Thermodynamics and Thermochemistry 2 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.
$$ \begin{aligned} \mathrm{C}(s) & \longrightarrow \mathrm{C}(g) ; & \Delta H & =172 \ \mathrm{H}{2}(g) & \longrightarrow 2 \mathrm{H}(g) ; & \Delta H & =104 \ \mathrm{H}{2}(g)+\frac{1}{2} \mathrm{O}{2}(g) & \longrightarrow \mathrm{H}{2} \mathrm{O}(l) ; & & \Delta H=-68.0 \ \mathrm{C}(s)+\mathrm{O}{2}(g) & \longrightarrow \mathrm{CO}{2}(g) ; & & \Delta H=-94.0 \end{aligned} $$
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 $x$ kcal be the $\mathrm{C}-\mathrm{C}$ bond energy and $y$ kcal be the $\mathrm{C}-\mathrm{H}$ bond energy per mole.
$$ \begin{aligned} \Rightarrow \quad 2 \mathrm{C}(\mathrm{gr})+3 \mathrm{H}{2}(g) \longrightarrow \mathrm{C}{2} \mathrm{H}_{6} & (g) ; \ \Delta H^{\circ} & =-2 \times 94-3 \times 68+372 \ & =-20 \mathrm{kcal} \end{aligned} $$
$\Rightarrow \quad-20 \mathrm{kcal}=2 \times 172+3 \times 104-\mathrm{BE}\left(\mathrm{C}{2} \mathrm{H}{6}\right)$
$\Rightarrow \mathrm{BE}\left(\mathrm{C}{2} \mathrm{H}{6}\right)=676 \mathrm{kcal}$
Similarly, $3 \mathrm{C}(\mathrm{gr})+4 \mathrm{H}{2}(\mathrm{~g}) \longrightarrow \mathrm{C}{3} \mathrm{H}_{8}(g)$;
$$ \Delta H^{\circ}=-3 \times 94-4 \times 68+530 $$
$$ =-24 \mathrm{kcal} $$
$\Rightarrow \quad-24 \mathrm{kcal}=3 \times 172+4 \times 104-\mathrm{BE}\left(\mathrm{C}{3} \mathrm{H}{8}\right)$
$\Rightarrow \mathrm{BE}\left(\mathrm{C}{3} \mathrm{H}{8}\right)=956 \mathrm{kcal}$
Also, $\quad \mathrm{BE}\left(\mathrm{C}{2} \mathrm{H}{6}\right)=676 \mathrm{kcal}=x+6 y$
$$ \mathrm{BE}\left(\mathrm{C}{3} \mathrm{H}{8}\right)=956 \mathrm{kcal}=2 x+8 y $$
Solving Eqs. (i) and (ii) gives
$$ \begin{aligned} & y=99 \mathrm{kcal}(\mathrm{C}-\mathrm{H}) \mathrm{BE} \ & x=82 \mathrm{kcal}(\mathrm{C}-\mathrm{C}) \mathrm{BE} \end{aligned} $$
