Find the bond enthalpy (in kJ/mol) of ‘three centre two electron bond’ in from the following data: ; — Thermodynamics and Thermochemistry Chemistry Question
Question
Find the bond enthalpy (in kJ/mol) of ‘three centre two electron bond’ in $\text{B}_2\text{H}_6$ from the following data: $\Delta_f H^\circ[\text{BH}_3\text{(g)}] = 100\text{ kJ/mol}$; $\Delta_f H^\circ[\text{B}_2\text{H}_6\text{(g)}] = 36\text{ kJ/mol}$; $\Delta H_{\text{Atomization}}[\text{B(s)}] = 565\text{ kJ/mol}$; $\Delta H_{\text{Atomization}}[\text{H}_2\text{(g)}] = 436\text{ kJ/mol}$.
💡 Solution & Explanation
Atomization of $\text{BH}_3$: $\Delta H = \Delta H_f(\text{B,g}) + 3\Delta H_f(\text{H,g}) - \Delta H_f(\text{BH}_3) = 565 + 3(218) - 100 = 1119\text{ kJ/mol}$. Since $\text{BH}_3$ has 3 terminal B-H bonds, $E_{\text{terminal}} = 1119 / 3 = 373\text{ kJ/mol}$. Atomization of $\text{B}_2\text{H}_6$: $2(565) + 6(218) - 36 = 2402\text{ kJ/mol}$. $\text{B}_2\text{H}_6$ possesses 4 terminal and 2 bridge bonds: $4(373) + 2(E_{\text{bridge}}) = 2402 \implies 1492 + 2E_{\text{bridge}} = 2402 \implies 2E_{\text{bridge}} = 910 \implies E_{\text{bridge}} = 455\text{ kJ/mol}$.