Problem 31 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Diborane is a potential rocket fuel which undergoes combustion according to the reaction. [2000 - 2 Marks] \(\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) From the following data, calculate the enthalpy change for the combustion of diborane. \(2 \mathrm{~B}(\mathrm{~s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) \quad \Delta H=-1273 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell) \quad \Delta H=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad \Delta H=44 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(2 \mathrm{~B}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g}) \quad \Delta H=36 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The enthalpy change for the combustion of diborane is \(-2035 \mathrm{~kJ/mol}\).

Step 1: Write the Target Reaction

The target reaction is the combustion of diborane as given in the problem: \(\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). We need to find the enthalpy change (\(\Delta H_{\text{rxn}}\)) for this reaction.

Step 2: Analyze Given Reactions and Data

We are given several thermochemical equations with their enthalpy changes (\(\Delta H\)). These are:1. \(2 \mathrm{~B}(\mathrm{~s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})\) with \(\Delta H = -1273 \mathrm{~kJ/mol}\).2. \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) with \(\Delta H = -286 \mathrm{~kJ/mol}\).3. \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) with \(\Delta H = 44 \mathrm{~kJ/mol}\).4. \(2 \mathrm{~B}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})\) with \(\Delta H = 36 \mathrm{~kJ/mol}\).

Step 3: Transform \\(\Delta H\) of Each Step

We need to mathematically manipulate the given reactions to add up to the target reaction, using Hess’s Law, which states the total \(\Delta H\) for a chemical reaction is the sum of the \(\Delta H\) values from its steps. This also involves potentially reversing the reaction direction and altering stoichiometric coefficients.

Step 4: Reverse Reaction 4

To incorporate diborane as a reactant in the target reaction, reverse reaction 4:\[\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g}) \rightarrow 2 \mathrm{~B}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g})\]This reversal changes the enthalpy sign:\(\Delta H = -36 \mathrm{~kJ/mol}\).

Step 5: Use Reactions 1 and 2

To form \(\mathrm{B}_{2} \mathrm{O}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}(\ell)\): - Reaction 1: \(2 \mathrm{~B}(\mathrm{~s}) + \frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})\, \Delta H = -1273 \mathrm{~kJ/mol}\) - To form \(3 \mathrm{H}_{2} \mathrm{O}(\ell)\), use Reaction 2 multiplied by 3:\[3 \left(\mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\right)\]\(\Delta H = 3(-286 \mathrm{~kJ/mol}) = -858 \mathrm{~kJ/mol}\).

Step 6: Convert \\(\mathrm{H}_{2} \mathrm{O}(\ell)\) to \\(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

To convert \(3 \mathrm{H}_{2} \mathrm{O}(\ell)\) to \(3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), use reaction 3 multiplied by 3:\[3 \left(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right)\]\(\Delta H = 3(44 \mathrm{~kJ/mol}) = 132 \mathrm{~kJ/mol}\).

Step 7: Calculate Total \\(\Delta H_{rxn}\\)

Sum the enthalpy changes from each adapted reaction to get \(\Delta H_{\text{rxn}}\):- Reverse of Reaction 4: \(-36 \mathrm{~kJ/mol}\)- Reaction 1: \(-1273 \mathrm{~kJ/mol}\)- 3 x Reaction 2: \(-858 \mathrm{~kJ/mol}\)- 3 x Reaction 3: \(+132 \mathrm{~kJ/mol}\)Total \(\Delta H_{\text{rxn}} = -36 - 1273 - 858 + 132 = -2035 \mathrm{~kJ/mol}\).

Key Concepts

Hess's Law

Hess's Law is an essential principle in thermochemistry, making calculations of enthalpy changes in chemical reactions straightforward. It states that the total enthalpy change for a chemical reaction is the same, irrespective of the route taken, whether it occurs in a single step or multiple steps.

This principle relies on the law of conservation of energy, emphasizing that energy within a system is neither created nor destroyed. Thus, by adding the enthalpy changes from intermediate steps, you can determine the total enthalpy change for a complex reaction. In practice, this involves manipulating known reactions and their enthalpy changes, rearranging them to match the target reaction.

• Sum the enthalpy changes (\(\Delta H\) values) of known reactions adjusted to form a series of steps equivalent to the desired reaction.
• Multiply or flip reactions as needed, considering the corresponding changes in \(\Delta H\).

Understanding Hess's Law enables students to derive the enthalpy change for reactions, like the combustion of diborane, using pre-measured data from other reactions.

Thermochemical Equations

Thermochemical equations are a powerful tool, embedding chemical reactions with enthalpy changes. They indicate not only the reactants and products but also the heat absorbed or released during the reaction.

A thermochemical equation includes specific symbols and coefficients, representing the precise physical state of substances involved (e.g., gas, liquid, solid) and stoichiometry, which must balance both atoms and charge. The enthalpy change (\(\Delta H\)) is stringent, tailored to conditions outlined, like temperature and pressure.

For example, in the reaction:\[2 \mathrm{B} (s) + \frac{3}{2} \mathrm{O}_{2} (g) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3} (s)\]
The \(\Delta H\) value is given per mole of reaction, allowing calculations of energy changes for any amount of reactants involved. Thermochemical equations not only help calculate energy changes using Hess’s Law, but they also visually exhibit energy transformations within the reaction.

Combustion Reaction

A combustion reaction is a high-energy process often involving a fuel and an oxidant, typically oxygen, producing heat and creating products like carbon dioxide, water, or other oxides.

These reactions are exothermic, meaning they release energy and often appear in the burning of fuels. For example, the combustion of diborane is:\[\mathrm{B}_{2} \mathrm{H}_{6} (g) + 3 \mathrm{O}_{2} (g) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3} (s) + 3 \mathrm{H}_{2} \mathrm{O} (g)\] Diborane's role as a rocket fuel candidate highlights its high-energy release upon combustion.

• An exothermic process, demonstrating vast amounts of energy released as heat.
• Produces stable oxides like \(\mathrm{B}_{2} \mathrm{O}_{3} (s)\) and water vapor.

Understanding these reactions helps assess fuel efficiency and energy output, pinpointing why specific compounds, like diborane, are investigated in advanced fuel applications.

Diborane

Diborane (\(\mathrm{B}_{2} \mathrm{H}_{6}\)) is an intriguing inorganic compound, boasting unique chemical properties. It consists of boron and hydrogen, outlining its role as a hydride of boron.

It is colorless, unstable, and highly reactive with a well-earned reputation for its vigorous combustion when exposed to oxygen, making it interesting within space exploration contexts. However, its high reactivity also demands caution when handling, as it can be quite hazardous.

• Potential as a high-energy rocket fuel.
• Explored for various industrial applications due to its high reactivity.

Despite its promise, the instability of diborane means it poses safety challenges, balancing its potential uses against risks. Its behavior during combustion provides key insights into its energy release, pivotal for energy and propulsion applications.