Q75P - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

1. Balanced equation for the formation of BN is given below:

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

   2. $\mathrm{\Delta }{\mathrm{H}°}_{\mathrm{rxn}}=27 {\mathrm{kJmol}}^{-1}$

   3. 5.3 kg of borax is needed to produce 1.09 of BN.

Formation of BN

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. 

Calculation of enthalpy of reaction for BN formation

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, $\mathrm{\Delta }{\mathrm{H}°}_{\mathrm{rxn}}=1.30\times {10}^2 \mathrm{kJ}$

Amount of borax needed

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.