chemi notes

EXOTHERMIC AND ENDOTHERMIC REACTIONS

iSEK. MEN. KEB. SULTAN ISMAIL, JOHOR BAHRU.PHYSICAL CHEMISTRY/ UPPER SIX/ 2013TOPIC : CHEMICAL ENERGETICS

EXOTHERMIC AND ENDOTHERMIC REACTIONS

1. In all chemical reactions, both bond formations and bond cleavages occur.2. Bond formation releases heat energy to the reaction mixture or environment. This is manifested with a temperature rise of the reaction mixture.3. Bond cleavage absorbs heat energy from the reaction mixture or environment. As a result of this, the temperature of the reaction mixture falls.4. If in a chemical reaction the total heat energy released due to bond formations is larger than the total heat energy absorbed from the reaction mixture due to bond cleavages, then the overall chemical reaction is a heat energy loss reaction. Such a chemical reaction is termed as an exothermic reaction.5. On the other hand, if the total heat energy absorbed due to bond cleavages is larger than the heat energy released due to bond formations, then the overall chemical reaction is a heat energy absorbing reaction. Such a chemical reaction is known as the endothermic reaction.6. Exothermic reaction causes temperature rise of the reaction mixture while endothermic reaction causes temperature fall of the reaction mixture.7. Hence, this change in temperature of the reaction mixture can be used as a parameter to determine the amount of heat energy change (E) or enthalpy change (H) that occurs in a chemical reaction.

ENTHALPY AND ENTHALPY CHANGE

1. All chemical and physical reactions are accompanied by changes in energy content. The study of these energy changes is called energetic.2. The enrgy stored in compound is called the energy content or the enthalpy of the compound. It is not possible to measure enthalpies directly, but it is possible to measure both enthalpy change associated with a chemical change. The enthalpy change represents the difference between the enthalpies of the reactants and products.H = Hp - Hr where H = change in enthalpy Hp = enthalpy of the products Hr = enthalpy of the reactants3. Enthalpy is the energy content of a substance and is symbolized as H. The energy content is in the form of kinetic energy and potential energy.4. The enthalpy of a substance depends on : (a) the temperature under which the experiment is carried out (b) the external pressure (c) the physical state of the substances involved (d) the concentration of solutions (e) the type of allotrope of the reactants.5. When a chemical reaction occurs, reactants and products are formed. The enthalpy of the reactants and the products are not the same. Hence, an enthalpy change, H occurs during a chemical reaction.

A + B C + D reactants productsH = H products H reactants

If H products > H reactants, then H > 0 (positive)This means the reacting species draw heat energy from the surroundings and increase the enthalpy, causing them to change to products. This causes the temperature of the reaction mixture and its surroundings to fall. This is called the endothermic reaction. If H products < H reactants, then H < 0 (negative)This means the reacting species release heat energy to the surroundings and decrease the enthalpy, causing them to change to products. This causes the temperature of the reaction mixture and its surroundings to rise. This is called the exothermic reaction.6. Enthalpy change, H is referred to as the heat energy change that occurs in a chemical reaction at constant pressure. The enthalpy change only depends on the initial and final energy states of the reaction mixture or the system. The normal unit of H is kJmol-1.

Question

Why does an enthalpy change occur during a chemical reaction ?

Answer This is because during a chemical reaction, old bonds in reactants are broken and new bonds in products are formed. Hence, the state of the reactants and products are altered resulting in an enthalpy change.

The Standard Conditions for Thermochemical Measurements

1. The standard conditions of temperature and pressure for thermochemical measurements are 298K (250C) and 1 atm (or 101 kPa) respectively.2. Any enthalpy change measured under these conditions is described as a standard heat of reaction and given the symbol H.3. The value of an enthalpy change depends on :(a) the temperature under which the experiment is carried out (b) the external pressure (c) the physical state of the substances involved (d) the concentration of solutions (e) the type of allotrope of the reactants.4. The standard heat of reaction is defined as the amount of heat absorbed or evolved when the amounts of reactants as stated in the reaction equation react together under standard conditions (i.e. at a pressure of 1 atm, at a temperature of 298k, with substances in their normal physical states under these conditions and solutions having concentration of 1.0moldm-3) to give products in their standard states.5. The following are two important rules for manipulating thermochemical equations : (a) When a thermochemical equation is multiplied by any factor, the value of H for the new equation is obtained by multiplying the value of H in the original equation by that same factor.(b) When a chemical equation is reversed, the value of H is reversed in sign.(c) For example, consider the thermochemical equation for the synthesis of ammonia :N2(g) + 3H2(g) 2NH3(g)H = -92 kJ(i) Doubling the reactants, the H would be 2 (-92kJ) = -184 kJ2N2(g) + 6H2(g) 4NH3(g)H = -184kJ(ii) Reserving the equation gives the dissociation of two moles of ammonia into its elements :2NH3(g) N2(g) + 3H2(g)H = +92 kJ

Calorimetry1. Calorimetry is the experimental technique to determine the enthalpy change of a reaction.2. It consists of a well insulated container called a calorimeter inside which the reaction occurs.3. The energy evolved during the reaction is transferred to water or to the reaction mixture itself, and the change in temperature is noted.4. The heat change (Q) can then be calculated using the following equation :Q = mct Where m = mass of the solution c = specific heat of the solution t = change in temperature.

termometer

stirer

insulation

Reacting system

Standard Enthalpy Of Combustion1. The standard enthalpy of combustion of an element or a compound is the enthalpy change when 1.0 mole of the substance is completely burnt in oxygen under the standard conditions at 298 K and 1.0 atm.2. The symbol for the standard enthalpy of combustion is Hc . All Hc values are negative. For example,C2H2(g) + O2(g) 2CO2(g) + H2O(l)Hc = -1310 kJmol-1

Standard Enthalpy of Formation1. The standard enthalpy of formation of a compound is the heat evolved or absorbed when 1.0 mole of the compound is formed from its elements in their normal physical states under the standard conditions of 298 K and 1.0 atm.

2. The symbol for standard enthalpy of formation is Hf.C (s) + O 2(g) CO2(g)Hf = -393.5 kJmol-1 C (s) + 2H2(g) CH4 (g)Hf = -75 kJmol-1

3. In many cases, the standard enthalpy of formation cannot be measured directly because the elements will not combine directly under experimental conditions. For example, carbon and hydrogen do combine directly to form methane. Hence, the standard enthalpy of formation cannot be measured directly.

4. When the standard enthalpy of formation of a compound cannot be found directly, it is often possible to find its value indirectly by the application of Hesss law.

Calculating enthalpies of reaction from enthalpies of formationThe enthalpy change of a chemical reaction equals the sum of the enthalpies of formationof all the products minus the sum of the enthalpies of formation of all the reactants.

Example :The enthalpies of formation of carbon dioxide, ethane and water are -393 kJmol-1, -85kJmol-1and -286 kJmol-1respectively. Calculate the enthalpy of combustion ethane.

SEK. MEN. KEB. SULTAN ISMAIL, JOHOR BAHRU.PHYSICAL CHEMISTRY/ UPPER SIX/ 2013TOPIC : CHEMICAL ENERGETICS

Hess Law

1. The total enthalpy change of a chemical reaction depends on the difference between the enthalpy of the products and that of the reactants. It does not depend on how the reaction is completed.H = [ H (products) ] [ H (reactants) ]2. This concept is given in Hess Law which states that the overall enthalpy change of a reaction is dependent of its pathway.3. Consider the thermochemical cycle in the diagram :

H1

XW

H5

H3H2

H4ZY

The cycle above shows three different ways to convert substance W to Z. By Hess Law :H1+ H3 = H2+ H4 = H5

4. Hess law is especially useful to calculate enthalpy changes that cannot measured directly from experiments.SEK. MEN. KEB. SULTAN ISMAIL, JOHOR BAHRU.PHYSICAL CHEMISTRY/ UPPER SIX/ 2013TOPIC : CHEMICAL ENERGETICS

Hess Law

1. Many reactions have enthalpy changes that cannot be found directly from a single experiment. Hess law is a method for the indirect determination of enthalpy changes.2. Hess law states that the overall enthalpy change in a chemical reaction is constant, and independent of the path taken, provided the initial and final stages remain unchanged.3. For example, for the reaction A B, according to Hess law, the enthalpy change (Hr) will be the same, whether by a one-step reaction A B; or by a series of steps A X Y B

YAHrBor

H2Xenthalpy H1H3

H3H1A

HrB XH2Y

Hr = H1 + H2 + H34. Different methods based on Hess law can be used to calculate enthalpy changes, depending on the information provided, as shown in the following example:

(a) Method 1 (Constructing an enthalpy diagram)

Question : Calculate the standard enthalpy change of combustion of methane from the following data :Standard enthalpy change of formation (in KJ mol-1);CH4 = -74.4 ; CO2 = -393.5 ; H2O = -285.8

Method 2 (Combining equations)

Question : The standard enthalpy changes of formation of SnO2 , CO and CO2 are -581 kJ mol-1; -111 kJ mol-1 ; and -394 kJ mol-1 respectively.Calculate the enthalpy change of the reaction :SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)

Method 3 (Using standard enthalpies of formation)The enthalpy change of a reaction can be calculated using the formula :Hreaction = Hf (products) - Hf (reactants)

Question : Use the data below to calculate the standard enthalpy change for the reaction :N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)

compoundHf(kJ mol-1)CO(g)-110CO2(g)-393N2O(g)+81N2O4(g)+9.7


Bond Enthalpy

1. Bond enthalpy is the heat change required to break one mole of bonds in the molecules of a gaseous element or compound so that the resulting gaseous species exert no forces on each other. (a) this energy refers to a specific bond in a molecule, but if a molecule has more of the same type of bond (e.g. the four C-H bonds in CH4), then different bond enthalpies can occur. For example :CH4(g) CH3(g) + H(g)H = + 427 kJ mol-1CH3(g) CH2(g) + H(g)H = + 371 kJ mol-1

So, it is much more useful to know the average amount of energy needed to break a particular bond. In this case, the process of breaking all the bonds in methane ending up with gaseous atoms. This process could be written as :CH4(g) C(g) + 4H(g)H = + 1648 kJ mol-1

The enthalpy change for this reaction is +1646 kJ mol-1, so the mean (or average) bond enthalpy is + 1648/4 = =412 kJ mol-1

(b) Bond enthalpy is always endothermic as energy must be put in to break any chemical bond. (Making a bond is always exothermic as it is the opposite of breaking a bond)

(c) The smaller the bond energy, the weaker the bond and the easier it is to break.

2. Using bond enthalpies to calculate the enthalpy change of a reaction.

(a) Example 1Use the bond enthalpies given to calculate the enthalpy change for the hydrogenation of ethene.C2H4(g) + H2(g) C2H6(g)BondBond enthalpy (kJ mol-1)H H436C H413C C346C = C611

(i) Construct an enthalpy diagram:(ii) Calculate the total energy required to break bonds of reactants, H2(iii) Calculate the total energy required to break bonds of products, H3(iv) Applying Hess law : H1 =

(b) Example 2

Use the data given to find the enthalpy change for each of the following reaction.(i) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g)(ii) N2(g) + 3H2(g) 2NH3(g)

BondBond enthalpy (kJ mol-1)C H+413C = O+740O = O+497O H+463C = C+612C C+347N N+945H H+436N H+391


Standard Enthalpy Change of Formation and Stability

1. Generally, substances with large negative standard enthalpy change of formation are more stable than those with large positive standard enthalpy change of formation.

2. The standard enthalpy change of formation of water is -286 kJ mol-1H2(g) + Hf = -286 kJ mol-1 This indicates that water is energetically more stable than its elements hydrogen and oxygen.

3. On the other hand, the standard enthalpy change of formation of nitrogen trichloride, NaCl3 , is + 230 kJ mol-1 + Hf = +230 kJ mol-1 Hence, nitrogen trichloride is energetically less stable than its elements nitrogen and chlorine.

4. The table below lists the standard enthalpy change of formation of the hydrides of chlorine, bromine and iodine.CompoundHClHBrHI

Hf / kJ mol-1-92.0-36.0+26.5

5. From the standard enthalpy change of formation above we can predict that HI is least stable of the three hydrides followed by HBr and HCl. This indeed is the case if we look at the percentage dissociation of the hydrides into their constituent elements at different temperatures.Temperature /0CPercentage dissociation

HClHBrHI

200--13

10000.010.425

20000.44.230


Standard Enthalpy of Neutralisation Between Strong Acids and Strong Bases1. From the table given, the standard enthalpy of neutralization between strong acids and strong bases is a constant (-57.3 kJmol-1).

2. This is because all strong acids and strong bases dissociate completely in water to form aqueous ions :NaOH (aq) Na+(aq) + OH-(aq)

3. Neutralisation between strong acids and strong bases involves the same reaction, which is the combination of H+(aq) and OH-(aq)to form H2O. The other ions present are just spectator ions (in bold face) that do not take part in the reactions.Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)Na+(aq) + Cl-(aq) + H2O (l)

K+(aq) + OH-(aq) + H+(aq) + NO3-(aq)K+(aq) + NO3-(aq) + H2O (l)

The overall reaction is :H+(aq) + OH-(aq) H2O (l)Hence, the heat released is the same.

4. However, the standard enthalpy of neutralization between sulphuric acid and sodium hydroxide is more exothermic than expected.

H= -66.8 kJmol-15. However, the heat released when 1 mole of water is formed from the above reaction is only 57.3 kJmol-1. So, where the extra 9.5 kJmol-1comes from ?

6. This extra heat released is the enthalpy change of dilution of sulphuric acid.

7. When aqueous sodium hydroxide is added to sulphuric acid, the acid is diluted in the process.

8. The heat released due to dilution is significant. This make the standard enthalpy change of dilution more exothermic. [The H for the reaction = heat of neutralization + heat of dilution]9. Heat of dilution of other acids is insignificant and is ignored.

SEK. MEN. KEB. SULTAN ISMAIL, JOHOR BAHRU.PHYSICAL CHEMISTRY/ UPPER SIX/ 2011TOPIC : THERMOCHEMISTRY

Standard Enthalpy of Neutralisation Between Weak Acids or Weak Bases

1. The standard enthalpy of neutralization involving weak acids or weak bases is less than 57.3 kJmol-1. Example :

2. This is because weak acids are only partial dissociated in water.

On addition of a strong base such as NaOH, the OH- from NaOH will react with the H+ from the dissociation of .

3. Removal of H+will cause the above equilibrium to shift to the right where more molecules would dissociate. However, the dissociation process requires absorption of heat energy.

As a result, the total heat released is less than expected.

4. Similarly, for the case of ammonia, a weak base :

On addition of a strong acid such as HCl, the H+from HCl will react with from ammonia to form water.

5. Removal of causes the above equilibrium to shift to the right, and more ammonia molecules dissociate. However, this dissociation requires absorption of energy.

As a result, the total heat released is less than expected.

Standard Enthalpy of Neutralisation Involving Hydrofluoric Acid, HF

1. Hydrofluoric acid, HF, is a weak acid that dissociates partially in water. HF (aq)H+(aq) + F- (aq)

2. Hence, we would expect the standard enthalpy of neutralization between HF and a strong base such as NaOH to be less than 57.3 .

3. However, the experimental standard enthalpy of neutralization between NaOH and HF is more negative than expected.

This is in contradiction to our discussions above.

4. This is because the dissociation of hydrogen fluoride molecule is an exothermic process.

5. Dissociation of HF involves the breaking of the H-F bond and energy needs to be absorbed. However, the F- ions, due to its small size, has high hydration energy.

F-(g) + aq F-(aq) This makes the whole process of dissociation exothermic.

HF (aq)H+(aq) + F- (aq) As a result, the standard enthalpy change of neutralization is more negative than expected.

Standard Enthalpy Of Solution

1. The heat evolved or absorbed when 1.0 mole of solute dissolves in water to form an infinitely dilute solution is known as the standard enthalpy of solution. For example, NaCl (s) + water NaCl (aq)H = standard enthalpy of solution

2. A solution is said to be at infinite dilution if the volume of water used is sufficient so that addition of more water produces no further heat change.

Question :

1. When 10.6 g of anhydrous sodium carbonate are dissolved in 100cm3 of water, the maximum temperature rise recorded is 50C. Calculate the enthalpy of solution. (The specific heat capacity of water is 4.2 Jg-1 k-1)

H = mc = 100 x 4.2 x 5 = 2100 J Number of moles of Na2CO3 (Mr = 106) used = Therefore standard enthalpy of solution = 2100 x J Na2CO3(s) + water (infinite volume) = Na2CO3(aq) Hneut = - 21.0 kJ mol-1SEK. MEN. KEB. SULTAN ISMAIL, JOHOR BAHRU.PHYSICAL CHEMISTRY/ UPPER SIX/ 2011TOPIC : THERMOCHEMISTRY

Standard Enthalpy Change of Hydration (Hh)1. The standard enthalpy change of hydration is defined as the enthalpy change or heat energy at constant pressure when 1 mol of gaseous ions dissolves in infinite amount of water (that is, an infinitely dilute solution is formed with no further enthalpy change) at standard conditions.

2. For a cation X+ (g) + aq X+(aq) ;Hh= standard enthalpy change of hydration

Water molecules evelope a cation through ion-dipole bonds or attractions. Hence, the cation is hydrated . This denoted as X+(aq)hydrated aqueous

For a anion Y-(g) + aq Y-(aq) ;Hh= standard enthalpy change of hydration

Water molecules evelope an anion through ion-dipole bonds or attractions. Hence, the cation is hydrated . This denoted as Y-(aq)hydrated aqueousNote : aq represents water. Hence, the reaction equation of hydration can also be

waterwater written as :

X+ (g) X+(aq)andY-(g) Y-(aq)3. Since hydration occurs through the formation of ion-dipole bonds or attractions between the polar water molecules and charged ions, the standard enthalpy change of hydration is always exothermic.

4. The magnitude of the enthalpy change of hydration depends on the charge density of the ion.

5. A larger charge density will exert a stronger ion-dipole attraction and hence, produce a larger magnitude of enthalpy change of hydration.

6. Charge density of an ion depends on the magnitude of the charge carried as well as on its ionic radius.

7. Charge density is directly proportional to the charge carried by the ion but inversely proportional to the ionic radius. Hence, mathematically,

8. An ion with a large magnitude of charge and a small ionic radius has a very high charge density. Hence, has a large magnitude of enthalpy change of hydration.

Example :

Al3+ ions have a high charge density due to a very large magnitude of charge (+3) anda small ionic radius. Hence, the magnitude of its standard enthalpy change ofhydration is large.H+ ions have a small magnitude of charge but at the same time has a very small ionicradius. Hence, its charge density is still considered to be large. It standard enthalpychange of hydration is expected to be large. In fact, the standard enthalpy change of

hydration of H+ ions is -1090 .

Question :The magnitude of the standard enthalpy change of hydration of magnesium ions is

1920 .(a) Write a thermochemical equation depicting the standard enthalpy change of hydration.(b) Predict the standard enthalpy change of hydration of calcium ions.Solution(a) Standard enthalpy change of hydration is always exothermic with Hh being negative.Mg2+(g) + aq Mg2+(aq)Hh = -1920 kJ mol-1(b) - the charge on both calcium and magnesium ions is the same. - the ionic radius of the Ca2+ ion is bigger than Mg2+ ion. - hence, the charge density of Ca2+ ion < Mg2+ ion - consequently, the magnitude of the standard enthalpy change of hydration of Ca2+ ion is expected to be smaller than the magnesium ions.SEK. MEN. KEB. SULTAN ISMAIL, JOHOR BAHRU.PHYSICAL CHEMISTRY/ UPPER SIX/ 2013TOPIC : CHEMICAL ENERGETICS

Standard Enthalpy Change of Atomisation , Hatm

1. The standard enthalpy change of atomisation of an element is the enthalpy change when one mole of free gaseous atom is formed from the element under standard conditions.2. The standard enthalpy change of atomisation of sodium is +107 kJmol-1. This refers to the following process :Na (s) Na (g)H= +107 kJmol-13. The standard enthalpy change of atomisation of chlorine is +121kJmol-1. This refers to:

H= +121 kJmol-14. The standard enthalpy change of atomisation of the Noble gases (He, Ne, Ar, Kr, Xe and Rn) is zero, because all of them exist as monatomic gases at standard conditions.5. The standard enthalpy change of atomisation of diatomic gases (e.g. O2, Cl2andN2,) is equal to half the value of their bond energies.

6. The processes that are involved in the atomisation of a solid metal comprise the following steps (sodium is used as an example) :(a) Heat is absorbed by one mole of the solid to increase its temperature from 298 K to its melting point (371 K).(b) The enthalpy of fusion is absorbed (+2.6kJmol-1) to change one mole of the solid to liquid at its melting point.(c) Heat is absorbed to increase the temperature of one mole of the liquid sodium to its boiling point (1156 K).(d) The enthalpy of vaporisation is absorbed (+96.8kJmol-1) to change one mole of liquid sodium to vapour at its boiling point.(e) This is summarized in the diagram below :

Na (g)enthalpy of vaporisation (+96.8 kJmol-1)Na (l)at boiling point(1156 K)

Enthalpy change of atomisationHeat absorbedNa (l)at 371 Kenthalpy of fusion (+2.6 kJmol-1)Na (s)at melting point(371 K)Heat absorbedNa (s)at 298 K

Lattice Energies For Simple Ionic Crystals1. The lattice energy (HL)of a given ionic compound is the enthalpy change on forming one mole of the compound from its isolated gaseous ion. For example,Na+(g) + Cl-(g) NaCl (s)HL= -781 kJmol-1

2. All lattice energies are negative, that is, heat energy is released when oppositely charged ions are brought together.

3. Lattice energy is a measure of the strength of interior attractions. An ionic compound with small and highly charged ions has the strongest electrostatic attraction and thus has the highest lattice energy.

4. Lattice energy is proportional to the product of the charges on the ions and inversely proportional to the sum of the ionic radii of the cation and anion.

Lattice energy

This means that lattice energy Increases as the ionic charge increases Increases as the sum of the ionic radii of the cation and anion decreasesThe sum of the ionic radii of the cation and anion is also known as the interior distance.

5. The energy required to separate one mole of the ionic compound into its constituent ions in the gas phase has the same magnitude but opposite in sign to the lattice energy. Thus,Na+(g) + Cl-(g) NaCl (s)HL= -781 kJmol-1and the reverse reaction is endothermicNaCl (s) Na+(g) + Cl-(g)H= +781 kJmol-16. Applications of lattice energiesLattice energy can be used to explain : The melting points of ionic compounds The presence of covalent character in some ionic compounds.

Melting Points and Lattice EnergiesLattice energy is a measure of the strength of interionic attractions. The greater the lattice energy, the higher the melting point.

Determining The Lattice Energy of Ionic Compounds

1. The lattice energy of an ionic compound cannot be determined directly from experiment because, it would involve bringing one mole of Na+ gaseous ions and one mole of Cl- gaseous ions together and combining them to form NaCl solid.

2. Instead, the lattice energy has to be calculated from other known enthalpy changes (that can be determined from experiments) and by using Hess Law.

3. The thermochemical cycle used to determine the lattice energy is called the Born-Haber cycle.

4. Example : sodium chlorideThe standard enthalpy change of formation of sodium chloride = -411kJmol-1. This refers to following process :

Na (s)+ H= -411 kJmol-1This reaction can be assume as comprising the following hypothetical steps. The actual reaction does not follow these individual steps. However, it makes our calculation easier to follow. (a) Atomisation of sodiumNa (s) Na (g)H= +108 kJmol-1(b) Ionisation of sodium :Na (gs) Na+(g) + eH= +494 kJmol-1 (c) Atomisation of chlorine

H= +121 kJmol-1(d) Formation of Cl- ion : first electron affinity.

Cl (g) + e H= -364 kJmol-1(e)

Combination of Na+(g) to form Na+(g) + Cl-(g) NaCl (s)Hlattice

5. The Born Haber cycle is then constructed as follows :

Na+(g) + Cl (g)-364+121Na+(g) + Cl-(g)

Na+(g) + +494

H(lattice)Na (g) + +108

Na (s) +

Na+Cl-(s)

Base on Hess law :(+108) + (+494) + (+121) + (-364) + H(lattice) = -411

H(lattice) = -770 kJmol-1

6. Sometimes, the Born-Haber can be drawn as shown below :

Na+(g) Cl-(g)

+494-364Na (g) Cl(g)H(lattice)+108+121-411

Na (s)Na+Cl-(s)

Lattice energy of sodium chloride from the Born-Haber cycle.

HlNa+(g) + Cl-(g)NaCl (s)

HeaNa+(g) + Cl (g)

Hie

Na (g) + Cl (g)Hf

Ha[Cl2(g)]

Na (g) +

Ha[Na (s)]

Na (s) +

Using Hess law Hl = -Hea - Hie - Ha[Cl2(g)]- Ha[Na (s)] + HfHl = lattice energy of crystalline sodium chlorideHea = first electron affinity of chlorineHie = first ionization energy of sodiumHa[Cl2(g)] = standard enthalpy change of atomisation of chlorineHa[Na (s)] = standard enthalpy change of atomisation of sodiumHf = standard enthalpy change of formation of sodium chloride

Question :

The following enthalpies are given :Standard enthalpy of atomisation of magnesium, Ha= +146 kJmol-1Standard enthalpy of atomisation of chlorine, Ha= +122 kJmol-1First ionization energy of magnesium, Hie= +740kJmol-1Second ionization energy of magnesium, Hie= +1450kJmol-1First electron affinity of chlorine, Hea= -348kJmol-1Standard enthalpy change of formation of magnesium chloride, Hf= -566kJmol-1Construct a Born- Haber cycle and determine the lattice energy of magnesium chloride

22 | GSS

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