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
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
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
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
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