Problem 39 - A Complete Resource Book in Chemistry for JEE Main

Question

Bond energy of \(\mathrm{N}-\mathrm{H}, \mathrm{H}-\mathrm{H}\), and \(\mathrm{N} \equiv \mathrm{N}\) bonds are \(\mathrm{Q}_{1}\), \(\mathrm{Q}_{2}\) and \(\mathrm{Q}_{3} ; \Delta \mathrm{H}\) of \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\) is (a) \(\mathrm{Q}_{3}+3 \mathrm{Q}_{2}-2 \mathrm{Q}_{1}\) (b) \(2 Q_{1}-Q_{3}-2 Q_{2}\) (c) \(\mathrm{Q}_{3}+3 \mathrm{Q}_{2}-6 \mathrm{Q}_{1}\) (d) \(\mathrm{Q}_{1}+\mathrm{Q}_{2}-\mathrm{Q}_{3}\)

Step-by-Step Solution

Answer

The correct answer is (c): \(\Delta \mathrm{H} = \mathrm{Q}_{3} + 3\mathrm{Q}_{2} - 6\mathrm{Q}_{1}\).

Step 1: Understanding the Reaction

The reaction given is the formation of ammonia (\(\mathrm{NH}_{3}\)) from nitrogen \(\mathrm{N}_{2}\) and hydrogen \(\mathrm{H}_{2}\): \(\mathrm{N}_{2}+3\mathrm{H}_{2} \longrightarrow 2\mathrm{NH}_{3}\). In this reaction, nitrogen molecules \(\mathrm{N}_{2}\) and hydrogen molecules \(\mathrm{H}_{2}\) are broken into individual nitrogen and hydrogen atoms to form ammonia molecules.

Step 2: Bond Breaking Calculations

For the reactants, the energy required to break all necessary bonds must be calculated. Breaking 1 mole of \(\mathrm{N}\equiv\mathrm{N}\) bond needs \(\mathrm{Q}_{3}\) kJ/mol, and breaking 3 moles of \(\mathrm{H}-\mathrm{H}\) bonds requires \(3\mathrm{Q}_{2}\) kJ/mol.

Step 3: Bond Formation in Products

When forming the products, two \(\mathrm{NH}_{3}\) molecules have a total of 6 \(\mathrm{N}-\mathrm{H}\) bonds. Each of these bonds releases \(\mathrm{Q}_{1}\) kJ/mol upon formation. Thus, forming these 6 bonds releases \(6\mathrm{Q}_{1}\) kJ/mol.

Step 4: Calculate the Total Enthalpy Change (\(\Delta H\))

The enthalpy change \(\Delta \mathrm{H}\) is calculated by subtracting the total bond formation energy from the total bond breaking energy. Therefore, \(\Delta \mathrm{H} = (\mathrm{Q}_{3} + 3\mathrm{Q}_{2}) - 6\mathrm{Q}_{1}\). Thus, the correct answer is option (c).

Key Concepts

Understanding Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a crucial concept in chemistry that reflects the total heat absorbed or released during a chemical reaction. It helps us understand the energy changes that occur when reactants transform into products. When bonds are broken, energy is absorbed, whereas, when bonds form, energy is released. The overall enthalpy change of a reaction is the balance between these energy alterations.
💡 Key points to remember about enthalpy change:

• Positive \( \Delta H \) indicates an endothermic reaction, where heat is absorbed from the surroundings.
• Negative \( \Delta H \) signifies an exothermic reaction, where heat is released to the surroundings.

In the context of ammonia formation, the enthalpy change aids us in understanding whether the reaction is endothermic or exothermic based on bond energies involved.

Ammonia Formation Reaction

The formation of ammonia (\( \text{NH}_3 \)) from nitrogen (\( \text{N}_2 \)) and hydrogen (\( \text{H}_2 \)) is a classic example of a chemical reaction that involves breaking and forming bonds. The balanced chemical equation for ammonia formation is \( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \). Here, nitrogen and hydrogen molecules are transformed into two molecules of ammonia.
This reaction is significant because ammonia is extensively used in fertilizers and various industrial processes. Understanding this reaction helps chemists design efficient processes to synthesize ammonia on a large scale.

Bond Breaking and Formation

Bond breaking and formation are fundamental processes in understanding chemical reactions. Each bond has a specific bond energy, which is the amount of energy needed to break that bond.
In the context of ammonia formation:

• Breaking one mole of \( \text{N}\equiv\text{N} \) triple bond requires energy \( \text{Q}_3 \).
• Breaking three moles of \( \text{H}-\text{H} \) single bonds needs energy equivalent to \( 3\text{Q}_2 \).
• Forming six \( \text{N}-\text{H} \) bonds in two \( \text{NH}_3 \) molecules releases \( 6\text{Q}_1 \) energy.

These processes are about the balance of breaking old bonds to form new ones, and calculating their respective energies helps us understand the overall enthalpy change.

N2 and H2 Reaction Mechanics

The reaction between \( \text{N}_2 \) and \( \text{H}_2 \) to form ammonia highlights the principle of bond energy calculations. Initially, nitrogen and hydrogen molecules must be dissociated into their respective atoms. This means breaking a strong \( \text{N}\equiv\text{N} \) triple bond and hydrogen's \( \text{H}-\text{H} \) bonds, which requires energy input.
After the dissociation, the atoms combine to form new \( \text{N}-\text{H} \) bonds. These new bonds release energy.

• The energy required to break the initial bonds and the energy released upon forming new bonds determines \( \Delta H \).
• For this specific reaction, \( \Delta H \) is calculated as \( (\text{Q}_3 + 3\text{Q}_2) - 6\text{Q}_1 \).

This calculation helps determine whether the overall reaction is energy-absorbing or energy-releasing, which is vital for applications like ammonia synthesis.