Question

Calculate the standard enthalpy of formation of \(CH_3OH(l)\) from the following data.
\(CH_3OH (l) +\frac 32 O_2(g)→CO_2(g)+2H_2O (l)\) ;  \( ∆_rH^Θ= –726 \ kJ mol^{–1}\),
\(C(graphite)+O_2(g)→CO_2(g)\);  \(∆_cH^Θ = –393 \ kJ mol^{–1}\),
\(H_2(g)+\frac 12 O_2(g)→H_2O (l)\);  \(∆_f H^Θ= –286 kJ mol^{–1}\)

Answer 1

The reaction that takes place during the formation of \(CH_3OH(l)\) can be written as:
\(C(s) + 2H_2O(g) + \frac 12 O_2(g) → CH_3OH(l)\) ……… (i)
The reaction (i) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) – equation (i)
\(∆_fH^Θ [CH_3OH(l)] = ∆_cH^Θ + 2∆_fH^Θ [H_2O(l)] – ∆_rH^Θ\)
= \((–393\ kJ mol^{–1}) + 2(–286\ kJ mol^{–1}) – (–726 \ kJ mol^{–1})\)
= \((–393 – 572 + 726) \ kJ mol^{–1}\)
∴ \(∆_fH^θ [CH_3OH(l)] = –239\ kJ mol^{–1}\)