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

ΔH^o = [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.  

                        N_2(g) + 3H_2(g) ——> 2NH_3(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 SF_6(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 = ΔH_f (MX)

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

Sample problem: What is the expression of lattice energy (U) of CaBr_­2 using BornHaber cycle?

= S + IE_­1 + IE₂ + D - 2E.A – U_CaBr2 

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 CaBr_­2?

Using Born Haber cycle?

 ΔH_f = S + IE_­1 + 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?

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