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Question

Calculate the standard enthalpy of formation of CH3OH(l) from the following data.
CH3OH(l)+32O2(g)CO2(g)+2H2O(l); ΔHo=726kJ mol1
C(graphite)+O2(g)CO2(g); ΔcH0=393kJ mol1
H2(g)+12O2(g)H2O(l); ΔtHo=286 kJ mol1.

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Solution

CH3OH+3/2O2CO2+2H2O ΔHo=726kJmol1
C+O2(g)CO2 ΔcHo=393kJmol1
H2+12O2H2O ΔtH=296kJmol
CH3OH enthalapy of formation
C+2H2+12O2CH3OH
CO2+2H2OCH3OH+32O2 726
C+O2CO2 393
2H2+O22H2O 296×2=592
C+2H2+2O2CH3OH+32O2 259kJmol1
C+2H2+12O2CH3OH 259kJmol1

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