Chapter 14, Problem 14.43P

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the formation of $\text{BN}$ has to be written.

Concept Introduction:

Balanced equation:

In a chemical equation, the number of atoms or elements that present in both reactant and product sides are equal means this equation is known as balanced equation.

Explanation of Solution

Boron nitride is formed with water when, Boron oxide is treated with ammonia.

The balanced equation for this react ion is,

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{ + 2NH}}_{\text{3}\left(\text{g}\right)}\stackrel{}{\to }{\text{2BN}}_{\text{(s)}}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

(b)

Interpretation Introduction

Interpretation:

Enthalpy change of a reaction of formation of Boron nitride has to be calculated.

Concept Introduction:

Hess's law:

Hess's law state that the different in enthalpy change of reactant to product is known as enthalpy change of a reaction.

The enthalpy change reactant to product is calculated by subtraction of standard enthalpy of formation (SEF) of reactant from SEF product.

    $\begin{array}{l}{\text{ΔH}}_{{}_{\text{reaction}}}^{\text{o}}\text{=}{\displaystyle \sum {\text{mΔH}}_{{}_{\text{product}}}^{\text{o}}}\text{-}{\displaystyle \sum {\text{nΔH}}_{{}_{\text{reactant}}}^{\text{o}}}\\ \text{where,}\\ {\text{ΔH}}_{{}_{\text{reaction}}}^{\text{o}}\text{=}\text{enthalpy change}\text{ofreaction}\\ {\text{ΔH}}_{{}_{\text{Product}}}^{\text{o}}\text{=}\text{standard enthalpy of formation of reactant}\\ {\text{ΔH}}_{{}_{\text{reactant}}}^{\text{o}}\text{=standard enthalpy of formation of product}\end{array}$

Explanation of Solution

Given,

${\text{ΔH}}_{\text{f}}\text{=}\text{-254}\text{kJ/mol}$ and ${\text{In}}^{\text{3+}}$ are,

Boron nitride is formed with water when, Boron oxide is treated with ammonia.

The balanced equation for this react ion is,

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{ + 2NH}}_{\text{3}\left(\text{g}\right)}\stackrel{}{\to }{\text{2BN}}_{\text{(s)}}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

From the standard data, the standard enthalpy of formation of Boron oxide and ammonia are substitute in the below equation we get enthalpy change of a reaction of formation of Boron nitride.

    $\begin{array}{l}\text{=}{\text{{2ΔH}}_{\text{f}}{\text{[BN}}_{\text{(s)}}\text{]}{\text{+3ΔH}}_{\text{f}}{\text{[H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{]}-{2ΔH}}_{\text{f}}{\text{[B}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{]}{\text{+2ΔH}}_{\text{f}}{\text{[NH}}_{\text{3}}{}_{\text{(g)}}\text{]}}\\ \text{=}\text{[2(-254}\text{kJ/mol)}\text{+}\text{(3mol)}\text{(-241}\text{.86}\text{kJ/mol)]}\\ \text{-[(2mol)(-1272}\text{kJ/mol)+(2mol)(-45}\text{.0kJ/mol)]}\\ \text{=}\text{130}\text{.322}\text{J}\\ \text{=}\text{1}{\text{.30×10}}^{\text{2}}\text{kJ}\end{array}$

Hence, the enthalpy change of a reaction is $\text{1}{\text{.30×10}}^{\text{2}}\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

The amount of borax is needed to produce $\text{1}\text{.0 kg of BN}$ at $\text{72\%}$ yield has to be calculated.

Concept Introduction:

Percentage yield:

Percentage yield is given by,

    $\text{\%Y}\text{=}\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Mole:

Mole of the solid substance is given by,

    $\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

The balanced equation for this react ion is,

    ${\text{B}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{ + 2NH}}_{\text{3}\left(\text{g}\right)}\stackrel{}{\to }{\text{2BN}}_{\text{(s)}}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Produce mass of $\text{ BN}$ is $\text{1}\text{.0 kg}$

The amount of borax is needed to produce $\text{1}\text{.0 kg of BN}$ at $\text{72\%}$ yield is,

    $\begin{array}{l}\text{=}\text{1}\text{.0 kg of BN}\left(\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\right)\left(\frac{\text{1mol}\text{BN}}{\text{24}\text{.82}\text{g}\text{BN}}\right)\left(\frac{\text{1}\text{mol}\text{B}}{\text{1}\text{mol}\text{BN}}\right)\left(\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7 }}{\text{×10H}}_{\text{2}}\text{O}}{\text{4}\text{mol}\text{B}}\right)\\ \text{×}\left(\frac{\text{381}\text{.38}\text{g}}{\text{1}\text{mol}}\right)\left(\frac{\text{100\%}}{\text{72}\text{.0\%}}\right)\\ \text{=}\text{5}{\text{.3×10}}^{\text{3}}\text{g}\text{borax}\end{array}$

Hence, the amount of borax is needed to produce $\text{1}\text{.0 kg of BN}$ at $\text{72\%}$ yield is $\text{5}{\text{.3×10}}^{\text{3}}\text{g}\text{borax}$.