Thermodynamics and Thermochemistry 2 Question 19

19. Compute the heat of formation of liquid methyl alcohol in kJmol1, using the following data. Heat of vaporisation of liquid methyl alcohol =38 kJ/mol. Heat of formation of gaseous atoms from the elements in their standard states :

H=218 kJ/mol,C=715 kJ/mol,O=249 kJ/mol.

Average bond energies:

(1997,5M)

CH=415 kJ/mol,CO=356 kJ/mol,

OH=463 kJ/mol

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Solution:

  1. Given : $\mathrm{CH}{3} \mathrm{OH}(g) \longrightarrow \mathrm{CH}{3} \mathrm{OH}(l) ; \quad \Delta H=-38 \mathrm{~kJ}$

C(g)+4H(g)+O(g)CH3OH(g)

ΔH=(3×415+356+463)

H=H1+H2=2064 kJ

C(g)C(g);

ΔH=715 kJ

$$ \begin{aligned} 2 \mathrm{H}{2}(g) \longrightarrow & 4 \mathrm{H}(g) ; \ & \Delta H=2 \times 2 \times 218=872 \mathrm{~kJ} \ \frac{1}{2} \mathrm{O}{2}(g) \longrightarrow & \mathrm{O}(g) ; \quad \Delta H=249 \mathrm{~kJ} \end{aligned} $$

Adding : C (gr) $+2 \mathrm{H}{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)$

ΔH=266 kJ/mol =1×1.250.082×300×20.8×(189.86300)J =116.4 J