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.(ΔH^o)

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

      ΔH₁^o (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:

(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

                        (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

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

                        C_Amorphous + 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

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

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

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

                        Subtracting equation (2) from equation (1):

                        C_D – C_A ——> 0;                                  ΔH = +3.3 kcal/mole

                        C_D ——> C_A                                          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)

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