Question

Given the following thermochemical equations: $$ \begin{aligned} 4 \mathrm{~B}(s)+3 \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{~B}_{2} \mathrm{O}_{3}(s) & & \Delta H^{\circ}=-2543.8 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & & \Delta H^{\circ}=-241.8 \mathrm{~kJ} \\ \mathrm{~B}_{2} \mathrm{H}_{6}(s)+3 \mathrm{O}_{2} \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ} &=-2032.9 \mathrm{~kJ}\end{aligned} $$ Calculate \(\Delta H^{\circ}\) for the decomposition of \(\mathrm{B}_{2} \mathrm{H}_{6}\) into its elements.

Short answer

Based on the given thermochemical equations and their corresponding enthalpy changes, the enthalpy change for the decomposition of B₂H₆ into its elements (2B(s) + 3H₂(g)) is 1236.3 kJ.

Step-by-step solution

Invert the first equation

In order to get the desired reaction, we will first invert the first equation. To do this, we will reverse the direction of the first given thermochemical equation and change the sign of its enthalpy change: $$ \begin{aligned} 2\mathrm{B}_{2}\mathrm{O}_{3}(s) &\longrightarrow 4\mathrm{B}(s) + 3\mathrm{O}_{2}(g) \\ & \Delta H^{\circ} = 2543.8\mathrm{~kJ} \end{aligned} $$

Invert the second equation and multiply by 3

Seeing that the desired reaction has \(3\mathrm{H}_{2}(g)\), we will invert the second equation and multiply it by 3, too: $$ \begin{aligned} 3\left( \mathrm{H}_{2}\mathrm{O}(g) \longrightarrow 3\mathrm{H}_{2}(g) + \frac{3}{2}\mathrm{O}_{2}(g) \right) \\ 3 \cdot \Delta H^{\circ} = 3 \cdot 241.8\mathrm{~kJ}\end{aligned} $$ This results in: $$ \begin{aligned} 3\mathrm{H}_{2}\mathrm{O}(g) &\longrightarrow 3\mathrm{H}_{2}(g) + \frac{3}{2}\mathrm{O}_{2}(g) \\ \Delta H^{\circ} = 725.4\mathrm{~kJ} \end{aligned} $$

Add the manipulated equations and their enthalpy changes

Now, we will sum the two manipulated equations and their corresponding enthalpy changes. This will give us the desired equation of decomposition and its enthalpy change: $$ \begin{aligned}& (2\mathrm{B}_{2}\mathrm{O}_{3}(s) \longrightarrow 4\mathrm{B}(s) + 3\mathrm{O}_{2}(g))\\ &+(3\mathrm{H}_{2}\mathrm{O}(g) \longrightarrow 3\mathrm{H}_{2}(g) + \frac{3}{2}\mathrm{O}_{2}(g)) \\& -(\mathrm{B}_{2}\mathrm{H}_{6}(s)+3\mathrm{O}_{2} \longrightarrow \mathrm{B}_{2}\mathrm{O}_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(g)) \end{aligned} $$ Adding the three equations: $$ \begin{aligned} \mathrm{B}_{2}\mathrm{H}_{6}(s) &\longrightarrow 2\mathrm{B}(s) + 3\mathrm{H}_{2}(g) \end{aligned} $$ Adding their enthalpy changes: $$ \Delta H^{\circ} = 2543.8\mathrm{~kJ} + 725.4\mathrm{~kJ} - 2032.9\mathrm{~kJ} = 1236.3\mathrm{~kJ} $$

The enthalpy change for the decomposition reaction

The enthalpy change for the decomposition of \(\mathrm{B}_{2}\mathrm{H}_{6}\) into its elements is: $$ \Delta H^{\circ} = 1236.3\mathrm{~kJ} $$

Key concepts

Understanding Enthalpy Change

When studying chemical reactions, the term 'enthalpy change' represents the heat released or absorbed by a system under constant pressure. It's denoted by \( \Delta H \) and measured in kilojoules (kJ).

In essence, enthalpy change is the total energy change in a chemical reaction. A negative \( \Delta H \) signifies an exothermic reaction where energy is released to the surroundings. Conversely, a positive \( \Delta H \) indicates an endothermic reaction, meaning the system absorbs energy.

For example, in the provided exercise, burning boron \( \mathrm{B} \) in oxygen \( \mathrm{O}_{2} \) to form boron trioxide \( \mathrm{B}_{2} \mathrm{O}_{3} \) releases energy, hence a negative \( \Delta H \) is observed. Thermochemical equations couple stoichiometry with enthalpy changes, providing critical insights about energy requirements or releases during reactions.

Chemical Reaction Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's fundamental for predicting the amounts of substances consumed and produced.

In stoichiometry, balancing equations ensures that the number of atoms for each element is the same on both sides. For energy considerations, stoichiometric calculations also involve the enthalpy change (\(\Delta H\)). With a balanced equation, you can determine the enthalpy change for a given amount of reactant or product.

In the problem at hand, the stoichiometry of the inverted and modified equations helps reveal the individual steps of the decomposition of \( \mathrm{B}_{2} \mathrm{H}_{6} \) into \( \mathrm{B} \) and \( \mathrm{H}_{2} \) with their associated enthalpy changes. These steps lead to the calculation of the total enthalpy change for the decomposition process.

Applying Hess's Law

Hess's Law states that the total enthalpy change in a chemical reaction is the same, regardless of the number of steps the reaction is carried out in. Simply put, enthalpy is a state function—it only depends on the initial and final states, not on the path taken.

In practice, Hess's Law allows us to calculate the \( \Delta H \) for a reaction that may not be easily measurable by using other reactions that sum up to the desired process. This is invaluable because it simplifies complex reactions into a series of more manageable steps with known enthalpy changes.

For the provided problem, Hess’s Law enables us to manipulate given equations (by reversing and multiplying while accordingly changing the sign and magnitude of \( \Delta H \) respectively), then combine them to derive the enthalpy change for a reaction that's not given directly—the decomposition of \( \mathrm{B}_{2} \mathrm{H}_{6} \) into \( \mathrm{B} \) and \( \mathrm{H}_{2} \) in this case. The meticulous alignment of reactants and products properly accounts for all substances involved, demonstrating the practicality of Hess's Law in solving thermochemical puzzles.