Thermodynamics And Thermochemistry Result Question 24
24. Using the data (all values are in kilocalories per mol at $25^{\circ} \mathrm{C}$ ) given below, calculate the bond energy of $\mathrm{C}-\mathrm{C}$ and $\mathrm{C}-\mathrm{H}$ bonds.
$\mathrm{C}(s) \longrightarrow \mathrm{C}(g) ; \hspace {10mm} \Delta H=172 4$
$\mathrm{H}_2(g) \longrightarrow 2 \mathrm{H}(g) ; \hspace {10mm} \Delta H=104 $
$\mathrm {H}_2(g)+\dfrac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \hspace {10mm} \Delta H=-68.0$
$\mathrm{C}(s)+\mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \hspace {10mm} \Delta H=-94.0$
Heat of combustion of $\mathrm{C}_2 \mathrm{H}_6=-372.0$
Heat of combustion of $\mathrm{C}_3 \mathrm{H}_8=-530.0$
(1990, 5M)
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Solution:
- Let $x $ $\mathrm{kcal}$ be the $\mathrm{C}-\mathrm{C}$ bond energy and $y \mathrm{kcal}$ be the $\mathrm{C}-\mathrm{H}$ bond energy per mole.
$\Rightarrow \quad 2 \mathrm{C}(\mathrm{gr})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) ; $
$\Delta H^{\circ}=-2 \times 94-3 \times 68+372 $ $=-20$ $ \mathrm{kcal} $
$\Rightarrow \quad -20 $ $\mathrm{kcal}=2 \times 172+3 \times 104-\mathrm{BE}\left(\mathrm{C}_2 \mathrm{H}_6\right) $
$\Rightarrow \quad \mathrm{BE}\left(\mathrm{C}_2 \mathrm{H}_6\right)=676 $ $\mathrm{kcal}$
Similarly, $3 \mathrm{C}(\mathrm{gr})+4 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})$;
$\Delta H^{\circ} =-3 \times 94-4 \times 68+530 $ $ =-24$ $ \mathrm{kcal} $
$\Rightarrow \quad -24 $ $\mathrm{kcal} =3 \times 172+4 \times 104-\mathrm{BE}\left(\mathrm{C}_3 \mathrm{H}_8\right) $
$\Rightarrow \quad \mathrm{BE}\left(\mathrm{C}_3 \mathrm{H}_8\right) =956$ $ \mathrm{kcal}$
Also,
$\mathrm{BE}\left(\mathrm{C}_2 \mathrm{H}_6\right)=676 $ $\mathrm{kcal}=x+6 y \quad$ …..(i)
$\mathrm{BE}\left(\mathrm{C}_3 \mathrm{H}_8\right)=956 $ $\mathrm{kcal}=2 x+8 y \quad$ …..(ii)
Solving Eqs. (i) and (ii) gives
$y=99 $ $\mathrm{kcal}(\mathrm{C}-\mathrm{H}) \mathrm{BE} $
$ x=82 $ $\mathrm{kcal}(\mathrm{C}-\mathrm{C}) \mathrm{BE}$
BETA
