Application of Hess Law

Hess’s law finds its application in determining the heat of changes for reactions for which experimental determination is not possible. The various applications of Hess’s law can be understood from following headings;

1.   Calculation of Enthalpy of Formation

The enthalpy of formation of certain compounds can be determined by Hess’s Law, where direct determination is not possible experimentally.

Example: Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, graphite and hydrogen are 890.2 KJ, 393.4 KJ and 285.7 KJ mol^–1 respectively.

Solution:

Given

(i).          CH₄  +  2O₂   ––––––––>  CO₂  +  2H₂O  ;  DH₁  =  – 890.2 KJ/mol^–1

(ii).         C   +   O₂        ––––––––>  CO₂  ;  DH₂  =  393.4 KJ/mol^–1    

(iii).        H₂  +  ½ O_2­    ––––––––>  H₂O   ;  DH₃  =  285.7 KJ/mol^-1

To find

C  +   2H₂   ––––––––>  CH₄  ;  DH  =  ?

Multiplying (iii) by 2 and then add to (ii) we get,

C  +  O₂ + 2 (H₂  +  ½ O₂)    ––––––––>  CO₂ + 2H₂O  ;  DH  =  DH₂  +  2DH₃  =   – 393.4  +  2(– 285.7)  

C  +  O₂ + 2 H₂  + O₂             ––––––––>  CO₂ + 2H₂O    ;  DH  =  – 393.4 – 571.4

C  +  2 H₂  +  2O₂                    ––––––––>  CO₂ + 2H₂O    ;  DH  =  – 964.8 –––––––– (iv)

Subtract (i) from (iv), we get.    

C  +  2H₂ – CH₄  ––––––––> 0             ;  DH  =  – 964 + 890.2 KJ         

C  +  2H₂                     ––––––––> CH₄        ;  DH  =   – 74.6 KJ      

2.     Calculation of Enthalpy of Allotropic Transformation

      The transformation of an element from one allotropic form to another form involves a small change of heat. This heat can be calculated using Hess’s Law.

Example:  Calculate enthalpy change of transformation of sulphur (rhombic) to sulphur (monoclinic). Given than combustion of rhombic sulphur and monoclinic sulphur as – 71.1 K Cal. and – 71.7 K Cal respectively.

Solution:

Given

 (i).     S _(Rhombic)   +    O₂  (g) ––––––––>   SO₂ (g)  ;  DH  =  – 71.1 K Cal.

 (ii).    S _(monoclinic) +    O₂ (g)  ––––––––>   SO₂ (g)  ; DH   =  – 71.7 K Cal.

Subtract (ii) from (i) we get           

S _(Rhombic)   –   S _(monoclinic)  ––––––––>     0   ;   DH    =    – 71.1  +  71.7           

S _(Rhombic)   ––––––––> S _(monoclinic)    ;   DH  =  0.6 K Cal.

3.            Calculation of Calorific Value

Example:   Calculate the heat of glucose and its calorific value from following data:

(i).          C _(graphite)  +  O₂ (g)   ––––––––>  CO₂ (g)  ;  DH  =  – 395 KJ

(ii).         H₂ (g)  +     ––––––––>  H₂O (l)   ;  DH  =  – 269.4 KJ.

(iii).        6C _(graphite)  +  6H₂ (g)  +  3O₂ (g) ––––––––>  C₆H₁₂O₆ (s)  ;  DH  =  –1169.8 KJ.

Solution:

We have to find

C₆H₁₂O₆ (s) +  6O₂ (g)   ––––––––>   6CO₂ (g)   +   6H₂O (l)  ;  DH   =   ?

Multiplying (i) by 6 and (ii) by 6, then add.

 6C _(graphite) + 6O₂ (g) + 6H2 (g)  +  6 ´ ½ O₂ (g) ––––––––> 6CO₂ + 6H₂O (l) ; DH = –395 ´ 6 + (– 269.4 ´ 6)

(iv) 6C_graphite + 6H2 (g)  + 9O₂ (g)  ––––––––> 6CO₂ + 6H₂O (l)  ;  DH =–2370–1616.4 = – 3986.4 KJ

Subtract eq. (iii) from (iv), we get.

C₆H₁₂O₆ (s)  +  6O₂     ––––––––>  6CO₂  +  6H₂O (l)    ;     DH  =  2816.6 KJ.

Thus,

1 mole of C₆H₁₂O₆ liberates energy = – 2816.6 KJ.

i.e.   180 gm of C₆H₁₂O₆ liberates energy = – 2816.6 KJ.

\            1 gm of C₆H₁₂O₆ liberates energy =  -2816.6/180= 15.64 KJ.

Thus, calorific value of C₆H₁₂O₆ is 15.64 KJ.