If the heat of formation of H2O2 and H2O are –188 kJ/mole and –286 kJ/mole respectively then what will be enthalpy change for the following reaction
2H2O2 → 2H2O + O2

Dear Student,

The  reaction to be considered is 2H2O2  2H2O +O2Given data are :2H2 +O2  2H2O                              Hi=-2×286 kJ/mole  ..(i)2H2 + 2O2  2H2O2                          Hii=-2×188 Kj/mole  ..(ii)We have multiplied heat of formation by 2 as here 2 mol are formed. On reversing (ii), we get2H2O2  2H2 +2O2                          Hiii=+2×188 kJ/mole   ..(iii)Adding (i) and (iii), we can obtain the desired equationSo,  H= Hi + Hiii               =(-2×286)+(2×188)               =-572+376               =-196 kJ/mole

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