Question

Compounds of boron and hydrogen are remarkable for their unusual bonding (described in Section 14.5 ) and for their reactivity. With the more reactive halogens, for example, diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) forms trihalides even at low temperatures: $$\begin{array}{r}\mathrm{B}_{2} \mathrm{H}_{6}(g)+6 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{BCl}_{3}(g)+6 \mathrm{HCl}(g) \\\\\Delta H=-755.4 \mathrm{~kJ}\end{array}$$ What is \(\Delta H\) per kilogram of diborane that reacts?

Short answer

\text{ΔH per kg of } \(\text{B}_{2}\text{H}_{6}\) \ \text{is} \(-27334.29\) kJ.

Step-by-step solution

- Calculate Molar Mass of Diborane

First, calculate the molar mass of diborane \(\text{B}_{2}\text{H}_{6}\). Boron (B) has a molar mass of 10.81 g/mol and hydrogen (H) has a molar mass of 1.01 g/mol. Therefore, \(\text{Molar mass of B}_{2}\text{H}_{6} = 2 \times 10.81 + 6 \times 1.01 = 27.64 \text{g/mol}\).

- Convert Kilograms to Grams

Since the problem asks for \(\text{ΔH}\) per kilogram, convert 1 kilogram of \(\text{B}_{2}\text{H}_{6}\) to grams. \(\text{1 kg} = 1000 \text{grams}\).

- Calculate the Amount in Moles

Calculate the number of moles of \(\text{B}_{2}\text{H}_{6}\) in 1 kilogram using its molar mass. \(\text{Moles of B}_{2}\text{H}_{6} = \frac{1000 \text{grams}}{27.64\text{g/mol}} = 36.18 \text{moles}\).

- Determine Total Enthalpy Change

The balanced reaction has \(\text{ΔH} = -755.4 \text{kJ}\) for 1 mole of \(\text{B}_{2}\text{H}_{6}\) reacting. Therefore, for 36.18 moles, the total enthalpy change is \[36.18 \text{ moles} \times -755.4 \text{ kJ/mole} = -27334.29 \text{kJ}\].

- Result per Kilogram

Thus, the \(\text{ΔH}\) per kilogram of \(\text{B}_{2}\text{H}_{6}\) that reacts is \(-27334.29\) kJ.

Key concepts

Molar Mass Calculation

To solve this problem, the first step is calculating the molar mass of diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)). The molar mass is essential because it helps us convert between grams and moles, which is crucial for the subsequent calculations.
Boron (B) has a molar mass of 10.81 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol.
When we have a compound like \(\mathrm{B}_{2}\mathrm{H}_{6}\), we need to account for each atom within the molecule. Since there are two boron atoms and six hydrogen atoms:

• The mass contribution from boron: \(2 \times 10.81 = 21.62 \text{ g/mol}\)
• The mass contribution from hydrogen: \(6 \times 1.01 = 6.06 \text{ g/mol}\)

This gives us the total molar mass by adding both of these contributions together:
\(\text{Molar mass of } \mathrm{B}_{2}\mathrm{H}_{6} = 21.62 \text{ g/mol} + 6.06 \text{ g/mol} = 27.64 \text{ g/mol}\).

Enthalpy Change

Next, let's discuss enthalpy change (\(\Delta H\)). Enthalpy (\(H\)) is a measure of the total energy of a thermodynamic system. It includes internal energy plus the product of pressure and volume. The enthalpy change (\(\Delta H\)) in a reaction indicates how much heat is absorbed or released.
In the given reaction: \[\mathrm{B}_{2}\mathrm{H}_{6}(g) + 6 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{BCl}_{3}(g) + 6 \mathrm{HCl}(g) \]The enthalpy change (\(\Delta H\)) is -755.4 kJ.
This negative sign signifies that the reaction releases heat, making it an exothermic process.
For this reaction, -755.4 kJ of heat is released for every mole of \(\mathrm{B}_{2}\mathrm{H}_{6}\) that reacts.
To find the enthalpy change per kilogram of \(\mathrm{B}_{2}\mathrm{H}_{6}\), we'll first convert kilograms to grams (1 kg = 1000 grams). Then, we'll calculate the number of moles in 1000 grams using the molar mass of \(\mathrm{B}_{2}\mathrm{H}_{6}\): Number of moles = \(\frac{1000 \text{ grams}}{27.64 \text{ g/mol}} = 36.18 \text{ moles}\).
Thus, if 1 mole releases 755.4 kJ, then 36.18 moles would release: \(36.18 \times -755.4 = -27334.29 \text{ kJ}\).

Stoichiometry

Now, let's dive into stoichiometry, which is the calculation of reactants and products in chemical reactions.
In this problem, we utilize stoichiometry to determine how much energy is released per kilogram of diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)).
The balanced chemical equation provided is:
\(\mathrm{B}_{2}\mathrm{H}_{6}(g) + 6 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{BCl}_{3}(g) + 6 \mathrm{HCl}(g)\)
This means that one mole of \(\mathrm{B}_{2}\mathrm{H}_{6}\) reacts with six moles of \(\mathrm{Cl}_{2}\) to produce two moles of \(\mathrm{BCl}_{3}\) and six moles of \(\mathrm{HCl}\).
By understanding the molar relationships between reactants and products, we can accurately calculate quantities needed and predict yields.
When we calculated the moles of \(\mathrm{B}_{2}\mathrm{H}_{6}\) in one kilogram by accounting for its molar mass, we then used the enthalpy change for one mole to find the total energy released.
Stoichiometry thus connects the theoretical calculation to real-world measurements, ensuring accurate results in experiments and industrial applications.