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 \(1000^{\circ} \mathrm{C}\) (a) Write a balanced cquation for the formation of BN; water also forms. (b) Calculate \(\Delta H_{\text {ran }}^{\circ}\) for the production of BN \(\left(\Delta H_{i}^{\circ}\right.\) of \(\mathrm{BN}\) is \(-254 \mathrm{~kJ} / \mathrm{mol})\) (c) Boron is obtained from the mineral borax, \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}\). How much borax is needed to produce \(1.0 \mathrm{~kg}\) of BN, assuming \(72 \%\) yield?

Short answer

a) \( \text{B}_2\text{O}_3 + 2 \text{NH}_3 \rightarrow 2 \text{BN} + 3 \text{H}_2\text{O} \) b) \( \Delta H_{\text {ran }}^{\circ} = -508 \text{kJ} \) c) 10.67 kg borax

Step-by-step solution

Write the Balanced Equation

Identify the reactants and products. The problem states that BN is formed by heating diboron trioxide \(\text{B}_2\text{O}_3\) with ammonia \(\text{NH}_3\) and water is also formed as a by-product. The balanced chemical reaction is: \[ \text{B}_2\text{O}_3 + 2 \text{NH}_3 \rightarrow 2 \text{BN} + 3 \text{H}_2\text{O} \]

Calculate \( \Delta H_{\text {ran }}^{\circ} \)

Use the given standard enthalpy of formation for BN \( \( \Delta H_f^{\circ} \) = -254 kJ/mol \). Since 2 moles of BN are produced from the reaction: \[ \Delta H_{\text {ran }}^{\circ} = 2 \times (-254 \text{kJ/mol}) = -508 \text{kJ} \]

Determine Borax Needed for 1.0 kg of BN

First, convert the mass of BN to moles using its molar mass \( \text{BN} = 24.82 \text{g/mol} \): \[ \frac{1.0 \text{ kg}}{24.82 \text{ g/mol}} \times \frac{1000 \text{ g}}{1 \text{ kg}} = 40.28 \text{ mol BN} \] Since the reaction yield is only 72%, use the formula: \[ \text{Required moles of BN} = \frac{40.28}{0.72} = 55.94 \text{ mol BN} \] Since each mole of BN requires half a mole of \text{B}_2\text{O}_3 \, we find moles of \text{B}_2\text{O}_3: \[ \text{Required moles of } \text{B}_2\text{O}_3 = \frac{55.94}{2} = 27.97 \text{ mol } \text{B}_2\text{O}_3 \] Finally, convert moles of \text{B}_2\text{O}_3 to mass of borax using the stoichiometry: \[ \frac{27.97 \text{ mol } \text{B}_2\text{O}_3}{1} \times \frac{1 \text{ mol } \text{Na}_2\text{B}_4\text{O}_7 \cdot 10 \text{H}_2\text{O}}{1 \text{ mol } \text{B}_2\text{O}_3} \times \ 381.37 \text{ g/mol} = 10.67 \text{ kg borax} \]

Key concepts

balanced chemical equation

Understanding the concept of a balanced chemical equation is essential in chemistry. A balanced chemical equation shows the reactants and products in a chemical reaction, including their relative proportions. In the process of creating boron nitride (BN), we start with diboron trioxide \((B_2O_3)\) and ammonia \((NH_3)\). To ensure the equation is balanced, we verify that the number of atoms for each element is the same on both sides of the reaction. The balanced equation for the synthesis of BN is:
\[B_2O_3 + 2NH_3 \rightarrow 2BN + 3H_2O \]

• Two moles of ammonia react with one mole of diboron trioxide.
• The reaction produces two moles of boron nitride and three moles of water.
This balance ensures that the law of conservation of mass is maintained. A balanced equation is foundational for any further calculations in stoichiometry and related concepts.

enthalpy of reaction

The enthalpy of reaction \(\Delta H_{rxn}^{\circ}\) represents the change in heat content during a chemical reaction at constant pressure. It reflects the favorability of the reaction in terms of energy. For the synthesis of BN, we use the standard enthalpy of formation \(\Delta H_{f}^{\circ}\) given for BN: \(-254 \, \text{kJ/mol}\).
Since the reaction produces two moles of BN, we calculate the total enthalpy change as:
\[\Delta H_{rxn}^{\circ} = 2 \times (-254 \, \text{kJ/mol}) = -508 \, \text{kJ} \]

• Exothermic reactions have a negative \(\Delta H_{rxn}\), releasing heat.
• Endothermic reactions have a positive \(\Delta H_{rxn}\), absorbing heat.

A negative value in this context means the reaction releases 508 kJ of energy, making it exothermic.

molar mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between mass and moles in chemistry. In our BN example, the molar mass calculation helps determine how much material we need. For boron nitride (BN), the molar mass is the sum of the molar masses of boron (B) and nitrogen (N):

• Boron (B): 10.81 g/mol
• Nitrogen (N): 14.01 g/mol

Total molar mass of BN:
\[10.81 \, \text{g/mol} + 14.01 \, \text{g/mol} = 24.82 \, \text{g/mol} \]
This value allows us to convert the mass of BN produced into moles, which is a critical step in subsequent reaction yield calculations.

reaction yield

Reaction yield measures the efficiency of a chemical reaction, expressed as a percentage of the theoretical maximum amount of product that could be formed. In practice, yields are often less than 100% due to various factors like incomplete reactions or side reactions. For boron nitride, the exercise assumes a 72% yield. To determine how much borax is needed to produce 1.0 kg of BN, the yield factor is crucial:
First, convert the mass of BN to moles:
\[ \frac{1.0 \, \text{kg}}{24.82 \, \text{g/mol}} \times 1000 \, \text{g/kg} = 40.28 \, \text{mol BN}\ \]

• Adjust for the 72% yield:
  \[ \frac{40.28}{0.72} = 55.94 \, \text{mol BN}\ \]

This means we need to start with enough reactants to make 55.94 moles of BN to end up with 40.28 moles, considering the practical limitations.

stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced. For the BN synthesis, stoichiometry lets us determine the amount of borax needed. The balanced chemical equation shows the molar ratios:
\[ B_2O_3 + 2 NH_3 \rightarrow 2 BN + 3 H_2O \]
For 1 mole of \( B_2O_3\), two moles of \( NH_3\) yield two moles of \( BN \). Given our earlier adjustment to moles of \( BN \), we find the moles of \( B_2O_3\):

• 55.94 moles BN needs 27.97 moles of \( B_2O_3\).

Since boron is derived from borax \((Na_2B_4O_7 \cdot 10 H_2O)\), convert moles of \( B_2O_3\) to mass of borax:
\[ \frac{27.97 \, \text{mol} \ B_2O_3}{1} \times \frac{1 \, \text{mol} \ Na_2B_4O_7 \cdot 10 \, H_2O}{1 \, \text{mol} \ B_2O_3} \times 381.37 \, \text{g/mol} = 10.67 \, \text{kg borax} \]
This calculation ensures we have the correct amount of starting materials.