Wednesday, 25 May 2011

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    ––––––––>  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) 6Cgraphite + 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.

2 comments:

  1. Tanks but are there any more applications?????????????????????????????

    ReplyDelete
  2. Nice examples

    ReplyDelete