The energy level diagram for an exothermic reaction.

The energy level diagram for an endothermic reaction.

ENTHALPY CHANGES

①All chemical reactions involve breaking and forming of bonds. Heat energy is absorbed when bonds are broken and heat energy is released when bonds are formed.

②Enthalpy changes measured under standard conditions:

Temperature 250C / 298 KPressure 1 atm / 101 kPaConcentration (for aqueous solution)

1 mol dm-3

The substance must be in its most stable form and physical state (solid, liquid or gas)

③We cannot measure the absolute value of the enthalpy of system, but we can measure the change in enthalpy.The experimental technique for measuring enthalpy changes is called calorimetry. ④The symbol for the change in enthalpy is ΔH in the units of kJ. Hence for the reaction

A + B → C + D

The enthalpy change is given by ΔH.ΔH = The total enthalpy of the products – The total enthalpy of the reactants

= H2 – H1

Where H2 = (HC + HD) and H1 = (HA + HB)

⑤ If the reaction is exothermic, that is, heat is given out to surroundings, the enthalpy change for the reaction, ΔH has a negative value because the heat content of the products is less than the heat content of the reactants.

Notice that in an exothermic reaction: The energy is lost during the reaction is first transferred

to the reaction mixture. The temperature of the reaction mixture will increase. This energy is then transferred to the surroundings. But

once the reaction has finished, the temperature falls to room temperature again.

Examples of exothermic reaction are:a) All reactions involving burningb) Reactions between acids and metals

⑥ if the reaction is endothermic, that is , heat is absorbed from surroundings the enthalpy change for the reaction, ΔH has a positive value because the heat content of the products is higher than the heat content of the reactants.

Notice that in an exothermic reaction: The energy is absorbed during the reaction is first taken

from the reaction mixture. The temperature of the reaction mixture will decrease.

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Since the temperature of the surroundings is now higher than of the reaction mixture, energy flows into the reaction mixture from surroundings. Thus, once the reaction has completed, the temperature rises to room temperature again.

Examples of endothermic reactions:a) The energy may be supplied directly by a flame or electricity. An example is the thermal

decomposition of calcium carbonate into calcium oxide and carbon dioxide.b) In some endothermic changes, no direct external heating is needed. For example, when

ammonium nitrate (V), NH4NO3, dissolves in water, energy is transferred from the surroundings to the solution. This cause a large drop in temperature and the water may freeze to ice on the outside of beaker.

c) The most important endothermic reaction of all is photosynthesis in plants. The energy is supplied to the reactions in cells by sunlight.

STANDARD ENTHALPY CHANGE OF A REACTION

Is defined as the enthalpy change that accompanies a reaction, in the molar quantities that are expressed in a chemical equation under standard conditions, all reactants and products being in their standard states.

Example : H2(g) + 12 O2 (g) H2O(l) ΔHθ = -286 kJmol-1

1. ΔH = enthalpy change

2. This shows that 286 kJ of heat energy is evolved when 1 mole of H2 react with 12 mole of O2

according to the equation .3. Θ = means that all reactants and products are at standard state conditions, 101.3 kPa(1 atm)

and 298 K(25oC).4. Negative sign (-) means that the reaction is exothermic , heat energy is given out. If positive

sign (+) means , the reaction is endothermic and heat energy is absorbed.

There are three basic rules which we must apply when we use thermochemical equations for calculations.① The total amount of energy released or absorbed is directly proportional to the number of moles of the reactants used.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔHθ = -890 kJmol-1 Thus the combustion of 2 moles of methane produces 2 x 890 = 1790 kJ of heat.

②The enthalpy change (ΔH) for the reverse reaction is equal to the magnitude but opposite in sign to the enthalpy change for the forward reaction.

Na+(g) + Cl-(g) NaCl(s) ΔHθ = - 770 kJmol-1 Hence NaCl(s) Na+(g) + Cl-(g) ΔHθ = + 770 kJmol-1

③ The value of ΔH for a reaction is the same whether it occurs in the step or in a series of steps. This is Hess’ law.

Measuring energy transfers and enthalpy changes(a) Q = (mass of solution x specific heat capacity x temperature change) units: Joule

Or Q = m.c.ΔT Joule

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(b) The following assumptions are made:

(i) The density of the solution is 1.00 g per cm3.(ii) The specific heat capacity of the solution is 4.18 J g-1 oC-1

(iii) The polystyrene cup or calorimeter absorbs a negligible amount of heat.

STANDARD HEAT OF FORMATION

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy that accompanies the formation of 1 mole of the compound from its elements, with all substances in their standard states. Its symbol is ΔHf

O . The superscript symbol indicates that the process has been carried out under standard conditions. Standard States are as follows:

1. For a gas: standard state is a pressure of exactly 1atm2. For a substance present in a solution: a concentration of exactly 1 M at a pressure of 1

atm3. For a pure substance in a condensed state (a liquid or a solid): the pure liquid or solid4. For an element: the form in which the element is most stable in under 1 atm of

pressure and the specified temperature. (Usually 25 degrees Celsius or 298.15 K)

For example, the standard enthalpy of formation of carbon dioxide would be the enthalpy of the following reaction under the conditions above:

C(s,graphite) + O2(g) → CO2(g)

Note that all elements are written in their standard states, and one mole of product is formed. This is true for all enthalpies of formation.

The standard enthalpy of formation is measured in units of energy per amount of substance. Most are defined in kilojoules per mole (kJ mol−1), but can also be measured in calories per mole, joules per mole or kilocalories per gram (any combination of these units conforming to the energy per mass or amount guideline). In physics the energy per particle is often expressed in electronvolts which corresponds to about 100 kJ mol−1.

All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation.

The standard enthalpy of formation is equivalent to the sum of many separate processes included in the Born-Haber cycle of synthesis reactions. For example, to calculate the standard enthalpy of formation of sodium chloride, we use the following reaction:

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Na(s) + (1/2)Cl2(g) → NaCl(s)

This process is made of many separate sub-processes, each with their own enthalpies. Therefore, we must take into account:

Standard enthalpy change of formation Born-Haber diagram for lithium fluoride.

1. The standard enthalpy of atomization of solid sodium

2. The first ionization energy of gaseous sodium

3. The standard enthalpy of atomization of chlorine gas

4. The electron affinity of chlorine atoms5. The lattice enthalpy of sodium chloride

The sum of all these values will give the standard enthalpy of formation of sodium chloride.

Additionally, applying Hess's Law shows that the sum of the individual reactions corresponding to the enthalpy change of formation for each substance in the reaction is equal to the enthalpy change of the overall reaction, regardless of the number of steps or intermediate reactions involved. In the example above the standard enthalpy change of formation for sodium chloride is equal to the sum of the standard enthalpy change of formation for each of the steps involved in the process. This is especially useful for very long reactions with many intermediate steps and compounds.

Hess's law

❶ Hess' law states that the enthalpy change for any chemical or physical process is independent of the pathway or number of steps required to complete the process provided that the final and initial reaction conditions are the same. In the other words if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the

route by which the chemical change occurs, provided the initial and final condition are the same.

A B

ΔH 2θ C

ΔH 3θBy Hess’s law : ΔH 1

θ = ΔH 2θ + ΔH 3

θ

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Route 1 ΔH 1θ

Route 2

❷Hess' law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on the chemical equation of reactions using previously determined values for the enthalpies of formation.

❸Addition of chemical equations may lead to a net equation. If enthalpy change is known for each equation, the result will be the enthalpy change for the net equation.

If the net enthalpy change is negative (ΔHnet < 0), the reaction is exothermic and is more likely to be spontaneous; positive ΔH values correspond to endothermic reactions. Entropy also plays an important role in determining spontaneity, as some reactions with a positive enthalpy change are nevertheless spontaneous.

Hess' Law states that enthalpy changes are additive. Thus the ΔH for a single reaction can be calculated from the difference between the heat of formation of the products and the heat of formation of the reactants:

where the o superscript indicates standard state values.

STANDARD ENTHALPY OF COMBUSTION

① The standard enthalpy of combustion is the enthalpy change when one mole of a reactant completely reacts with oxygen under standard thermodynamic conditions (although experimental values are usually obtained under different conditions and subsequently adjusted). By definition, combustion reactions are always exothermic and so enthalpies of combustion are always negative, although the values for individual combustions may vary.

② The most common way of calculating the enthalpy change of combustion (or formation) is by using a Hess cycle or by using numerical based bond enthalpies. It is commonly denoted as or . When the enthalpy required is not a combustion, it can be

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denoted as . Enthalpies of combustion are typically measured using bomb calorimetry, and have units of energy (typically kJ); strictly speaking, the enthalpy change per mole of substance combusted is the standard molar enthalpy of combustion (which typically would have units of kJ mol−1).

③ Need to understand the difference between the two terms heat and temperature. While they are related, they are not the same thing!

TEMPERATURE

Measures the kinetic energies of molecules present in a substance. Independent of the amount of substance present.

HEAT

Measure of the total energy of a substance. Depends on the amount of substance present.

So a bucket full of water at 50ºC would have the same temp. as a 250ml beaker of water at the same temperature, but the heat content of the bucket would be bigger.

We need a new term.

Specific Heat Capacity, is the amount of heat required to raise the temperature of 1g of a substance by 1K.· Differs from substance to substance.· Water = 4.2 J g-1 K-1

· Ethanol = 2.4 J g-1 K-1

This new term gives us a new equation to learn:

Heat transfer = mass x specific heat capacity x temperature change

Heat transfer = m c T

Units are absolutely essential. Specific Heat Capacity is normally given symbol c in units J g-

1 K-1. As a result m must be in grams and DT can either be in ºC or K.

④ Calorimetry was invented to study the heat transfer to water in this way using a calorimeter.

Measurement of enthalpies of combustion.

This is basically a method identical to that alcohol's investigation!

Known mass of water placed in beaker (150g) Temperature noted (23 ºC) Final temperature noted (43ºC) Mass of ethanol needed to obtain heat increase (0.9g)

Therefore 

Heat gained by water in calorimeter = m x c x T 

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= 150g x 4.2 J g-1 K-1 x 20ºC 

= 12,600 J Heat produced by burning 0.9g of ethanol = 12.6kJ

A simple calorimeter.

Bomb Calorimeter

The Standard Enthalpy of Neutralisation, Hneut, 298, is the heat evolved when 1 mole of water forms from the reaction between an acid and an alkali, measured under standard conditions of temperature and pressure. Since the essential reaction is H+ ions and OH- ions reacting to form water, the value is approximately -57.2 kJ mol-1.

H+(aq) + OH-

(aq) H2O(l) Hneut, 298 = -57.2 kJ mol-1

Where the acid is weak, for example ethanoic acid, the value will be slightly less exothermic owing to the absorption of some energy in bringing about ionisation of the acid molecules.

Some enthalpy data patterns continued

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Since 0.9g is 0.9/46 mol ethanol:

Heat produced by burning 1 mol of ethanol = 12.6 kJ / (109/46) kJ mol-1= 644kJTherefore Hc [CH3CH2OH] = -644kJ mol-1

Answer is much smaller than the accepted value of -1371 kJ mol-1.

Heat is lost to surroundings despite precautions 

Some heat is given to the calorimeter (the glass beaker). We could allow for this if we calibrated the calorimeter. 

Incomplete combustion of reactants, leading to formation of soot and carbon monoxide.

Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as bomb calorimetry. In order to ensure complete combustion, the bomb is initially charged with pure oxygen above atmospheric pressure. The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture.

Enthalpies of Neutralisation and Enthalpies of Formation of oxides and hydrides

Neutralisation reaction: base + acid ==> salt + water ΔHneutralisation kJ mol-1

1. NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) -57.12. KOH(aq) + HCl(aq) ==> KCl(aq) + H2O(l) -57.2

3. NaOH(aq) + HNO3(aq) ==> NaNO3(aq) + H2O(l) -57.34. KOH(aq) + HNO3(aq) ==> KNO3(aq) + H2O(l) -57.3

5. Ba(OH)2(aq) + 2HNO3(aq) ==> Ba(NO3)2(aq) + 2H2O(l) -116.4 (-58.4 per H2O)6. NaOH(aq) + HF(aq) ==> NaF(aq) + H2O(l) -68.6

7.  NH3(aq) + HCl(aq) ==> NH4Cl(aq) -52.28. NaOH(aq) + CH3COOH(aq) ==> CH3COONaaq) + H2O(l) -55.2

9. NH3(aq) + CH3COOH(aq) ==> CH3COONH4aq) -50.410. KOH(aq) + HCN(aq) ==> KCN(aq) + H2O(l) -11.7

11.  NH3(aq) + HCN(aq) ==> NH4CN(aq) -5.4

Notes on the above examples from the data table above.

Beware on comparing values!

(i) NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)  ΔHθneutralisation = -57.1 kJ mol-1

(ii) Ba(OH)2(aq) + 2HNO3(aq) ==> Ba(NO3)2(aq) + 2H2O(l)  ΔHθneutralisation = -116.4 kJ mol-1

(iii) 1/2Ba(OH)2(aq) + HNO3(aq) ==> 1/2Ba(NO3)2(aq) + H2O(l)  ΔHθneutralisation = -58.2 kJ mol-1

Note! It looks as if the enthalpy of neutralisation of barium hydroxide is approximately double that of sodium hydroxide ie ~ twice as exothermic! Well yes it is! and no it isn't!

Yes - ~twice as much energy is released per mole of soluble base/alkali.No - however, on the basis of heat released per mole of water formed, they are actually very similar.In other words, which value you quote, depends on which point you want to make.Yet another example of carefully qualifying enthalpy values with respect to the context.

Comparing Weak & strong acids or bases (soluble-alkali)

Neutralisation reactions 1., 2., 3., 4. and 5. are reactions between ~fully ionised strong acids and strong bases and the only real chemical change is formation of a water molecule from hydrogen/oxonium ions and hydroxide ions.

H3O+(aq) + OH-

(aq) ==> 2H2O(l)  or more simply H+(aq) + OH-

(aq) ==> H2O(l)  ΔH = ~-57 kJ mol-1

All the other ions e.g. Na+, K+, Ba2+, Cl-, NO3-, SO4

2- etc. are spectator ions and take no part in the reaction.

The resulting salt solution has a pH of ~7.

Wherever one of the acids or bases is 'weak' the energy change is less exothermic compared to a pair of 'strong' reactants.

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The reason for this relates to the fact that if an acid or base is not fully ionised, then energy must be expended facilitate the ionisation. Weak acids or weak bases are on partially ionised (e.g. 2%) in aqueous solution. Or, you can argue theoretically another way in terms of salt hydrolysis i.e. the ions of the salt react with water to 'reform' some of the acid or the base. In other words, you don't really get complete neutralisation of the acid and base on a strictly stoichiometric molar basisReaction 7. is the neutralisation of the weak base ammonia and the strong hydrochloric acid.

NH4+

(aq) + H2O(l) H3O+(aq) + NH3(aq)

ammonium chloride solution has a pH of ~4

Reaction 8. is the neutralisation of the strong base sodium hydroxide and the weak organic carboxylic acid, ethanoic acid.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-

(aq)

sodium ethanoate solution (CH3COONa(aq)) solution has a pH of ~9

Reaction 9. is even less exothermic because this is the neutralisation of a weak base (ammonia) and a weak acid (ethanoic acid).

See the equations for reactions 7. and 8. above

Reaction 10. is the neutralisation of an extremely weak acid (HCN) and, despite KOH being a strong base, only about 1/5th of the energy of a strong acid - strong base neutralisation is released.

CN-(aq) + H2O(l) HCN(aq) + OH-

(aq)

potassium cyanide solution (KCN(aq)) has a pH of over 10.

Reaction 11. is the neutralisation of an extremely weak acid (HCN) and a weak base (NH3) in which only a 1/10th of the maximum possible energy is released.

See equations for reaction 7 and 10.

Reactions 6 is unusually high despite the fact that hydrofluoric acid is a weak acid.

Bond Enthalpy (bond dissociation energy) calculations for Enthalpy of Reaction

The bond enthalpy/energy is the energy required to break 1 mole of a specified bond for a gaseous species at 298K/25oC. (note gaseous species are specified)

i.e. A-B(g) ==> A(g) + B(g)    ΔHBE(A-B) = + ??? kJmol-1

The diagram above shows, in terms of a dot and cross diagram, the breaking of a C-Cl bond by homolytic bond fission.

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Note that:* bond breaking is always endothermic (+) and *bond formation is always exothermic (-).

How are bond enthalpies determined?1. Just as spectroscopy can be used to determine ionisation energies (see hydrogen spectrum), spectroscopic techniques can be used to determine bond energies i.e. it is possible to estimate the frequency of radiation required to cause bond fission.

2. Electron impact methods - in essence it is a method which measures the energy to fragment molecules.

3. Thermochemical calculations - the method most appropriate to this pages level of study.

3. is illustrated by the diagram below to calculate the C=O bond enthalpy in carbon dioxide.

+498 is the bond enthalpy of the oxygen molecule (O=O) or twice the enthalpy of atomisation ofoxygen.

+715 is the enthalpy of atomisation of carbon.

-393 is the enthalpy of combustion of carbon or the enthalpy of formation of carbon dioxide.

This becomes +393 in the Hess's Law cycle, arrow and sign reversed.

x is equal to twice the bond enthalpy of a C=O bond in a carbon dioxide molecule.

x is the only one of the four delta H values that cannot be determined by experiment.

However using Hess's Law

x = +393 +715 +498 = +1606 kJ mol-1

therefore the C=O bond energy in CO2 = 1606/2 = 803 kJ mol-1

A second example of using Hess's Law to calculate the average C-H bond enthalpy in the methane molecule

ΔHØf (methane) = -75 kJmol-1

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C(s) + 2H2(g) CH4(g)

ΔHθsub(C(s)) + 2 x ΔHθ

BE(H-H(g))

= (+715) + (2 x +436)

both endothermicC(g) +

 

4H(g)

-4 x avΔHθBE(C-H(g)) = -X

exothermic

formation of 4 C-H bonds

The cycle for the standard enthalpy of formation of methane, (ΔHθf), based on the enthalpy of

sublimation of carbon (graphite) and the H-H and C-H bond enthalpies.

From Hess's Law, add up the sequence of enthalpy changes via the lower 'staged route' to get the overall enthalpy change for the enthalpy of formation of methane from its elements in their

normal stable states.

ΔHθf (methane) = {ΔHθ

sub(C(s)) + 2 x ΔHθBE(H-H(g))} + {-4 x avΔHθ

BE(C-H(g))}  = -75 kJmol-1

-75 = +715 +872 -X

X = (715 + 872 +75) = 1662

avΔHθBE(C-H(g)) = 1662/4 = +415.5 kJmol-1

Bond enthalpy calculations using average bond enthalpies

Bond enthalpy calculations using a Hess's Law cycle can be used to calculate unknown enthalpy changes for a reaction.

BUT there are limitations to the results of these calculations since they are based on ..o (i) average bond enthalpies (i.e. typical vales for a particular bond) ando (ii) only valid gaseous reactants AND products (quite restrictive, this because

bond enthalpies are defined, measured and based on gaseous species only).

Ex 1. Calculating the enthalpy of a reaction

Given the following bond enthalpies in kJ mol-1: Cl-Cl = 242, H-H = 436, H-Cl = 431

Calculate the enthalpy of the reaction of forming 2 moles of hydrogen chloride from its elements in their standard states ... the Hess's Law cycle for this is ...

H2(g) + Cl2(g) 2HCl(g)

2H(g) + 2Cl(g)

For simple Q's like this, unless asked for, you can solve easily without drawing a cycle, either way ...

ΔHθreaction = {endothermic ΔH for H2 and Cl2 bonds broken} + {exothermic ΔH for HCl bonds

formed}

ΔHθreaction = {+436 +242} + {2 x -431}

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ΔHθreaction = +(436 + 242 -862) = -184 kJmol-1

Note that ΔHθreaction /2 = -92 kJmol-1 = ΔHθ

f(HCl(g))

Ex 2. Given the following bond dissociation enthalpies at 298K in kJmol-1 ...

 Bond C-C (single) C-H C=O in CO2 O-H O=OBond enthalpy +348 +412 +805 +463 +496

... calculate the enthalpy of combustion of butane. You construct a Hess's Law Cycle from the 'normal' combustion equation and 'atomise' the reactant molecules to give ALL of the 'theoretical' intermediate atoms. From this you can  theoretically calculate the enthalpy of the reaction:

ΔHcomb, 298 (butane) = ??? kJmol-1

(g) + 61/2O=O(g) 4O=C=O(g) + 5H-O-H(g)

(3 x ΔHBE(C-C))

+ (10 x ΔHBE(C-H))

+ (6.5 x ΔHBE(O=O))

= (3 x +348) + (10 x +412) + (6.5 x +496)

endothermic       

4C(g)

+ 10H(g)

+ 13O(g)

  

exothermic

-(8 x ΔHBE(C=O))

- (10 x ΔHBE(O-H))

= -(8 x 743) - (10 x 463)

ΔHcomb, 298 (butane) = 1044 + 4120 + 3224 -6440 -4630 = -2682 kJmol-1

The true thermodynamic value from very accurate calorimetry experiments is -2877 kJmol-1

So why the significant difference/error?

The difference is the theoretical value from the bond energy calculation is less exothermic by 195 kJ.

All enthalpy calculations done by this method will always be in error to some extent because the bond enthalpies quoted in data information are based on average values for that particular bond. Each 'A-B' bond will differ slightly depending on the 'ambient' electronic situation e.g.

The C-H bond in methane, , will be slightly different than in ethane,

etc.

Ex 3. example of bond enthalpy calculation method

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Liquid Hydrazine N2H4 and hydrogen peroxide H2O2 have both been used as rocket fuels.

Hydrazine has been used as monopropellant in rocket engines because it can catalysed to decompose into nitrogen and hydrogen gas very rapidly and exothermically.

(1)  N2H4 ==> N2 + 2H2

Hydrazine can also be used to power rockets in combination with hydrogen peroxide (a bipropellant fuel).

(2)  N2H4 + 2H2O2 ==> N2 + 4H2O

From the list of bond enthalpies in kJmol-1, given below, calculate the enthalpy changes for reactions (1) and (2)

 Bond N-H N N N-N H-H O-H O-OBond

enthalpy +388 +944 +163 +436 463 +146

For (1)  N2H4 ==> N2 + 2H2

H2N-NH2 N N + 2H-H

ΔHBE(N-N)

+ (4 x ΔHBE(N-H))

= (+163) + (4 x +388)endothermic       

2N + 4H

  

exothermic

-(ΔHBE(N N))

- (2 x ΔHBE(H-H))

= -(+944) -(2 x 436)

ΔHreaction(decomp N2H4) = +163 + (4 x 388) -944 - (2 x 436)

= +163 +1552 -944 -872 = -101 kJmol-1

For (2)  N2H4 + 2H2O2 ==> N2 + 4H2O

H2N-NH2 + 2H2O2 N N + 4H-O-HΔHBE(N-N)

+ (4 x ΔHBE(N-H))

+ (2 x ΔHBE(O-O))

+ (4 x ΔHBE(O-H))  

= (+163) + (4 x +388) + (2 x +146) + (4 x

+463)endothermic        2N + 8H +

4O

  

exothermic

-(ΔHBE(N N))

- (8 x ΔHBE(O-H))

= -(+944) -(8 x +463)

watch the - signs!

ΔHreaction(combustion N2H4) = +163 +1552 +292 +1852 -944 -3704

ΔHreaction(combustion N2H4) = -789 kJmol-1

which is considerably more exothermic than the catalysed thermal decomposition of hydrazine into nitrogen and hydrogen,

but the mixture is more costly and more dangerous to handle!

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ΔH ENTHALPIES OF ION HYDRATION, SOLUTION, ATOMISATION, LATTICE ENERGY, ELECTRON AFFINITY AND THE BORN-HABER CYCLE

What happens when an ionic compound dissolves in water and the energetics of why?

This section will discusses a dissolving enthalpy cycle - another application of Hess's Law - in which the process of dissolving an ionic compound such as a halide salt or a metal oxide, is broken down into theoretical stages to help understand the factors involved in deciding whether of not a substance will dissolve readily or not at all. For the moment, only enthalpy changes will be considered but eventually entropy changes must be discussed too!

This ΔH is called the ENTHALPY of SOLUTION

(i) ionic compound(s) + aq cations(aq)  +  anions(aq)

(ii) solid ionic compound

vapourised to give gaseous

ions - this ΔH is called the LATTICE

ENTHALPY(endothermic)

 

 

gaseous cations and

anions(exothermic)

(iii) hydration of the cations and

anions - these ΔHs are called the ENTHALPY of

HYDRATION of the ION

In terms of enthalpy changes: (i) = (ii) + (iii)

Enthalpy of solution ΔHsolution(compound) is defined as the heat absorbed or released when 1 mole of compound dissolves in water

Lattice enthalpy ΔHLE(compound) is defined as the energy released when 1 mole of a solid is formed from its separated gaseous ions

Enthalpy of hydration is defined as the energy released when 1 mole of gaseous ions interacts with water to give a solution of the hydrated ions

All three enthalpy terms are re-defined in more detail and explained in further detail as  the discussion proceeds.

The changes that occur in the dissolving process

Despite the strong ionic bonding forces in most salts or simple binary compounds like oxide or chloride crystals i.e. the strong electrostatic attraction between positive ions (cations) and negative ions (anions) many ionic compounds readily dissolve in water. Therefore, not surprisingly, a great deal of energy is required to separate the ions, but dissolving can still take place. So how can we explain this?

Water consists of highly polar molecules due to the great electronegativity difference between hydrogen and oxygen (O > H hence the polarity of the bond δ-O-Hδ+). When salts dissolve in water a process of solvation occurs in which the ions become solvated by association with the solvent molecules. If water is the solvent, the process is called hydration and is exothermic because it involves particles coming together via intermolecular forces or covalent bonds in the case of aqua-cation complex ions - so this is a compensating source of energy.

In some cases cations become hydrated via dative covalent bonds to form a complex ions e.g.

Li(H2O)4]+, Cu(H2O)4]2+, Mg(H2O)6]2+,  Al(H2O)6]3+ etc. because in the case of water, the most electronegative part of the highly polar water molecule (>Oδ-) will be attracted to the positive ion and, since the oxygen atom has two lone pairs of electrons, it is also the source of the dative covalent bond by donation of one of these pairs of electrons into a vacant metal ion orbital.

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In the case of anions, the positive ends of the water molecules (-Hδ+) will orientate themselves towards the negative anion and the water molecules become weakly associated with anion, but no covalent bonds are formed.

This solvation of the ions means the ions are effectively bigger particles which makes the distance between the positive and negative ion centres greater, and by the laws of electrostatics, the attractive forces is weakened and the hydration process is always exothermic.

PLEASE note that dissolving a solute in this situation cannot be simply regarded as a physical change. The ionic lattice is broken down, but NOT by melting, and chemical bonds are formed between the cation and water molecules to form hydrated ions or aqua-ions such as those listed above.

Diagram illustrating the dissolving-solvation-hydration process for sodium chloride crystals forming salt solution

For diagram simplicity, each ion is surrounded by four water molecules, though in reality, much more than this - see later.

The strong crystal lattice (giant ionic structure) is broken down by the solvation process so the salt dissolves and the ions are free to move around in the water solvent.

Its worth noting that this involves an increase in entropy - the solution is more disordered (more possible arrangements of the particles) compared to the pure liquid solvent and the highly ordered crystal lattice of very limited possible arrangements.

o This increase in entropy favours the dissolving process, (as well as a very exothermic value for the enthalpy of solution!).

However, when the ions become hydrated there is a decrease in entropy, which counts against a substance dissolving. This decrease in entropy is due to the orientation of the water molecules when they associate with the cations and anions i.e. a decrease in the possible ways the water molecules can be arranged.

o In other words the hydrated ions produce a more ordered arrangement of some of the water molecules.

o This decrease in entropy does not favour dissolving, (and neither does a very endothermic enthalpy of solution)

Therefore you need to consider entropy changes to fully explain why substances do/do not dissolve - especially when the enthalpy of solution is positive or only slightly negative.

In order to try to understand processes you can use a Hess's Law cycle to break the process down into theoretical steps, each of which can measured experimentally or theoretically calculated. The relative magnitude of each energy change can help understand why a substance will dissolve or be insoluble. However, this cycle only uses enthalpy values and excludes entropy changes - which I'll deal with later.

The connection between lattice enthalpy, enthalpies of ion hydration and enthalpy of solution

(i) The energy change for a substance dissolving in a solvent is called the enthalpy of solution.

The enthalpy of solution ΔHsolution(compound) is defined as the heat absorbed or released when 1 mole of compound (the solute) dissolves in a solvent to form an 'infinitely' dilute solution where no further heat change takes place at 298K e.g.

NaCl(s) + aq ==> Na+(aq) + Cl-

(aq)  ΔHsolution(sodium chloride) = +4 kJ mol-1

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Al2(SO4)3(s) + aq ==> 2Al3+(aq) + 3SO4

2-(aq)  ΔHsolution(aluminium sulfate) = -318 kJ mol-1

Examples of enthalpy of solution (kJ mol-1) are tabulated below and refer to an infinitely dilute solution

cation\anion OH- F- Cl- Br- I- CO32- NO3

- SO42-

Li+ -21.2 +4.5 -37.2 -49.1 -63.3 -17.6 -2.7 -30.2Na+ -42.7 +0.3 +3.9 -0.6 -7.6 -24.6 +20.5 -2.3K+ -55.2 -17.7 +17.2 +20.0 +20.5 -32.6 +34.9 +23.8

NH4+ - +5.0 +15.2 +16.2 +13.4 - +25.8 +6.2

Mg2+ +2.8 -17.7 -155 -186 -214 -25.3 -85.5 -91.2Ca2+ -16.2 +13.4 -82.9 -110 -120 -12.3 -18.9 -17.8Sr2+ -46.0 +10.9 -52.0 -71.6 -90.4 -3.4 +17.7 -8.7Al3+ - -209 -332 -360 -378 - - -318

Comments on the enthalpy of solution values

Quite a few of the values are relatively endothermic and perhaps at first sight dissolving seems unfavourable e.g. potassium nitrate (+34.9) and ammonium nitrate (+25.8), but both these salts are very soluble in water - you need to consider entropy too!

The value depends on the structure and strength of the ionic lattice and the hydration enthalpies of the constituent ions.

You can get simple trends, but they are not easy to explain at times e.g.o For Group 1 hydroxides and fluorides delta H solution gets more exothermic/less

endothermic BUT the chlorides, bromides and iodides get less exothermic/more endothermic on dissolving.

o Group 2 compounds show a mixture of trends i.e. down the group more endothermic or more exothermic.

o Aluminium compounds show the most exothermicity on dissolving, perhaps because of the high charge on the ion giving a large ion hydration enthalpy (see section below).

To derive an alternative route to form the solution employing Hess's Law you then can further consider:

(ii) The ionic crystal lattice is vapourised into its gaseous positive ion (cation) and gaseous negative ion (anion) constituents.

This is always a very endothermic process because you are separating strongly attracted oppositely charged ions.

This process is defined as happening in the reverse exothermic direction and is called the lattice enthalpy (lattice energy).

The lattice enthalpy ΔHLE(compound) is defined as the energy released when 1 mole of a solid is formed from its separated gaseous ions at 298K and is very exothermic e.g.

Na+(g) + Cl-

(g)  ==> NaCl(s)   ΔHLE(sodium chloride) = -774 kJ mol-1

2Al3+(g) + 3O2-

(g)  ==> Al2O3(s)   ΔHLE(aluminium oxide) = -15916 kJ mol-1

Factors affecting the lattice energy:

The greater the force of attraction the greater the energy needed to vapourise the lattice or the greater the energy released on forming the ionic lattice.

The size of the ion radius and the ion charges are the most important factors to consider.

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The smaller the ion radius or the greater the ion charge the greater the lattice enthalpy i.e. more exothermic.

The law of attraction of electric charges states that the force of attraction is proportional to the product of the charges divided by the distance between them

The attractive force is proportional to C+ x C- / d2, where d is the distance between the ion centres (atomic nuclei in simple 1 monatomic ions) and C+ and C- are the numerical charges on the cation and anion respectively and d is effectively the sum of the cation and anion radius for simple lattice structures.

Series and comments on examples of lattice enthalpies ΔHLE

LE quoted as -ΔHLE(compound) in kJ mol-1

Group 1 halide salts - decrease from F to I as the anion radius increases (same trend for Na, K, Rb and Cs) LiF, 1022 LiCl, 846 LiBr, 800 LiI, 744

Group 1 halide salts - decrease from Li to Cs as the cation radius increases (same trend for Na to Cs) LiF, 1022 NaF, 902 KF, 801 RbF, 767

Comparing sodium chloride and magnesium oxide. Neglecting radii differences, MgO has over 4x the lattice energy of sodium chloride because the ion charges are doubled i.e. 1+ x 1- compared to 2+ x 2-, a ratio of 1 : 4.

NaCl, 774 MgO, 3889    

The first three Period 3 oxides - from left to right, increasing cation charge AND decreasing cation radius with increasing interactions between oxide ions per cation

Na2O, 2488

MgO, 3889

Al2O3, 15916  

(iii) The gaseous ions then interact with water to give the hydrated ions in aqueous solution.

This is always a very exothermic process.

This energy change is called enthalpy of hydration.

The enthalpy of hydration is defined as the energy released when 1 mole of gaseous ions interacts with water to give a solution of the hydrated ions at 298K e.g.

Na+(g) + aq ==> Na+(aq)   ΔHhyd(Na+

(g)) = -364 kJ mol-1

Cl-(g) + aq ==> Cl-(aq)   ΔHhyd(Cl-

(g)) = -406 kJ mol-1

Factors affecting the ion hydration energy:

The size of the ion radius and the ion charge are the most important factors to consider here.

The smaller the ion radius or the greater the ion charge, the greater the ion hydration enthalpy i.e. more exothermic. The smaller the radius OR the higher the charge, the greater is the intensity-potential of the electric field so interaction with an oppositely charged particle (ion or polar molecule) is stronger and therefore more 'energetic' i.e. more exothermic.

Series and comments for the process Mn±(g) + aq

==> Mn±(aq) (always exothermic) - ion hydration

enthalpies ΔHhyd(cation/anion)quoted as ΔHhyd(ion(g)) kJ mol-1

The first three Period 3 cations - decreasing radius and increasing charge - double trend effect, so more dramatic increase from left to right

Na+(aq) -406 Mg2+

(aq) -1920

Al3+(aq) -

4690  

Group 1 cation trend, shows the effect of increasing radius down the group with decreasing enthalpy of ion hydration

Li+(aq) -519 Na+(aq) -406 K+

(aq) -322 Rb+(aq) -301

Group 7 halide anions, shows a similar effect of increasing radius F-

(aq) -506 Cl-(aq) -364 Br-(aq) -335 I-

(aq) -293

Comparing the iron(II) ion with the smaller and more Fe2+(aq), -1950 Fe3+

(aq), -4430

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highly charged iron(III) ion giving a considerable increase in hydration enthalpy

The Born Haber Cycle

This section looks at the application of Hess's Law to the theoretical formation of an ionic lattice from its constituent elements in their standard states i.e. most stable state at 298K (25oC), 101kP (1 atm).

Brown arrows = exothermic

Blue arrows = endothermic

Fig 1. The simplest way, and the easiest way to learn to construct and present a Born-Haber cycle - in this case for sodium chloride NaCl.

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Right Fig 2. The enthalpy level diagram for the Born-Haber cycle for sodium chloride.

Left Fig 3. A way of thinking to solve the cycle for one unknown enthalpy component in a Born-Haber cycle presented in the style of Fig 2.

All the delta H values are defined and their use explained below. You need to be able to recognise any of ΔH 1-6 and be familiar with the 'styles' of figs 1-2 and be able to complete/construct a cycle and solve a problem to obtain an unknown value.

In any enthalpy value with a superscript theta, (θ), i.e. ΔHθx = ??, the θ denotes a standard

enthalpy value, which usually means 1 atm. pressure (101 kPa) and a 298 K temperature (25oC). Where relevant, any such criteria need to be added to the definitions below if quoted as standard values.

ΔH1 = ΔHθf(NaCl) the enthalpy of formation of sodium chloride:

o Na(s) + 1/2Cl2(g) ==> NaCl(s)  ΔHf(NaCl) = -411 kJ mol-1

o The standard enthalpy of formation is defined as the energy released or absorbed when 1 mole of a compound is formed from its constituent elements in their normal stable states at 298K and 1atm.

ΔH2 = ΔHθatom(Na)  the enthalpy of atomisation of sodium:

o Na(s) ==> Na(g)  ΔHatom(Na) = +107 kJ mol-1

o The standard enthalpy of atomisation is defined as the energy absorbed when 1 mole of gaseous atoms is formed from the element in its normal stable state at 298K and 1atm.

other examples, but not needed here are ... 1/2O2(g) ==> O(g)  ΔHatom(O) = +249 kJ mol-1 1/2Br2(l) ==> Br(g)  ΔHatom(Br) = +112 kJ mol-1 1/2I2(l) ==> I(g)  ΔHatom(I) = +107 kJ mol-1 Note that the enthalpy of atomisation for gaseous species is related

to the 'gaseous' bond enthalpy value e.g. the enthalpy of atomisation of oxygen is half the bond enthalpy of

the O=O double bond, ΔH3 = ΔHθ

atom(Cl2) the enthalpy of atomisation of chlorine:o 1/2Cl2(g) ==> Cl(g)  ΔHatom(Cl) = +121 kJ mol-1

o Already defined above and is also half the bond enthalpy for chlorine molecules Cl2(g) ==> 2Cl(g)  ΔHBE(Cl2) = +242 kJ mol-1

so take care with how the data is presented. ΔH4 = ΔHθ

el.affin.(Cl)  the 1st electron affinity of chlorine:o Cl(g) + e- ==> Cl-

(g)  ΔHelec.affin.(Cl) = -355 kJ mol-1

o The standard enthalpy change for the first electron affinity is defined as the energy released or absorbed when one mole of gaseous neutral atoms gain one electron each to form one mole of singly charged negative gaseous ions at 298K and 1atm.

I wouldn't worry about the 2nd of chlorine, the Cl2- ion will be too endothermic to be form in a chemical change and is electronically very unstable.

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The 2nd electron affinity would be defined as the energy absorbed or released when 1 mole of singly charged negative ions gain one electron each to form 1 mole of doubly charged negative ions.

However in a Born-Haber Cycle for the formation of an ionic metal oxide e.g. MgO (shown below)

1st electron affinity of oxygen is exothermic O(g) ==> O-

(g)  ΔH1st elec.affin.(O) = -142 kJ mol-1

but the 2nd is very endothermic O-

(g) + e- ==> O2-(g)  ΔH2nd elec.affin.(O) = +844 kJ mol-1

because of the O-... e- repulsion, but must be considered in the cycle for MgO etc.

o The 2nd electron affinity (not needed here) would be defined as: The energy released or absorbed when one mole of gaseous singly charged

negative ions gain one electron each to form one mole of doubly charged negative gaseous ions at 298K and 1atm.

This would be needed for the Born-Haber cycle for magnesium oxide (Mg2+O2-).

ΔH5 = ΔHθ1st IE(Na)  the 1st ionization enthalpy of sodium:

o Na(g) ==> Na+(g) + e-  ΔH1st IE(Na) = +502 kJ mol-1

o The standard first ionisation energy is defined as the energy absorbed when the most loosely bond electron is removed from one mole of neutral gaseous atoms to form one mole of singly positively charged gaseous ions at 298K and 1atm.

o Again, in a B-H cycle for MgO etc. the 2nd ionisation energy must be taken into consideration and would be defined as:

The energy absorbed when the most loosely bond electron is removed from one mole of singly positively charged gaseous ions to form one mole of doubly positively charged gaseous ions at 298K and 1atm.

ΔH6 = ΔHθLE(NaCl)  the lattice enthalpy of sodium chloride:

o Na+(g) + Cl-

(g) ==> NaCl(s)  ΔHLE(NaCl) = -786 kJ mol-1

o The standard enthalpy change, known as the lattice enthalpy (lattice energy) is defined as the energy released when 1 mole of an ionic compound is formed from its constituent gaseous positive and negative ions at 298K and 1atm.

o Factors affecting the value of lattice enthalpy - lattice enthalpy previously discussed - brief summary below.

The greater the force of attraction between the ions, the greater the energy release, in coming together to the point of minimum potential energy in forming the most stable ionic crystal lattice.

From the laws of electrostatics, the force of attraction is proportional to ... +ve charge x -ve charge/(distance between the centres of the

charges)2, translating this into the ionic lattice situation, the force of attraction is

proportional to the .. charge on cation x charge on anion/(radius of cation + radius of

anion)2, so quite simply, the smaller the radius of an ion or the greater the

charge on an ion, the greater will be the lattice enthalpy because the electric field intensity is increases, hence the force of attraction, hence the greater amount of energy released when the ions come together.

From this simple trends can be noted e.g. in terms of ∆LE (in kJmol-1) ...

LiF (1022) > LiCl (846) > LiBr (800) > LiI (744), from L to R, the anion radius becomes larger,

Al2O3 (15916) > MgO (3889) > Na2O (2478), from L to R, the cation charge decreases (and the cation radius increases),

MgO (3889) is over 4 x NaCl (780), charges 2+ x 2- versus 1+ x 1-, ignoring ionic radii differences.

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