Boron nitride (BN) has a structure similar to graphite, but is a white insulator rather than a black conductor. It is synthesized by heating diboron trioxide with ammonia at about $\mathbf{1000}\mathbf{°}\mathit{C}$.

(a) Write a balanced equation for the formation of BN; water forms also.

(b) Calculate $\mathbf{\Delta }{\mathbf{H}\mathbf{°}}_{\mathbf{rxn}}$for the production of BN ($\mathbf{\Delta }{\mathbf{H}\mathbf{°}}_{\mathbf{f}}$of BN is $\mathbf{-}\mathbf{254}{\mathbf{kJmol}}^{\mathbf{-}\mathbf{1}}$).

(c) Boron is obtained from the mineral borax, ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{B}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{7}}\mathbf{.}\mathbf{10}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. How much borax is needed to produce 0.1kg of BN, assuming 72% yield?

BN, boron nitride, is formed by heating diboron trioxide with ammonia at${1000}^0\mathrm{C}.$ Thus, the balanced chemical equation is given below:

${\mathrm{B}}_2{\mathrm{O}}_3+2{\mathrm{NH}}_3\to 2\mathrm{BN}+3{\mathrm{H}}_2\mathrm{O}$

This can also be obtained by reacting boric acid with ammonia or urea in a nitrogen atmosphere.

It is given that

The enthalpy of reaction for BN formation can be calculated by the given formula:$\mathrm{\Delta }{\mathrm{H}°}_{\mathrm{f}}=-254{\mathrm{kJmol}}^{-1}$

$$\begin{gathered} \mathrm{\Delta }{\mathrm{H}°}_{\mathrm{rxn}}=\left\{2{\mathrm{\Delta H}}_{\mathrm{f}}\left[{\mathrm{BN}}_{\left(\mathrm{s}\right)}\right]+3{\mathrm{H}}_{\mathrm{f}}\left[{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\right]\right\}-\left\{2{\mathrm{\Delta H}}_{\mathrm{f}}\left[{\mathrm{B}}_2{\mathrm{O}}_{3\left(\mathrm{s}\right)}\right]+2{\mathrm{\Delta H}}_{\mathrm{f}}\left[{\mathrm{NH}}_{3\left(\mathrm{g}\right)}\right]\right\} \\ \begin{array}{c}\mathrm{\Delta }{\mathrm{H}°}_{\mathrm{rxn}}=\left[\left(2\mathrm{mol}\right)\left(-254{\mathrm{kJmol}}^{-1}\right)+\left(3\mathrm{mol}\right)\left(-285.83{\mathrm{kJmol}}^{-1}\right)\right]\\ -\left[\left(2\mathrm{mol}\right)\left(-46.11{\mathrm{kJmol}}^{-1}\right)+\left(-1273.0{\mathrm{kJmol}}^{-1}\right)\right]\\ =27{\mathrm{kJmol}}^{-1}\end{array} \end{gathered}$$

Thus, the enthalpy of reaction for BN formation,$\mathrm{\Delta }{\mathrm{H}°}_{\mathrm{rxn}}=1.30\times {10}^2\mathrm{kJ}$

Boron is obtained from diboron trioxide. This process includes three steps. When borax is reacted with sulfuric acid, boric acid is produced which further produces oxoborinic acid and then diboron trioxide. The reactions are given below:

${\mathrm{Na}}_2{\mathrm{B}}_4{\mathrm{O}}_7.10{\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_2{\mathrm{SO}}_4\to 4{\mathrm{H}}_3{\mathrm{BO}}_3+2{\mathrm{Na}}_2{\mathrm{SO}}_4+5{\mathrm{H}}_2\mathrm{O}$

Borax Boric acid

$\begin{array}{l}{\mathrm{H}}_3{\mathrm{BO}}_3\to {\mathrm{HBO}}_2+{\mathrm{H}}_2\mathrm{O}\\ {\mathrm{HBO}}_2\to {\mathrm{B}}_2{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}\end{array}$

1 Kg of BN produces 72% yield of boron, i.e. x kg boron.

So, at 100% yield, x=1.39 kg amount of boron.

Now, the amount of borax can be calculated as follows:

$\begin{array}{l}1.39\mathrm{kg}\mathrm{BN}\times 1000\mathrm{g}\times \frac{1\mathrm{mol}\mathrm{BN}}{24.818\mathrm{g}\mathrm{BN}}\times \frac{1\mathrm{mol}{\mathrm{B}}_2{\mathrm{O}}_3}{2\mathrm{mol}\mathrm{BN}}\\ \times \frac{2\mathrm{mol}{\mathrm{HBO}}_2}{1\mathrm{mol}{\mathrm{B}}_2{\mathrm{O}}_3}\times \frac{1\mathrm{mol} {\mathrm{H}}_3{\mathrm{BO}}_3}{1\mathrm{mol}{\mathrm{HBO}}_2}\times \frac{1\mathrm{mol}\mathrm{borax}}{4\mathrm{mol}{\mathrm{H}}_3{\mathrm{BO}}_3}\times \frac{381.37\mathrm{g}\mathrm{borax}}{1\mathrm{mol}\mathrm{borax}}=5340\mathrm{g}\mathrm{borax}\end{array}$

Thus, 5.3kg of borax is needed to produce 1.0 of BN.