THIRD LAW OF THERMODYNAMICS

 

The entropy of a pure crystalline substance increases with increase of temperature, because molecular motion increases with increase of temperature and vice - versa.

 

Or the entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature is approached. This is third law of thermodynamics.

 

THERMOCHEMISTRY

 

The branch of chemistry which deals with energy changes involved in chemical reactions is called thermochemistry. The energy change that occurs in a chemical reactions is largely due to change of bond energy.

  

Change of internal energy in a chemical reaction

 

Let us consider a chemical reaction taking place at constant temperature and at constant volume. In such a case, w = 0 and hence from the first law

 

      ΔU = qv

 

Where qv is the heat exchanged at constant volume, or heat or enthalpy of reaction at constant volume.

 

Change of Enthalpy in a chemical reaction

 

Let qP be the heat exchanged in the chemical reaction taking place at constant pressure, Then evidently,

 

ΔH = qP = Heat or Enthalpy of reaction at constant pressure.

 

Exothermic and Endothermic reaction

 

Reaction that give out heat, i.e. which are accompanied by evolution of heat, are called exothermic reaction. In such reactions ΔH is negative. On the other hand, reaction that intake heat, i.e. which are accompanied by absorption of heat are called endothermic reactions. In these reactions ΔH is positive.

 

Solved example. Fill in the blanks with appropriate word in following:

 

                        (i) Combustion of reactions are usually ……………………..

 

                        (ii) Combustion of F2 in oxygen is ……………………..

 

Solution:       (ii) Exothermic

 

                     (iii) Endothermic

 Enthalpy of reaction

 

It is the enthalpy change taking place during the reaction when the number of moles of reactants and products are same as the stoichiometric coefficient indicates in the balanced chemical equation. The enthalpy change of the reaction depends upon the conditions like temperature, pressure etc under which the chemical reaction is carried out. Therefore, it is necessary to select the standard state conditions. According to thermodynamics conventions, the standard state refers to 1 bar pressure and 298 K temperature. The enthalpy change of a reaction at this standard state conditions is called standard enthalpy of the reaction.(ΔHo)

 

Different types of enthalpy

 

(i)   Enthalpy of formation: Enthalpy change when one mole of a given compound is formed from its elements.

 

            H2(g) + 1/2O2(g) ———> 2H2O(l),                   ΔH = –890.36 kJ / mol

  

Exercise.

 

      Calculate  for chloride ion from the following data:

 

      1/2 H2 (g) + 1/2 Cl2 (g) ———> HCl (g)        ΔH = –92.4 KJ

 

      HCl (g) + nH2O ——> H+ (aq) + Cl– (aq)      ΔH = –74.8 KJ

 

      ΔH1o (H+(aq)) = 0.0 KJ

 

(ii) Enthalpy of combustion: Enthalpy change when one mole of a substance is burnt in oxygen.

 

      CH4 + 2O2(g) ———> CO2 + 2H2O(l),      ΔH = –890.36 kJ / mol

  

Exercise.

 

      The heat liberated on complete combustion of 7.8 g benzene is 327 KJ. This heat has been measured at constant volume and at 27°C. Calculate heat of combustion of benzene at constant pressure at 27°C.

  

(iii) Enthalpy of Neutralization: Enthalpy change when one equivalent of an acid is neutralized by a base or vice – versa in dilute solution. This is constant and its value is –13.7 kcal for neutralization of any strong acid by a base since in dilute solutions they completely dissociate into ions.

 

            H+ (aq) + OH– (aq) ——> H2O(l),                  ΔH = –13.7 kcal

 

      For weak acids and bases, heat of neutralization is different because they are not dissociated completely and during dissociation some heat is absorbed. So total heat evolved during neutralization will be less.

 

      e.g. HCN + NaOH ——> NaCN + H2O,               ΔH = –2.9 kcal

 

      Heat of ionization in this reaction is equal to (–2.9 + 13.7) kcal = 10.8 kcal

  

Solved example.  Heat of neutralization of a strong acid by a strong base is equal to ΔH of

 

                        (A) H+ + OH-—> H2O                       

 

                        (B) H2O + H+ —> H3O+

 

                        (C) 2H2 + O2 = 2H2O                       

 

                        (D) CH3COOH+ NaOH = CH3COONa + H2O

  

Solution:       (A) Since heat of neutralization of strong acid and strong base is equal to the heat of formation of water.

 

                        i.e., NaOH + HCl —> NaCl + H2O + Q

 

                        Were Q = heat of neutralization

 

                        => Na+ + OH– + H+ + Cl– —> Na++Cl– + H2O + Q              

 

                        => H+ + OH– —> H2O + Q

  

(iv) Enthalpy of hydration: Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combines with the requisite no.of mole of water to form a specific hydrate. For example, the hydration of anhydrous copper sulphate is represented by

 

            CuSO4(s) + 5H2O (l) ——> CuSO45H2O(s),                        ΔH° = –18.69 kcal

  

SOLVED EXAMPLE. Ionisation energy of Al = 5137 kJ mole–1 (ΔH) hydration of Al3+ = – 4665 kJ mole–1. (DH)hydration for Cl– = – 381 kJ mole–1. Which of the following statement is correct

 

                        (A) AlCl3 would remain covalent in aqueous solution

 

                        (B) Only at infinite dilution AlCl3 undergoes ionisation

 

                        (C) In aqueous solution AlCl3 becomes ionic

 

                        (D) None of these

 

Solution:       If AlCl3 is present in ionic state in aqueous solution, therefore it has Al3+ & 3Cl–ions

 

                        Standard heat of hydration of Al3+ & 3Cl- ions

 

                        = – 4665 + 3 × (–381) kJ mole–1 = -5808 kJ/mole

 

                        Required energy of ionisation of Al = 5137 kJ mole–1

 

                        ∴ Hydration energy overcomes ionisation energy

 

                        ∴ AlCl3 would be ionic in aqueous solution

 

                        Hence (C) is the correct answer.

 

(v) Enthalpy of Transition:  Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.

 

            C (graphite) ——> C(diamond),                                         ΔH° = 1.9 kJ/mol

 

Solved example. The heat of transition for carbon from the following is

 

                        CDiamond + O2(g) ——> CO2(g)  ΔH = – 94.3 kcal

 

                        CAmorphous + O2(g) ——> CO2(g)  ΔH = – 97.6 kcal

 

                         (A) 3.3 kJ / mol                                 (B) 3.3 kcal / mol

 

                         (C) –3.3 kJ / mol                              (D) – 3.3 kcal / mol

 

Solution:       Given

 

                        CD + O2(g) ——> CO2(g)                          ΔH = –94.3 kcal/mole    …(1)

 

                        CA + O2(g) ——> CO2(g)                          ΔH = – 97.6 kcal/mole   …(2)

 

                        ———————————————————————————

 

                        Subtracting equation (2) from equation (1):

 

                        CD – CA ——> 0;                                  ΔH = +3.3 kcal/mole

 

                        CD ——> CA                                          ΔH = +3.3 kcal/mole           

 

                        (B)

  

Solved example. From the reaction P(white) ——> P (Red):  ΔH = - 18.4 kJ, It follows that

 

                        (A) Red P is readily formed from white P

 

                        (B) White P is readily formed from red P

 

                        (C) White P can not be converted to red P          

 

                        (D) White P can be converted into red P and red P is more stable

  

Solution:                   (D)

 HESS’S LAW

 

This law states that the amount of heat evolved or absorbed in a process, including a chemical change is the same whether the process takes place in one or several steps.

 

Suppose in a process the system changes from state A to state B in one step and the heat exchanged in this change is q. Now suppose the system changes from state A to state B in three steps involving a change from A to C, C to D and finally from D to B. If q1, q2 and q3 are the heats exchanged in the first, second and third step, respectively then according to Hess’s law

 

            q1 + q2 + q3 = q

 

Hess’s law is simply a corollary of the first law of thermodynamics. It implies that enthalpy change of a reaction depends on the initial and final state and is independent of the manner by which the change is brought about.

 

Sample problem.

 

 

                        In this case express ΔH in terms of ΔH1, ΔH2, ΔH3 .

 

Solution:       ΔH = ΔH1 + ΔH2 + ΔH3

 

Sample problem.    H2O (l) ® H2(g) + O2(g)    DH = + 890.36 kJ / mole

 

                        What is DH for H2O (l) from its constituent elements

 

Solution:       H2O(l) —> H2(g) + 1/2 O2 (g)     ΔH = + 890.36 kJ / mole

 

                        H2(s) + 1/2 O2 (g) —> H2O(l)  ΔH = – 890.36 kJ / mole

 

                        ∴ ((ΔHf))H2O  = –890.36 kJ / mole

 

Exercise .

 

(i) C + 1/2 O2 —> CO2(g)   ΔH = –94 Kcals

 

      C + 1/2 O2 —> CO(s)    ΔH = –26.4 Kcals

 

     CO + 1/2 O2 —> CO2(g)    ΔH =?

 

(ii)       What is heat evolved using neutralisation of HCN by a strong base? Heat of ionization of HCN is 10.8 Kcal.

 APPLICATION OF HESS’S LAW

 

1.   Calculation of enthalpies of formation

 

There are large number of compounds such as C6H6, CO, C2H6 etc whose direct synthesis from their constituent element is not possible. Their ΔH0

f values can be determined indirectly by Hess’s law. e.g. let us consider Hess’s law cycle for CO2 (g) to calculate the ΔH0

f of CO(g) which can not determined otherwise.

 

      C(s) + O2 (g) ———> CO(g) + 1/2 O2(g)       ΔH1 = ?

 

      CO(g) +1/2 O2 (g) ———> CO2 (g)               ΔH2 = –283 KJ/mole

 

      C(s) + O2 (g) ———> CO2 (g)                      ΔH1 = –0393 KJ/mole

 

 According to Hess’s law,

 

      ΔH3 = ΔH1 + ΔH2 or ΔH1 = ΔH3 – ΔH2

 

      = –393 – (–283)  => –110 KJ/mole

 

2.   Calculation of standard Enthalpies of reactions            

 

From the knowledge of the standard enthalpies of formation of reactants and products the standard enthalpy of reaction can be calculated using Hess’s law.

 

      According to Hess’s law

 

      

 

      

 

      ΔHo = [sum of statndard enthalpies of formation of products] – [sum of standard enthalpies of formation of reactants]

 

3.   In the calculation of bond energies

 

BOND ENERGY

 

Bond energy for any particular type of bond in a compound may be defined as the average amount of energy required to dissociate one mole, viz Avogadro’s number of bonds of that type present in the compound. Bond energy is also called the enthalpy of formation of the bond.

 

Calculation:      

 

For diatomic molecules like H2, O2, N2, HCl, HF etc, the bond energies are equal to their dissociation energies. For polyatomic molecules, the bond energy of a particular bond is found from the values of the enthalpies of formation. Similarly the bond energies of heteronuclear diatomic molecules like HCl, HF etc can be obtained directly from experiments or may be calculated from the bond energies of homonuclear diatomic molecules.

 

Sample problem.    Calculate the bond energy of HCl. Given that the bond energies of H2 and Cl2 are 430 KJmol-1 and 242 KJ mol-1 respectively and ΔH0

f for HCl is -91 KJ mol-1.

 

Solution:       H2(g) ——> 2H(g)              ΔH = +430 KJmol–1              ...(i)

 

                     Cl2(g) ——> 2Cl(g)             ΔH = +242 KJmol–1              ...(ii)

 

                     HCl(g) ——> H(g) + Cl(g)    ΔH = ?                                 ...(iii)

 

                     For the reaction (iii)

 

 

                    ΔH = 427 KJ mol-1

 

Sample problem.    Given that

 

                        2H2(g) + O2(g) ——> H2O(g) , DH = –115.4 kcal the bond energy of H–H and O = O bond respectively is 104 kcal and 119 kcal, then the O–H bond energy in water vapour is

 

                         (A) 110.6 kcal / mol                               (B) –110.6 kcal

 

                        (C) 105 kcal / mol                                   (D) None

 

Solution:             We know that heat of reaction

 

                        ΔH = ΣB.E. (reactant) – ΣB.E (product)

 

                        For the reaction,

 

                        2H–H(g) + O = O (g) ——> 2H – O–H(g)

 

                        ΔH = –115.4 kcal, B.E. of H–H = 104 kcal

 

                        B.E. of O=O = 119 kcal

 

                        Since one H2O molecule contains two O–H bonds

 

                        –115.4 = (2 × 104) + 119 – 4 (O–H) bond energy

 

                        ∴4 (O–H) bond energy = (2 × 104) + 119+115.4

 

                        i.e., O–H bond energy = (2×104) + 119 + 115.4 / 4 = 110.6 kcal mol–1

 

                        Hence, (A) is correct. 

 

Sample problem. Given the bond energies of N º N, H – H and N – H bonds are 945, 436 and 391 kJ/mol respectively, the enthalpy of the reaction.        

 

                        N2(g) + 3H2(g) ——> 2NH3(g) is

 

                        (A) – 93kJ                                          (B) 102kJ

 

                        (C) 90kJ                                              (D) 105kJ

 

Solution:             (A)

 

Exercise.

 

      Estimate the average S – F bond energy in SF6. The values of standard enthalpy of formation of SF6(g), SF(g) and F(g) are 1100, 275 and 80 KJ/mole respectively.

 

LATTICE ENERGY OF AN IONIC CRYSTAL (BORN–HABER CYCLE)

 

      

                  Born - Haber cycle

 

The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from its gaseous ions, is known as Lattice energy.

 

Step 1:      Conversion of metal to gaseous atoms

 

                  M(s) ——> M(g) ,                    ΔH1 = sublimation

 

Step 2:      Dissociation of X2 molecules to X atoms     

 

                  X2(g) ——> 2X (g),                    ΔH2 = Dissociation energy

 

Step 3:      Conversion of gaseous metal atom to metal ions by losing electron                 

 

                  M(g) ——> M+ (g) + e–,            ΔH3 = (Ionization energy)

 

Step 4:      X(g) atoms gain an electron to form X– ions      

 

                  X(g) + e– ——> X–(g),              ΔH4 = Electron affinity  

 

Step 5:      M+ (g) and X– (g) get together and form the crystal lattice

 

                  M+ (g) + X– (g) ——> MX(s)      ΔH5 = lattice energy

 

                  Applying Hess’s law we get

 

                  ΔH1 + 1/2 ΔH2 + ΔH3 + ΔH4 + ΔH5 = ΔHf (MX)

 

                  On putting the various known values, we can calculate the lattice energy.

 

Sample problem.    What is the expression of lattice energy (U) of CaBr2 using BornHaber cycle?

 

Solution:

 

                         = S + IE1 + IE2 + D - 2E.A – UCaBr2

 

Sample problem.    What is the relation between ΔH and ΔE in this reaction?

 

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

 

Solution:       ΔH = ΔE + ΔnRT

 

                        Δn = no. of mole of products - no. of moles of reactants = 1– 3 = –2

 

                        ΔH = ΔE – 2RT

 

Sample problem.    What is the expression of lattice energy (U) of CaBr2?

                        Using Born Haber cycle?

 

Solution:

 

                          ΔHf = S + IE1 + IE2 + D - 2E.A –

 

Sample problem. The lattice energy of solid NaCl is 180 kcal/mole. The dissolution of the solid in water in the forms of ions is endothermic to the extent of 1 kcal/mol. If the solution energies of Na+ and Cl– are in the ratio 6:5, what is the enthalpy of hydration of Na+ ion?

 

                        (A) - 85.6 kcal/mol                            (B) -97.5 kcal/mol

 

                        (C) 82.6 kacl/mol                              (D) +100 kcal/mol

 

Solution:       (B)

 

Exercise .     

 

(i) For a system C(s) + O2(g) = CO2(g) which of the following is correct:

 

      (a) ΔH = ΔE  (b) ΔH >ΔE       (c) ΔH <ΔE (d) ΔH = 0

 

(ii) C6H6(l) + 15/2 O2(g) —> 3H2O + 6CO2(g) ΔH = –3264.4 kJ mole–1.

 

      What is the energy evolved when 7.8 gm of benzene is burnt in air?