Question

Consider the reaction $$ \mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-2035 \mathrm{kJ} $$ Calculate the amount of heat released when 54.0 \(\mathrm{g}\) of diborane is combusted.

Short answer

The heat released when 54.0 g of diborane is combusted is -3973 kJ.

Step-by-step solution

Calculate the amount of moles of diborane

To calculate the amount of moles, we will use the formula: Moles = Mass (g) / Molar mass (g/mol) Diborane has a molar mass of (2 × 10.81) + (6 ×1.01) = 27.62 g/mol. So, Moles of diborane = 54.0 g / 27.62 g/mol

Find the moles of B₂O₃ and H₂O produced

From the balanced reaction: 1 mol B₂H₆ produces 1 mol B₂O₃ and 3 mol H₂O. But we calculated a different number of moles for B₂H₆ in Step 1, so using the stoichiometric ratios: Moles of B₂O₃ = moles of B₂H₆ produced Moles of H₂O = 3 × moles of B₂H₆ produced

Calculate the heat released

The enthalpy change provided is the heat released for the balanced equation. We can rewrite this as: Heat = ΔH × moles of B₂H₆ For this reaction, ΔH = -2035 kJ/mol, Plug in the values to calculate the heat released: Heat released = (-2035 kJ/mol) × moles of B₂H₆ from step 1

The final answer

The final result is the calculated heat released from the reaction, obtained by solving the equation in step 3: Heat released = (-2035 kJ/mol) × (54 g / 27.62 g/mol) Heat released = (-2035 kJ/mol) × (1.954 mol) Heat released = -3973 kJ Thus, the heat released when 54.0 g of diborane is combusted is -3973 kJ (negative sign indicating that the heat is released).

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that relates to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the balance and proportion of elements involved in the process. In the given chemical reaction of diborane combusting in oxygen, stoichiometry helps us understand how the moles of reactants convert into moles of products. Using the balanced equation,

• 1 mole of \( \text{B}_2\text{H}_6 \) reacts with 3 moles of \( \text{O}_2 \)
• Results in 1 mole of \( \text{B}_2\text{O}_3 \) and 3 moles of \( \text{H}_2\text{O} \)

These relationships indicate that for every mole of diborane combusted, an equal number of moles of \( \text{B}_2\text{O}_3 \) and three times that of water are formed. This mole ratio is crucial when calculating the expected outcomes of the reaction, allowing precise calculations of amounts generated or used.

Molar Mass

Molar mass is an essential concept that tells us the mass of one mole of a substance, allowing for the conversion between grams and moles. It is determined by summing the atomic masses of all atoms in a molecular formula. For diborane, \( \text{B}_2\text{H}_6 \),

• The atomic mass of Boron (B) is approximately 10.81 amu
• The atomic mass of Hydrogen (H) is approximately 1.01 amu

Using this, the molar mass of diborane is calculated as:\[(2 \times 10.81) + (6 \times 1.01) = 27.62 \text{ g/mol}\] This value lets us convert the mass of diborane provided in the exercise into moles:\[\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}}\]Thus, for 54.0 g of diborane, the moles are:\[\frac{54.0 \text{ g}}{27.62 \text{ g/mol}}\] Providing the precise number of moles needed in the stoichiometric calculations.

Heat Released

The concept of heat released refers to the amount of energy liberated when a reaction occurs, commonly expressed in kilojoules (kJ). Enthalpy change (\( \Delta H \)) signifies this energy transfer, and it is negative for exothermic reactions where heat is released. In the case of the combustion of diborane, the enthalpy change is given as\[\Delta H = -2035 \text{ kJ/mol}\] This means that for every mole of diborane combusted, 2035 kJ of heat is released. By knowing the number of moles of diborane,

• Heat released = \( \Delta H \times \text{moles of } \text{B}_2\text{H}_6 \)
• For 1.954 moles calculated, the heat released is \(-2035 \text{ kJ/mol} \times 1.954 \text{ mol} \)

This results in a total of -3973 kJ, indicating a substantial release of energy during the combustion process, with the negative sign emphasizing the exothermic nature of the reaction.