Question

Compute the enthalpy of formation of liquid methyl alcohol in kJ mol${}^{-1}$, using the following data.

Enthalpy of vaporisation of liquid $C{H}_{3}OH=38$ kJ / mol.

Enthalpy of formation of gaseous atoms from the elements in their standard states are
$H\to 218$ kJ / mol ; $C\to 715$ kJ / mol ; $O\to 249$ kJ / mol.

Average Bond energies
$C-H\to 415$ kJ / mol ; $C-O\to 356$ kJ / mol; $O-H\to 463$ kJ / mol

• A. -169 kJ mol${}^{-1}$
• B. -200 kJ mol${}^{-1}$
• C. -246 kJ mol${}^{-1}$
• D. -266 kJ mol${}^{-1}$

Answer 1

The correct option is D -266 kJ mol${}^{-1}$

The following information are given:

(1) ${CH}_{3}OH$(l) $⟶$ ${CH}_{3}OH$(g); $H_1$= 38kJ/mol

(2) $1/2$$H_2$(g) $⟶$ $H$(g); $H_2$= 218 kJ/mol

(3) $C$(s) $⟶$ $C$(g); $H_3$= 715 kJ/mol

(4) $1/2$$O_2$(g) $⟶$ $O$(g); $H_4$= 249 kJ/mol

(5) $CH$(g) $⟶$ $C$(g) + $H$(g); $H_5$= 415kJ/mol

(6) $CO$(g) $⟶$ $C$(g) + $O$(g) ;$H_6$ = 356 kJ/mol

(7) $OH$(g) $⟶$ $O$(g) + $H$(g); $H_7$ = 463 kJ/mol

(8) The enthalpy of the reaction ${CH}_{3}OH$(g) $⟶$ 4$H$(g) + $C$(g) + $O$(g) can be reconstituted as $H_8$= 3$\times$$H_5$ + $H_6$ + $H_7$ = (3$\times$415 + 356 + 463) = 2064 kJ/mol

(9) The formation of liquid methyl alcohol is represented as

$C$(s) + 2$H_2$(g) + $1/2$$O_2$(g) $⟶$ ${CH}_{3}OH$(l)

and $H_9$ can be calculated form the above and enthalpies values as:

$H_9$ = $H_3$ + 4$\times$$H_2$ + $H_4$ -$H_8$ - $H_1$

$H_9$= ( 715 + 4$\times$218 + 249 - 2064 - 38) kJ/mol

$H_9$ = -266 kJ/mol

Hence, the correct option is $D$