Thermodynamics And Thermochemistry Result 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.
$\begin{aligned} 2 \mathrm{~B}(s)+\frac{3}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{B}_2 \mathrm{O}_3(s) ; & \quad \Delta H=-1273 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; & \quad \Delta H=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_2 \mathrm{O}(g) ; & \quad \Delta H=44 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ 2 \mathrm{~B}(s)+3 \mathrm{H}_2(g) \longrightarrow \mathrm{B}_2 \mathrm{H}_6(g) ; & \quad \Delta H=36 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$
$(2000,2 M)$
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Answer:
Correct Answer: 16. $(-2035 \mathrm{~kJ})$
Solution:
- $\Delta H_r^{\circ}=\Delta H_f^{\circ}\left(\mathrm{B}_2 \mathrm{O}_3\right)+3 \Delta H_f^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)-\Delta H_f^{\circ}\left(\mathrm{B}_2 \mathrm{H}_6\right)$
$\Delta H_f^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)(g)=\Delta H_f^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)(l)+44=-242 \mathrm{~kJ}$
$\begin{aligned} \Rightarrow \quad \Delta H_r^{\circ} & =-1273-3 \times 242-36 \\ & =-2035 \mathrm{~kJ} \end{aligned}$
BETA
