

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

given that: Hof (Fe2O3) = - 822 kJ.mol-1

Hof (CO) = - 111 kJ.mol-1

Hof (CO2) = - 394 kJ.mol-1

Ho = Hof(products) - Ho

f(reactants)

Ho = {(-394 x 3) + 0)} - {(-111 x 3) + (-822)}

Ho = - 27 kJ.mol-1

BOND ENTHALPYThe bond dissociation enthalpy of a diatomic molecule refers to the enthalpy change for the process:

A B(g) A(g) + B(g)Note that all the species are in the gaseous state.

In polyatomic molecules (those containing 3 or more atoms), it is usual to use mean bond enthalpy.Consider the following examples, which show that the energy required to break a particular bond depends on its specific environment.

H2O(g) OH(g) + H(g) H = +495 kJ.mol-1OH(g) O(g) + H(g) H = +428 kJ.mol-1

The mean bond enthalpy of the O-H bond is +463 kJ.mol-1

CH4(g) CH3(g) + H(g) H = +423 kJ.mol-1CH4(g) C(g) + 4H(g) H = +1664 kJ.mol-1

The mean bond enthalpy of the C-H bond is +412 kJ.mol-1

CALCULATION OF H FROM MEAN BOND ENTHALPIESWhen a chemical change takes place, existing chemical bonds are broken and new bonds are formed. Bond breaking requires an input of energy, therefore the process is endothermic. The formation of a bond releases energy and is therefore exothermic. The overall enthalpy change in the reaction will be the difference between the energy required for the endothermic step and the energy released by the exothermic step.

Mean bond enthalpies can be used to calculate the enthalpy change in simple reactions. Since mean bond enthalpies are used, the value obtained is only an approximation.

Bond enthalpy calculations apply only to reactions carried out in the gaseous state.


Mean bond enthalpy is the energy required to break a particular covalent bond, averaged over a large number of compounds.

2 x Hodiss

Hoat

Hof

Ho

Example: Use the data given below to calculate a value for the bond dissociation enthalpy of the C-H bond.

Hof (CH4) = - 74 kJ.mol-1

Hoat(C) = +720 kJ.mol-1

Hodiss(H2) = +431 kJ.mol-1

CH4(g) C(g) + 4H(g)

C(s) + 2H2(g)

By Hess’s Law:Ho = -Ho

f + Hoat + 2 x Ho

diss = - (-74) + (+720) + (+431 x 2)

= + 1656 kJ.mol-1

Ho = 4 x Hdiss (C-H)

Hdiss (C-H) = +1656 = + 414 kJ.mol-1 4

H = Hdiss(bonds broken) – Hdiss(bonds formed)Example: Calculate the enthalpy change in the reaction:

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

Bonds broken Bonds formed1x N N 6 x N-H3 x H-H

Hdiss(N N) = +945 kJ.mol-1 Hdiss(N-H) = +391 kJ.mol-1Hdiss(H-H) = +436 kJ.mol-1

H = (+945) + (+436 x 3) - (+391 x 6)

= - 93 kJ.mol-1

FREE ENERGY & SPONTANEOUS CHANGEA spontaneous change is one which has a natural tendency to occur if left to itself. For example:

the water in a lake will freeze spontaneously if the air temperature above it is below 0oC.

ice cubes placed in a drink will melt spontaneously iron in the presence of oxygen and water will rust spontaneously a battery will spontaneously run down

However, the spontaneous change will only occur in one direction unless conditions are changed. It is often difficult to reverse a spontaneous change.

the water in the lake will not melt unless the temperature of the air above it rises

water needs the electrical energy used by a fridge to convert it into ice cubes iron oxide needs the conditions produced in a blast furnace to turn it back to

iron a battery can be re-charged using electrical energy from the mains

The quantity which decides whether a particular change can occur spontaneously is the free-energy change, which is given the symbol G. For a change to occur spontaneously:.

Free energy changes combine the effects, in a particular reaction, of changes in enthalpy and changes in entropy.


G must be negative or zero

G = H - TS

You will notice that the above equation takes the form of y=mx+c (the equation of a straight line). A plot of ∆G against T (in K) gives a straight line graph with intercept ∆H and gradient -∆s.

200 300 400 500 600 700 800 900

-30

-20

-10

0

10

20

30

40

50

Temperature (K)

∆G (k

J/m

ol)

1. Changes in EnthalpyThe sign of an enthalpy change does not allow a prediction to be made about the feasibility of a change.

There is a natural tendency for enthalpy to fall in a reaction, and many exothermic changes occur spontaneously. For example, the reaction between potassium and water occurs spontaneously and liberates a lot of heat:

2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

However, spontaneous endothermic changes are also known, for example the reaction of sodium hydrogencarbonate with dilute hydrochloric acid:

NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)

The fact that endothermic changes can occur spontaneously clearly indicates that a factor other than enthalpy change must also be involved. This additional factor is the change in entropy.


2. EntropyEntropy can best be thought of as a measure of the disorder of a chemical system; the more disordered the system, the higher the entropy. There is a natural tendency for the entropy of a system to increase.

The particles in a solid are ordered and their movement is restricted to vibration about a mean position. The particles in a gas are disordered and they move about randomly. Therefore, the entropy of a gas is much higher than that of a solid.

When a solid is heated, the amplitude of the vibrating particles increases, and therefore the entropy increases. When the solid reaches its melting point, the entropy increases at constant temperature as the ordered solid becomes a less ordered liquid.

When the liquid is heated, the kinetic energy of the particles increases, and therefore the entropy increases. When the liquid reaches its boiling point, the entropy increases at constant temperature as the less ordered liquid becomes a chaotic gas. The entropy change from liquid to gas is greater than that from solid to liquid.


entropy increases

solid liquid gas

ordered less ordered chaotic

m.p.

b.p.

entropy

temperature

gas

liquid

solid

Entropy decreases as temperature decreases, so that at absolute zero (0K), most substances are solids consisting of perfectly ordered particles which have ceased to vibrate. They therefore have zero entropy. This means that there is a definite starting point for entropy, and substances can be assigned an absolute standard entropy vaue (So). Compare this with enthalpy, where only changes can be measured, and where it is necessary to assign to elements the arbitrary standard enthalpy of zero.

The table below lists some standard entropies.

2. Changes in Entropy

a) Chemical ChangesThe entropy change during a chemical reaction is calculated from the expression:


The units of entropy are:J.K-1.mol-1

So / J.K-1.mol-1

CH4(g) 186 H2O(l) 69.9C(s,diamond) 2.4 NaCl(s) 72.4C(s,graphite) 5.7 NaHCO3(s) 102Na2CO3(s) 136 Mg(s) 32.5Cl2(g) 223 H2(g) 131O2(g) 205 CO2(g) 214H2O(g) 189 CH3OH(l) 127C2H5OH(l) 161 C6H6(l) 173C6H12(l) 204

So = So(products) - So(reactants)

Example: Calculate the standard entropy change for the combustion of methane:CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

So = So(products) - So(reactants)

So(products) = 214 + 2 x 69.9 = 353.8 J.K-1.mol-1

So(reactants) = 186 + 2 x 205 = 596 J.K-1.mol-1So = 353.8 - 596 = -242.2 J.K-1.mol-1

There is a fall in the entropy of the system (i.e. it becomes more ordered), because three moles of gas react to form one mole of gas and two moles of liquid. Liquids are more ordered than gases.

b) Physical ChangesWhen a system is in equilibrium between two physical states, for example water and steam at 373K and 105Pa, there is no spontaneous tendency for the equilibrium mixture to change. Unless the external conditions are changed, the mixture will neither spontaneously liquefy nor vaporise. Since there is no tendency to move spontaneously in either direction, the free energy change, G is zero.

Since G = H - TS, when G = 0 H = TS

Consider water boiling in a kettle. We can calculate the change in entropy (S) when water boils as follows.

Say a 2.4kW kettle (which delivers 2.4kJ of energy per second) turns 100g of water to steam in 100s.


S = H T

Calculate the enthalpy change of vaporisation (Hvap) when one mole of water (at 100oC) is vaporised.

Heat energy delivered by kettle in 100s = 2.4 x 100 = 240 kJ

Moles of water vaporised = mass/Mr = 100/18 = 5.55

Hence enthalpy of vaporisation = 240/5.55 = 43.2 kJ/mol

Since this happens at 373K

Svap = Hvap = 43.2 x 1000 = +115.8 J.K-1.mol-1 Tb 373

Nb Hvap has been converted into J.mol-1 (x1000) in order that s be calculated in the normal entropy units of J.K-1.mol-1

FREE ENERGY CHANGES

1. H is negative; S is positive

Since G is always negative, a change of this type is feasible at any temperature.

Examples: NH4NO3(s) 2H2O(l) + N2O(g)

C(s) + O2(g) CO2(g)

2. H is positive; S is positive

G becomes zero where the line crosses the x-axis (temperature). At temperatures above this point, G is negative and the change is feasible. Below this temperature, G is positve and the change is not feasible.

Examples: H2O(l) H2O(g) boiling

CaCO3(s) CaO(s) + CO2(g)

NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)3. H is negative; S is negative


G

H

TS

Temperature /KEnergychange 0

-

+

change not feasible

change feasible

TS

G

H

Temperature /KEnergychange 0

-

+

change not feasible

TS

G+

G becomes zero where the line crosses the x-axis (temperature). At temperatures below this point, G is negative and the change is feasible. Above this temperature, G is positve and the change is not feasible.

Examples: H2O(g) H2O(l) condensation

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

4. H is positive; S is negative

G is always positive, so a change of this type is never feasible.


G

TS

G

H

Temperature /K

Energychange 0

-

+

Calculating the temperature at which a reaction becomes feasible.Referring to the diagrams on the previous two pages, a change just becomes feasible when G = 0.

Since G = H - TS, when G = 0 H = TS

It is important in this calculation to ensure that H and S are in compatible units. H is usually given in kJ.mol-1, therefore, S must be converted to kJ.K-1.mol-1

by dividing by 1000. The answer will be in K.


T = H S

Example: Use the data below to calculate the temperature at which the following reaction becomes feasible.

CaCO3(s) CaO(s) + CO2(g)

Hf /kJ.mol-1 S /J.K-1.mol-1CaCO3(s) -1207 88.7CaO(s) -635 40CO2(g) -394 214

Ho = Ho(products) - Ho(reactants) = (-635-394) – (-1207) = + 178 KJ.mol-1

So = So(products) - So(reactants) = (40 + 214) – (88.7) = + 165.3 J.K-1.mol-1

Go = Ho - TSo

when the reaction just becomes feasible, G = 0

The temperature at which this occurs is given by T = Ho So

T = 178 x 1000 = 1077K 165.3

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