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The enthalpy at 298 K of the reactionH2O2l →H2Ol + 12O2g is -23.5 kcal mol-1 and the enthalpy of formation of H2O2l is -44.8 kcal mol-1. The enthalpy of formation of H2Ol is

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Detailed Solution

∆Hreaction=∆Hfo H2O - ∆Hfo H2O2 -23.5=x--44.8 or x = -23.5 - 44.8 = -68.3

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