Use the → Table of enthalpies of formation to to calculate the standard enthalpy of formation of methanol from carbon monoxyde and hydrogen: $CO(g)$ $+$ $2H_2(g)$ $\longrightarrow$ $CH_3OH(g)$ Find the bond energy of the $CO$ bond in carbon monoxide using the → Table of bond energies. Compare the "strength" of this bond to that of a $C=O$ or $C-O$ bond!
$CO(g)+2H_2(g)$ $\longrightarrow$ $CH_3OH(g)$ $\Delta H$ $ =$ $\Delta H_f(CH_3OH(g))$ $-$ $ \Delta H_f(CO(g))$ $=$ $ -90.13kJ$
$\Delta H$ $=$ $E_{bond}(CO)$ $ +$ $ 4E_{bond}(H-H)$ $ - $ $(3E_{bond}(C-H)$ $+$ $E_{bond}(C-O)$ $+$ $E_{bond}(O-H))$ Let be $x$ $=$ $E_{bond}(CO)$: $-90,13$ $=$ $x$ $+$ $4\cdot 436$ $-$ $(3\cdot 414$ $+$ $352$ $+$ $465) $ $E_{bond}(CO)$ $=$ $x$ ≈ $ 225kJ$
This bond is "weaker" than a $C-O$ bond ($E_{bond}(C-O)$ $=$ $ 352kJ$ or a $C=O$ bond ($E_{bond}(C=O)$ $=$ $707\;kJ$