Problem 27 - Chemistry An Atoms First Approach

Question

Boron hydrides were once evaluated for possible use as rocket fuels. Complete and balance the following equation for the combustion of diborane. $$\mathbf{B}_{2} \mathrm{H}_{6}(g)+\mathbf{O}_{2}(g) \longrightarrow \mathbf{B}(\mathrm{OH})_{3}(s)$$

Step-by-Step Solution

Answer

The short answer for the balanced equation for the combustion of diborane (B₂H₆) with oxygen gas (O₂) to form boron hydroxide (BOH₃) is: \(B_{2}H_{6}(g) + 3O_{2}(g) \longrightarrow 2B(OH)_{3}(s)\)

Step 1: Count atoms on both sides

First, count the number of each element (atoms) on the reactants and products side. Reactants side: B₂H₆ + O₂ 2 Boron (B) atoms, 6 Hydrogen (H) atoms, and 2 Oxygen (O) atoms. Products side: B(OH)₃ 1 Boron (B) atom, 3 Hydrogen (H) atoms, and 3 Oxygen (O) atoms.

Step 2: Balance the Boron atoms

To balance the Boron atoms, we see that there are 2 Boron atoms on the reactants side and 1 Boron atom in the products side. Therefore, we can multiply the entire BOH₃ by 2 to obtain 2 Boron atoms in the products side. Now the equation will appear as: \(B_{2}H_{6}(g) + O_{2}(g) \longrightarrow 2B(OH)_{3}(s)\) Now both sides have 2 Boron atoms.

Step 3: Balance the Hydrogen atoms

On the reactants side, we have 6 Hydrogen atoms, and on the products side, we have 6 Hydrogen atoms (since 2 * B(OH)₃ = 6 Hydrogen atoms). Therefore, Hydrogen atoms are already balanced.

Step 4: Balance the Oxygen atoms

Now we need to balance the number of Oxygen atoms in the reaction. On the reactants side, there are 2 Oxygen atoms, whereas, on the products side, there are 6 Oxygen atoms (2 * B(OH)₃ = 6 Oxygen atoms). Thus, we can balance the Oxygen atoms by multiplying O₂ by 3 on the reactants side. The complete balanced equation now becomes: \(B_{2}H_{6}(g) + 3O_{2}(g) \longrightarrow 2B(OH)_{3}(s)\) Now we have a balanced equation with 2 Boron atoms, 6 Hydrogen atoms, and 6 Oxygen atoms on both sides of the equation.

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is akin to solving a puzzle where each side of the equation must have the same number of atoms of each element. The combustion of diborane (B₂H₆) illustrates why this balance is essential for accurately representing chemical reactions.

When balancing equations, we follow a systematic approach, starting with individual elements and comparing their quantities on both sides of the equation. For the diborane combustion equation, we begin with the boron atoms and adjust coefficients to achieve balance. This meticulous step-by-step process ensures no element is unaccounted for.

After balancing Boron (B) atoms, a noticeable imbalance in Oxygen (O) atoms propels us to adjust the oxygen-containing reactants and products accordingly. It's not mere arithmetic; it's about understanding the stoichiometry, which includes realizing that O₂ molecules contain two oxygen atoms—a crucial detail in achieving a balanced reaction.

In the end, we arrive at a coherent, balanced equation, signifying that the law of conservation of mass is satisfied. The balanced equation serves as a blueprint for the reaction and is pivotal for further stoichiometric calculations.

Stoichiometry

The realm of stoichiometry is where chemistry meets mathematics; it's the quantitative relationship between reactants and products in a chemical reaction. By accurately balancing the equation for the combustion of diborane, we set the stage for precise stoichiometric calculations.

This discipline enables us to predict the amount of reactants needed and the quantity of products formed, vital for applications ranging from laboratory experiments to industrial processes. For instance, if we know the amount of diborane we have, stoichiometry tells us exactly how much oxygen is required for complete combustion and how much boron trihydroxide (B(OH)₃) will be synthesized.

It's crucial to grasp that stoichiometry is not just about the numbers; it's about the ratios. Each coefficient in the balanced equation represents a 'mole ratio', which can be thought of as a recipe for the chemical reaction. In the diborane reaction, the mole ratio of diborane to oxygen to boron trihydroxide is 1:3:2. With this mole ratio, we can scale up the reaction to any desired quantity, always preserving the same proportions.

Boron Hydrides

Boron hydrides, such as diborane (B₂H₆), belong to a fascinating class of compounds characterized by complex structures and diverse reactivity. They are known for their electron-deficiency, leading to unusual bonding schemes that intrigue chemists.

Diborane is an example of a binary boron hydride, composed only of boron and hydrogen. These compounds are of particular interest because they can form multi-center bonds, where two or more atoms share a pair of electrons. Despite this electron deficiency, they often display remarkable stability until reacted with specific reagents, such as oxygen, as we see in the combustion process.

Understanding the properties and reactions of boron hydrides is not just academic; they have practical implications. Originally studied for potential as rocket fuels due to their high energy content, boron hydrides also serve as reagents in organic synthesis and as precursors to other boron-containing materials. The combustion of diborane, resulting in the production of boron trihydroxide, exemplifies how the energy contained in the B-H bonds can be harnessed, albeit for a reaction that generates a non-fuel product.