calculate the standard enthalpy of formation of ethene (C2H4)from the following equation-

C2H4(g) + 3O2(g)-----》 2CO2 +2H2O enthalpy of formatiom of reaction is -1323 kJ/m

Enthalpy of formation of CO2,H2O and O2 are -393.5 ,-249,0 respectively

First let us consider the combustion reaction given that is,

 

C2H4(g) + 3O2(g)---à 2CO2(g) +2H2O(g)

 

Let us first calculate enthalpy of formation of CO2 and H2O on RHS of reaction from the given data.

 = 2(-393.5)+2 (,-249)

  = -(787 + 498)

   = -1285 kJ/mol.

 

This means -1358.6 kJ of energy released in the formation of CO2 and H2O.

 

In the LHS of the reaction one mole of C2H4 is formed first and then reacted with 3 moles of oxygen to give products. But the enthalpy of formation of O2 is given 0, hence

  =  + 3(0)

   =

 

Enthalpy of formation of ethene is given by

 

 

  • -9

delta f H =  delta H product - delta H reactants

                =  2(-393.5)+2(-249) - (-1323)

               = - 787 - 498 + 1323  

               =  38 

  • 9
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