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Question
Calculate the standard enthalpy of formation of \[\ce{CH3OH_{(l)}}\] from the following data:
\[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)} }\]; ΔrH° = − 726 kJ mol-1
\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; ΔcH° = −393 kJ mol−1
\[\ce{H2_{(g)} + 1/2 O_{(g)} -> H2O_{(l)}}\]; ΔfH° = −286 kJ mol−1
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Solution
Given: Given equations are,
\[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)} }\]; ΔrH° = − 726 kJ mol-1 ...(i)
\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; ΔcH° = −393 kJ mol−1 ...(ii)
\[\ce{H2_{(g)} + 1/2 O_{(g)} -> H2O_{(l)}}\]; ΔfH° = −286 kJ mol−1 ...(iii)
To find: The standard enthalpy of formation (ΔfH°) of \[\ce{CH3OH_{(l)}}\]
Calculation:
Required equation is, \[\ce{C_{(graphite)} + 2H2_{(g)} + 1/2 O2_{(g)} -> CH3OH_{(l)}}\]
Multiply equation (iii) by 2 and add to equation (ii),
\[\ce{2H2_{(g)} + O2_{(g)} -> 2H2O_{(l)}}\]; ΔfH° = −572 kJ mol−1
\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; ΔcH° = −393 kJ mol−1
________________________________________________________________________
\[\ce{C_{(graphite)} + 2H2_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}}\], ΔrH° = −572 −393 = −965kJ mol−1 ...(iv)
Reverse equation (i) and add to equation (iv),
\[\ce{CO2_{(g)} + 2H2O_{(l)} -> CH3OH_{(l)} + 3/2O2_{(g)}}\], ΔrH° = − 726 kJ mol-1
\[\ce{C_{(graphite)} + 2H2_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}}\], ΔrH° = −965
_____________________________________________________________________________
\[\ce{C_{(graphite)} + 2H2_{(g)} + 1/2O2_{(g)} -> CO2_{(g)} + CH3OH_{(l)}}\], ΔfH° = ΔrH° = 726 −965 = −239kJ mol−1
∴ The standard enthalpy of formation (ΔfH°) of \[\ce{CH3OH_{(l)}}\] from the given data is −239kJ mol−1
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