Calculate the enthalpy of formation of Benzene represented by following reaction:
6C + 2H2 gives C6H6
The standard enthalpy of combustion of Benzene is -3266.0 kJ and standard enthalpy of formation of CO2 and H2O are -393.1 kJ and -286.0 kJ respectively.

According to the given information
(i) C6H6 (l)+ 153O2 (g)  6CO2(g) +  3H2O(l)                H=-3266.0 kJ(ii) C(s) + O2 (g)  CO2 (g)                                                  H=-393.1 kJ(iii) H2 (g)  + 12O2 (g)  H2O(l)                                          H= -286.0 kJThe required equation is:6C(s)  + 3H2(g) C6H6(l)                                                    H= ?
Method to solve the problem:
The heat required equation can be obtained by algebraic method.

Multiply reaction (ii) by 6 and reaction (iii) by 3 and write (i) in reverse form. Add all the three equation.

6C(s)+ 6O2(g)  6CO2(g);                                    H=-393.1×6 kJ=2358.6 kJ3H2(g) + 32O2(g)  3H2O(l);                               H=-286.0×3 kJ=858 kJ6CO2(g) + 3H2O(l) C6H6(l) + 152O2(g);        H= + 3266.0 kJOn adding the three equations,6C(s) + 3H2(g)   C6H6(l);                                     H= + 3266.0 kJ-2358.6 kJ-858 kJ                                                                                       H= + 49.4 kJ
Hence, the enthalapy of formation of benzene is 49.4 kJ.


  • 45
there is slight mistake there reaction
plz take it as
6C + 3H2O gives C6H6.
  • 16
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