The enthalpy at 298 K of the reactionH2O2(l)→H2O(l)+12O2(g), is -23.5 kcal mol-1 and the enthalpy of formation of H2O2(l) is -44.8 kcal mol-1. The enthalpy of formation of H2O(l) is
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a
-68.3 kcal mol-1
b
68.3 kcal mol-1
c
-91.8 kcal mol-1
d
91.8 kcal mol-1
answer is A.
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Detailed Solution
ΔHreaction =ΔHf∘H2O2−ΔHf∘H2O2−23.5=x−(−44.8) or x=−23.5−44.8=−68.3