If the heat of formation of H2O2 and H2O are 188 kJ/mole and 286 kJ/mole respectively then what - Chemistry - Thermodynamics

Question

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

Vartika Jain · Accepted Answer

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

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

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