Diborane is a potential rocket fuel which undergoes combustion according to the reaction:<br><br>Fro — Thermodynamics and Thermochemistry Chemistry Question
Question
Diborane is a potential rocket fuel which undergoes combustion according to the reaction:<br>$\text{B}_2\text{H}_6\text{(g)} + 3\text{O}_2\text{(g)} \to \text{B}_2\text{O}_3\text{(s)} + 3\text{H}_2\text{O(g)}$<br>From the following data, calculate the enthalpy change for the combustion of diborane.<br>$2\text{B(s)} + 3/2 \text{O}_2\text{(g)} \to \text{B}_2\text{O}_3\text{(s)}; \Delta H = -1273\text{ kJ/mol}$<br>$\text{H}_2\text{(g)} + 1/2 \text{O}_2\text{(g)} \to \text{H}_2\text{O(l)}; \Delta H = -286\text{ kJ/mol}$<br>$\text{H}_2\text{O(l)} \to \text{H}_2\text{O(g)}; \Delta H = 44\text{ kJ/mol}$<br>$2\text{B(s)} + 3\text{H}_2\text{(g)} \to \text{B}_2\text{H}_6\text{(g)}; \Delta H = 36\text{ kJ/mol}$
💡 Solution & Explanation
$\Delta H_f(\text{B}_2\text{O}_3) = -1273$. $\Delta H_f(\text{H}_2\text{O, g}) = -286 + 44 = -242$. $\Delta H_f(\text{B}_2\text{H}_6) = 36$. $\Delta H_c = \Delta H_f(\text{B}_2\text{O}_3) + 3\Delta H_f(\text{H}_2\text{O, g}) - \Delta H_f(\text{B}_2\text{H}_6) = -1273 + 3(-242) - 36 = -1273 - 726 - 36 = -2035\text{ kJ/mol}$.