Question

Calculate the standard enthalpy of formation of \[\ce{CH3OH_{(l)}}\] from the following data:

\[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)} }\]; Δ_rH° = − 726 kJ mol^-1 

\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; Δ_cH° = −393 kJ mol⁻¹

\[\ce{H2_{(g)} + 1/2 O_{(g)} -> H2O_{(l)}}\]; Δ_fH° = −286 kJ mol⁻¹

Answer 1

Given: Given equations are,

\[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)} }\]; Δ_rH° = − 726 kJ mol^-1    ...(i)

\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; Δ_cH° = −393 kJ mol⁻¹   ...(ii)

\[\ce{H2_{(g)} + 1/2 O_{(g)} -> H2O_{(l)}}\]; Δ_fH° = −286 kJ mol⁻¹   ...(iii)

To find: The standard enthalpy of formation (Δ_fH°) 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),

\[\ce{2H2_{(g)} + O2_{(g)} -> 2H2O_{(l)}}\]; Δ_fH° = −572 kJ mol⁻¹

\[\ce{C_{(graphite)} + O2_{(g)} -> CO2_{(g)}}\]; Δ_cH° = −393 kJ mol^{−1
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\[\ce{C_{(graphite)} + 2H2_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}}\], ΔrH° = −572 −393 = −965kJ mol⁻¹   ...(iv)

Reverse equation (i) and add to equation (iv),

\[\ce{CO2_{(g)} + 2H2O_{(l)} -> CH3OH_{(l)} + 3/2O2_{(g)}}\], Δ_rH° = − 726 kJ mol^-1 

\[\ce{C_{(graphite)} + 2H2_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}}\], ΔrH° = −965
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\[\ce{C_{(graphite)} + 2H2_{(g)} + 1/2O2_{(g)} -> CO2_{(g)} + CH3OH_{(l)}}\], Δ_fH° = ΔrH° = 726 −965 = −239kJ mol⁻¹

∴ The standard enthalpy of formation (Δ_fH°) of \[\ce{CH3OH_{(l)}}\] from the given data is −239kJ mol⁻¹