Question

Calculate the standard enthalpy of formation of liquid methanol from the following data:

1. \[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]     ∆H° = – 726 kJ mol^–1
2. \[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\]          ∆_cH° = – 393 kJ mol^–1
3. \[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\]          ∆_fH° = – 286 kJ mol^–1 

Answer 1

Given: Given equations are,

\[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]     ∆H° = – 726 kJ mol^–1    ......(a)

\[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\]          ∆_cH° = – 393 kJ mol^–1     .....(b)

\[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\]          ∆_fH° = – 286 kJ mol^–1    .....(c)

To find: The standard enthalpy of formation (∆_fH°) of CH₃OH_(l)

Calculation: Required equation is,

\[\ce{C_{(graphite)} + 2H_{2(g)} + \frac{1}{2} O_{2 (g)} -> CH3OH_{(l)}}\]

Multiply equation (c) by 2 and add to equation (b),

\[\ce{2H_{2(g)} + O_{2(g)} -> 2H2O_{(l)}}\],	∆rH° = – 572 kJ mol–1
\[\ce{C_{(graphite)} + O_{2(g)} -> CO_{2(g)}}\],	∆cH° = – 393 kJ mol–1
\[\ce{C_{(graphite)} + 2H_{2(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]

∆_rH° = – 572 – 393 = – 965 kJ mol^–1 ….(d)

Reverse equation (a) and add to equation (d),

\[\ce{CO_{2(g)} + 2H2O_{(l)} -> CH3OH_{(l)} + \frac{3}{2} O_{2(g)}}\],	∆rH° = 726 kJ mol–1
\[\ce{C_{(graphite)} + 2H_{2(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]	∆rH° = – 965 kJ mol–1
\[\ce{C_{(graphite)} + 2H_{2(g)} + \frac{1}{2} O_{2(g)} -> CH3OH_{(l)}}\]

∆_fH° = ∆_rH° = 726 – 965 = – 239 kJ mol^–1

The standard enthalpy of formation (∆_fH°) of CH₃OH_(l) from the given data is – 239 kJ mol^–1