Thermodynamics and Thermochemistry 2 Question 16
16. Diborane is a potential rocket fuel which undergoes combustion according to the reaction
$$ \mathrm{B}{2} \mathrm{H}{6}(g)+3 \mathrm{O}{2}(g) \longrightarrow \mathrm{B}{2} \mathrm{O}{3}(s)+3 \mathrm{H}{2} \mathrm{O}(g) $$
From the following data, calculate the enthalpy change for the combustion of diborane.
(2000, 2M)
$$ \begin{aligned} 2 \mathrm{~B}(s)+\frac{3}{2} \mathrm{O}{2}(g) \longrightarrow \mathrm{B}{2} \mathrm{O}{3}(s) ; & \Delta H=-1273 \mathrm{~kJ} \mathrm{~mol}^{-1} \ \mathrm{H}{2}(g)+\frac{1}{2} \mathrm{O}{2}(g) \longrightarrow \mathrm{H}{2} \mathrm{O}(l) ; & \Delta H=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \ \mathrm{H}{2} \mathrm{O}(l) \longrightarrow \mathrm{H}{2} \mathrm{O}(g) ; & \Delta H=44 \mathrm{~kJ} \mathrm{~mol}^{-1} \ 2 \mathrm{~B}(s)+3 \mathrm{H}{2}(g) \longrightarrow \mathrm{B}{2} \mathrm{H}_{6}(g) ; & \Delta H=36 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
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Solution:
- $\Delta H_{r}^{\circ}=\Delta H_{f}^{\circ}\left(\mathrm{B}{2} \mathrm{O}{3}\right)+3 \Delta H_{f}^{\circ}\left(\mathrm{H}{2} \mathrm{O}\right)-\Delta H{f}^{\circ}\left(\mathrm{B}{2} \mathrm{H}{6}\right)$
$$ \begin{aligned} \Delta H_{f}^{\circ}\left(\mathrm{H}{2} \mathrm{O}\right)(g) & =\Delta H{f}^{\circ}\left(\mathrm{H}{2} \mathrm{O}\right)(l)+44=-242 \mathrm{~kJ} \ \Rightarrow \quad \Delta H{r}^{\circ} & =-1273-3 \times 242-36 \ & =-2035 \mathrm{~kJ} \end{aligned} $$
