Question

(a) Calculate the standard enthalpy of formation of gaseous diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) using the following thermochemical information: \(\begin{array}{clrl}4 \mathrm{~B}(\mathrm{~s})+3 \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{~B}_{2} \mathrm{O}_{3}(s) & \Delta H^{\circ}=-2509.1 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}= & -571.7 \mathrm{~kJ} \\\ \mathrm{~B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-2147.5 \mathrm{~kJ}\end{array}\) (b) Pentaborane \(\left(\mathrm{B}_{5} \mathrm{H}_{9}\right)\) is another boron hydride. What experiment or experiments would you need to perform to yield the data necessary to calculate the heat of formation of \(\mathrm{B}_{5} \mathrm{H}_{9}(l) ?\) Explain by writing out and summing any applicable chemical reactions.

Short answer

(a) Using Hess's Law, we manipulate the given reactions to represent the formation reaction of gaseous diborane (B2H6): \(B(s) + \dfrac{3}{2}H_2(g) \longrightarrow B_2H_6(g)\) Summing the adjusted reactions and their enthalpies, the standard enthalpy of formation for B2H6(g) is: \(\Delta H^{\circ}_{B_2H_6} = 1006.5\, kJ\) (b) To calculate the heat of formation of B5H9(l), we would need to perform a calorimetric experiment for the reaction: \(5B(s) + \dfrac{9}{2}H_2(g) \longrightarrow B_5H_9(l)\) This experiment would measure the heat of reaction, which is the enthalpy change for the reaction, allowing us to find the heat of formation of B5H9(l).

Step-by-step solution

Target Reaction

\(B(s) + \dfrac{3}{2}H_2(g) \longrightarrow B_2H_6(g)\)

Reverse reaction 1

We must reverse the first reaction so that B(s) is on the left side, and 2B2O3(s) is on the right side of the equation: \(2B_2O_3(s) \longrightarrow 4B(s) + 3O_2(g)\) Now, we need to multiply the enthalpy for this reversed reaction by -1. \(\Delta H^{\circ} = 2509.1\, kJ\)

Adjust the coefficients of reaction 2

We must multiply reaction 2 by 3/4 to allow cancellation of 3O2(g): \(3/4(2H_2(g) + O_2(g) \longrightarrow 2H_2O(l))\) Now, we need to multiply the enthalpy for this reaction by 3/4. \(\Delta H^{\circ} = -428.8\, kJ\)

Adjust the coefficients of reaction 3

We must multiply reaction 3 by 1/2 so that 3H2O(l) is on the right side of the equation: \(1/2(B_2H_6(g) + 3O_2(g) \longrightarrow B_2O_3(s) + 3H_2O(l))\) Now, we need to multiply the enthalpy for this reaction by 1/2. \(\Delta H^{\circ} = -1073.8\, kJ\)

Add all the adjusted reactions and their enthalpies

\(2B_2O_3(s) \longrightarrow 4B(s) + 3O_2(g) ; \Delta H^{\circ} = 2509.1\, kJ\) \(3/4(2H_2(g) + O_2(g) \longrightarrow 2H_2O(l)) ; \Delta H^{\circ} = -428.8\, kJ\) \(1/2(B_2H_6(g) + 3O_2(g) \longrightarrow B_2O_3(s) + 3H_2O(l)) ; \Delta H^{\circ} = -1073.8\, kJ\) Now, we can add the equations and their enthalpies to get the overall reaction and enthalpy change.

Overall Reaction

\(B(s) + \dfrac{3}{2}H_2(g) \longrightarrow B_2H_6(g)\) \(\Delta H^{\circ}_{B_2H_6} = 2509.1\, kJ - 428.8\, kJ - 1073.8\, kJ\) \(\Delta H^{\circ}_{B_2H_6} = 1006.5\, kJ\) (b) What experiment or experiments would you need to perform to yield the data necessary to calculate the heat of formation of B5H9(l)? We need to be able to write a balanced chemical equation for the reaction that forms B5H9(l) using elements in their standard states. An example would be: \(5B(s) + \dfrac{9}{2}H_2(g) \longrightarrow B_5H_9(l)\) To calculate the heat of formation, we need to know the enthalpy change for this reaction. We don't have any reactions involving B5H9(l) in the given data. We would need to perform a calorimetric experiment to measure the heat of reaction.

Experiment(s)

Perform a calorimetric experiment for the reaction: \(5B(s) + \dfrac{9}{2}H_2(g) \longrightarrow B_5H_9(l)\) By conducting this experiment, we will be able to measure the heat of reaction, which is the enthalpy change for the reaction, and thus find the heat of formation of B5H9(l).

Key concepts

Thermochemical Equations

Thermochemical equations are chemical equations that include enthalpy change as part of the equation itself. This means that alongside the balanced chemical equation of reactants and products, the enthalpy change (\( \Delta H \)) is indicated, providing insight into the energy aspect of the reaction.

For example, the reaction given in the exercise, \( 4 \text{B}(\text{s}) + 3 \text{O}_{2}(\text{g}) \longrightarrow 2 \text{B}_{2} \text{O}_{3}(\text{s}), \Delta H^{\circ} = -2509.1 \text{ kJ} \), explicitly mentions the energy released during the formation of boron trioxide from boron and oxygen.

A thermochemical equation ensures clarity about whether a reaction is exothermic (releases energy) or endothermic (absorbs energy) by displaying the direction and magnitude of energy flow.

• Exothermic reactions have a negative \( \Delta H \)
• Endothermic reactions have a positive \( \Delta H \)

This notation helps chemists and engineers ascertain energy requirements or production for industrial processes.

Enthalpy Change

Enthalpy change, \( \Delta H \), is a crucial concept in thermodynamics, especially when discussing chemical reactions. It represents the heat absorbed or released during a chemical process at constant pressure.

When \( \Delta H < 0 \), a reaction is termed exothermic, indicating that heat is released to the surroundings, as seen in our original exercise where reactions release heat. On the other hand, if \( \Delta H > 0 \), the reaction is endothermic, meaning it absorbs heat from its surroundings.

Calculating enthalpy changes often involves combining multiple reactions, sometimes reversing and scaling thermochemical equations, as demonstrated in the solution. This process is known as Hess's Law, which allows us to find the enthalpy change of a target reaction by summing the enthalpy changes of known reactions leading to the same final and initial states.

Being able to compute \( \Delta H \) provides critical insights into the feasibility and energy dynamics of chemical processes.

Calorimetry

Calorimetry is a scientific technique used to measure the amount of heat exchanged during chemical or physical processes. This method is essential for determining enthalpy changes, especially when data is not available from literature or cannot be determined using Hess's Law.

In a calorimetric experiment, a reaction is carried out in a calorimeter, a device that isolates the system to accurately assess heat exchange. Measuring the temperature change of the system allows for calculation of the heat transfer using the formula:
\[q = mc\Delta T\]Where:

• \( q \) is the heat exchanged
• \( m \) is the mass of the substance
• \( c \) is the specific heat capacity
• \( \Delta T \) is the change in temperature

This technique is applicable when determining the heat of formation for compounds like pentaborane \((\text{B}_5\text{H}_9)\), as stated in the exercise, by directly measuring the heat change when forming the compound.

Boron Hydrides

Boron hydrides, such as diborane \((\text{B}_2\text{H}_6)\) and pentaborane \((\text{B}_5\text{H}_9)\) , are compounds of boron and hydrogen featuring notable structural and energetic properties. They are often studied for their potential applications in chemistry and energy, such as rocket propulsion and hydrogen storage.

These hydrides have complex bonding and structures, making them interesting and somewhat challenging to analyze and synthesize. The enthalpy of formation for such compounds gives insight into their stability and reactivity, which is vital for applications like fuel or in material science.

Studying boron hydrides also contributes to understanding the unique chemistry of boron, including its ability to form multi-center bonds, which are quite different from those seen in more common organic and inorganic compounds.