Types of Enthalpy reaction

Types of Enthalpy reaction are listed below:-

(i) Enthalpy of Formation

It may be defined as, “The quantity of heat evolved or absorbed when one mole of the compound is formed from its elements”. It is expressed as $\Delta Hf$

E.g.

$\dfrac{1}{2} H_2 (g) + \dfrac{1}{2}I_2 (s) \to HI(g) \hspace{5mm} \Delta H_f = +6.2 kcal$

$H_2 (g) + \dfrac{1}{2} O_2(g) \to H_2 O (l) \hspace{5mm} \Delta H_f = -68 kcal$

$\dfrac{1}{2} N_2 (g) + \dfrac{3}{2} H_2 (g) \to NH_3 (g) \hspace{5mm} \Delta H_f = -11 kcal$

It is important to note that the thermochemical equation should be balanced in such a way that it represents the formation of one mole of the substance only.

The value of heat of formation at 298 K and 1 atm pressure is called standard heat of formation. $(\Delta H^0)\Delta H^0$ for free state of elements and standard state is taken as Zero.

(ii) Enthalpy of Combustion

It may be defined as, “The quantity of heat evolved when one mole of the substance is completely oxidized”.

Eg:

$(Delta H = -ve)$ always exothermic

$\underset{diamond}{C} + O_2 \to CO_2 + 95.5 kcal$

$CH_4 (g) + 2O_2 (g) \to CO_2 (g) + 2H_2 O (l) \hspace{2mm} \Delta H = -213 kcal$

$C(diamond) + O_2 (g) \to CO_2 (g) \hspace{10mm} \Delta H = -95.5 kcal$

$C(graphite) + O_2 (g) \to CO_2 (g) \hspace{10mm} \Delta H = -94.0 kcal$

$H_2 (g) + \dfrac{1}{2} O_2 (g) \to H_2O (l) \hspace{10mm} \Delta H = -68.4 kcal$

The heat of combustion is very useful in:

(a) Calculating the heat of formation which is otherwise not possible in some cases.

(b) Calculating the calorific value of fuels.

$\text{calorific value} = \dfrac{\Delta H}{Atomic wt. or molar mass}$

E.g.: Combustion of butane evolves -2878.8 k] per mole of heat so it’s calorific value (C.V.) is

$C.V = \dfrac{-287.8}{58} = -49.62 kJ$

Substance Calorific value in

Keroscene 48

L.P.G. j 55

Milk 3.11

Butter 30.5

(iii) Enthalpy of Solution

It may be defined as, “The quantity of heat evolved or absorbed when one mole of a solute is dissolved completely in large excess of water, so that further dilution of solution does not produce any heat change”.

E.g.

$KCl(s) + nH_2O \to KCl(aq) \Delta H = +4.40 kcal$

$HCl (g) + nH_2O \to HCl (aq) \Delta H = -39.2 kcal$

(iv) Enthalpy of Neutralisation

It may be defined as, “the quantity of heat evolved when one equivalent (or equivalent mass) of an acid is completely neutralised by one equivalent (or equivalent If mass) of a base in dilute solution”.

E.g.

$HNO_3 (aq) + NaOH (aq) \to NaNO_3 (aq) + H_2O (l) \Delta H = -13.69 kcal$

$HCl (aq) + NaOH (aq) \to NaCl(aq) + H_2O (l) \Delta H = -13.68 kcal$

The heat of neutralisation of strong acid and a strong base is taken as 13.7 kcal. or 57kJ.

On the basis of electrolytic dissociation theory, it has been clearly explained that this heat of neutralisation is merely the heat of formation of water from $H^+$ of an acid and $OH^-$ of a base.

$H^+(aq) + OH^- (aq) \to H_2O (l) \Delta H = -13.7 kcal$

The heat of neutralisation in case of a weak acid or a weak base is somewhat less than 13.7 kcal because some energy is used up in dissociating the weak electrolyte. The difference in the values gives the dissociation energy of the weak acid or a weak base.

$NH_4OH(w. base) + HCl(Strong acid) \leftrightharpoons NH_4Cl + H_2O \Delta H = -12.3 kcal$

Here 1.4 kcal heat is absorbed in the dissociation of weak base $NH_4OH$

$CH_3COOH (weak acid (aq)) + NH_4OH(weak base(aq)) \leftrightharpoons CH_3COOHNH_4 (aq) + H_2O(l) \Delta H = -11.9 kcal$

Here 1.8 kcal heats are absorbed in the dissociation of both the weak electrolytes.

Heat of neutralisation is measured in lab by using pothythene or polystyrene bottles.

(v) Enthalpy of Neutralisation

It may be defined as, “The quantity of heat absorbed when one mole of a substance completely dissociates into its ions”.

E.g.

$H_2O(l) \to H^+ + oh^- \Delta H = 13.7 kcal$

(vi) Enthalpy of Dilution

It may be defined as, “The quantity of heat evolved or absorbed when solution containing one mole of a solution is diluted from one concentration to another”.

$KCl (S) + 20H_2O \to KCl(20 H_2O) \Delta H_1 = +3.8 kcal$

$KCl(s) + 200 H_2O \to KCl(200 H_2O) \Delta H_2 + 4.44 kcal$

$\therefore$ Heat of dilution = $\Delta H_2- \Delta H_1 = 4.44- 3.80 = 0.64 kcal$

(vii) Enthalpy of Precipitation

It may be defined as, “The quantity of heat evolved in the precipitation of one mole of a sparingly soluble substance on mixing dilute solutions of suitable electrolytes” .

E.g.

$Ba^{2+} (aq) + SO^{2-}_4 (aq) \to BaSO_4 (s), \Delta H = -4.66 kcal$

(viii) Enthalpy of Hydration

It may be defined as, “The quantity of heat evolved or absorbed? when one mole of an anhydrous or a partially hydrated salt is combined with the required number of moles of water to form a specific hydrated substance”.

E.g.

$CuSO_4 (s) + 5H_2O (l) \to CuSO_4. 5H_2O (s) \Delta H = -18.7 kcal$

$CaCl_2(s) + 6H_2O (l) \to CaCl_2. 6H_2O (s) \Delta H = -18.8 kcal$

(ix) Enthalpy of fusion

It is change in enthalpy during conversion of 1 mole of a substance from solid to liquid state at its melting point (mostly endothermic)

$H_2O (solid) \overset{\Delta}{\underset{m.p.} \rightarrow} H_2O(liquid)- 1.44 kcal$

$\Delta H_{fus.} = 1.44 (kcal)$

(x) Enthalpy of vaporisation

It is the change in enthalpy when 1 mole of a substance is converted from liquid to gaseous state at it’s boiling point.

$H_2O(l) \underset{B.P}{\overset{\Delta} \rightarrow} H_2O- 10.5 kcal$ mostly endothermic.

(xi) Enthalpy of Sublimation

It is change in enthalpy when I mole of a substance is converted directly from solid to Vapour (gaseous) state.

$I_2(s) \overset{\Delta}{\rightarrow} I_2- 14.9 kcal$

$\Delta H = 14.9 kcal$

$\Delta H_{sub} = \Delta H_{vap.} + \Delta_{fus.}$

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