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Question
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
- \[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}ΔH^° = - 726 kJ mol^{-1}}\]
- \[\ce{C_{(s)} + O2_{(g)} → CO2_{(g)}Δ_cH^° = – 393 kJ mol^{-1}}\]
- \[\ce{H2_{(g)} + 1/2 O2_{(g)} -> H2O_{(l)}Δ_fH^° = - 286 kJ mol^{-1}}\]
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Solution
Required equation
\[\ce{C_{(s)} + 2H2_{(g)} +1/2 O2_{(g)} -> CH3OH_{(l)} Δ_FH^° = ?}\]
Reversing the equation (i), multiply equation (iii) by (i) and add all equations.
Equation (i) + (ii) + 2 × (iii)
\[\ce{\cancel{CO_2}_{(g)} + \cancel{2H_2O}_{(l)} -> CH3OH_{(l)} + \cancel{3/2} O2_{(g),} ΔH^° = - 726 kJ mol^{-1}}\]
\[\ce{C_{(s)} + O2_{(g)} -> \cancel{CO_2}_{(g),} Δ_cH^° = - 393 kJ mol^{-1}}\]
\[\ce{2H2_{(g)} + O2_{(g)} -> \cancel{2H_2O}_{(l),} Δ_fH^° = - 572 kJ mol^{-1}}\]
\[\ce{C_{(s)} + 2H2_{(g)} + 1/2 O2_{(g)} -> CH3OH_{(l)}, ΔH = - 239 kJ mol^{-1}}\]
The enthalpy of formation of CH3OH(l) is – 239 kJ mol–1.
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