Types of Heat (enthalpy) of Reactions

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The heat or enthalpy changes taking place during the chemical reactions are expressed in different ways depending upon the nature of the reaction. The various types of enthalpies of reactions are:

Heat of Formation or Enthalpy of Formation

The heat evolved or absorbed when 1 mole of a substance is formed from its elements is called heat of formation. It is denoted by DH_f.

For example, heat of formation of carbon dioxide and methane may be expressed as:

These equations should always be written for one mole as per the definition of the substance to be formed. If for balancing, we require the coefficient 2, 3.... etc., in the equation then DH_f values should also be multiplied by the same number as discussed earlier.

Standard heat of formation

The heat of formation DH_f depends upon the condition of temperature, pressure and physical state (gas, liquid or solid) of the reactants and the products. Therefore the heat change accompanying the formation of one mole of a compound from its elements when all the substances are in their standard states (1 atm pressure and 298 K), is called the standard heat of formation. It is expressed as DH_f^o.

Since no heat changes are involved in the formation of elements form themselves in their standard states, the standard enthalpy of formation of all elements is zero.

For example, the standard enthalpy of formation (DH_f^o) for H₂O_(l)  is - 286 kJ mol^-1 i.e., when one mole of liquid water is formed from its elements H_2(g) and O_2(g) at 298 K and 1 atm pressure, then 286 kJ mol^-1 of heat is released. The negative value of DH_f^o indicates the formation of a stable compound.

Standard heat of reaction from standard heats of formation

The knowledge of standard heats of formation of various substances can be used to calculate the heats of reactions under standard conditions. The standard heat of any reaction (DH°) is equal to the difference between the DH_f^o of all the reactants i.e.,

DH° = Sum of the standard heats - Sum of the standard heats of formation of products of formation of reactants

i.e., DH^o = SDH_f^o _(products) - SDH_f^o _(reactants)

For a reaction,

DH° = SDH_f^o _(products) - SDH_f^o _(reactants)

= [cDH_f^o (C) + dDH_f^o (D)] - [aDH_f^o (A) + bDH_f^o (B)]

The heat of formation of all elements in their standard state is zero. On this basis, it is evident that the heat of formation of a compound is the heat of the compound.

Heat of formation = H_f (compound) - H_f (elements)

Problems

8. The heat change for the reaction,

is -92.2 kJ. Calculate the heat of formation of ammonia.

Solution

The heat of formation of ammonia is the heat change for the formation of 1 mole of ammonia from its elements, i.e.,

The heat change for the reaction

is DH = -92.2 kJ. This equation corresponds to formation of two moles of ammonia. Thus,

9. Calculate the heat change for the reaction

The heat of formation of CH_4(g), CO_2(g) and H₂O_(l) are -74.8 kJmol^-1 -393.5 kJmol^-1 and -285.8kJmol^-1respectively.

Solution

DH° for the reaction,

is: DH° = DH_f^o_(products) -DH_f^o_(reactants)

{DH_f^o[CO₂(g)] + 2DH_f^o[H₂O(l)]} - {DH_f^o[CH₄(g)] + 2DH_f^o[O₂(g)]}

DH_f^o[CO₂(g)] = -393.5kJmol^-1, DH_f^o[H₂O(l)] = - 285.8kJmol^-1

DH_f^o[CH₄(g)] = - 74.8kJmol^-1, DH_f^o[O₂(g)] = 0 (by convention)

DH° = {-393.5 + 2(-285.8)} - {-74.8 + 2 x 0)}

= 965.1 + 74.8 = - 890.3kJmol-1

10. Calculate DH_f^ofor chloride ion from the data given below:

DH^o_HCl = - 92.30 kJ mol^-1

DH^o=- 75.14 kJ

DH^o_f of H⁺ ion = 0.00 kJ.

Solution

Consider the solution process of HCl(g)

Now, DH_sol = SH^o_f (Products) - SH^o_f (reactants)

= DH^o_f (Cl^-)+ DH^o_f (H⁺)] - [DH^o_f HCl(g)]

or -75.14 kJ = DH^o_f (Cl^-)+ 0 - (-92.30)

or D H^o_f (Cl^-) = -75.14 - 92.30 = - 167.44 kJ.