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) \(Q_{1}+Q_{2}-Q_{3}\)

Step-by-Step Solution

Answer

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

Step 1: Write the balanced equation

The balanced chemical equation for the reaction is \(\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3\). This reaction involves breaking and forming bonds.

Step 2: Identify bonds broken and formed

In the reaction, one \(\mathrm{N} \equiv \mathrm{N}\) bond and three \(\mathrm{H}-\mathrm{H}\) bonds are broken, and six \(\mathrm{N}-\mathrm{H}\) bonds are formed.

Step 3: Use bond energies to calculate \(\Delta \mathrm{H}\)

The enthalpy change \(\Delta \mathrm{H}\) can be calculated using the bond energies: - Bonds broken: \(\mathrm{Q}_3 + 3\mathrm{Q}_2\) - Bonds formed: \(6\mathrm{Q}_1\)The formula for \(\Delta \mathrm{H}\) is computed as: \[\Delta \mathrm{H} = (\text{Energy of bonds broken}) - (\text{Energy of bonds formed}) = (\mathrm{Q}_3 + 3\mathrm{Q}_2) - 6\mathrm{Q}_1\]

Step 4: Compare with given options

From the calculation, \(\Delta \mathrm{H} = (\mathrm{Q}_3 + 3\mathrm{Q}_2) - 6\mathrm{Q}_1\). This matches with option (c).

Key Concepts

Bond Energy

Bond energy is a fundamental concept in chemical thermodynamics. It refers to the amount of energy required to break one mole of a particular bond in a molecule. This energy is always expressed in kilojoules per mole (kJ/mol). Understanding bond energy helps us predict the stability of molecules and the energy changes during chemical reactions.

• For the reaction \( ext{N}_2 + 3 ext{H}_2 ightarrow 2 ext{NH}_3\), the significant bond energies are those of \( ext{N} ext{H}\), \( ext{H} ext{H}\), and \( ext{N} ext{N}\).
• Breaking these bonds involves absorbing energy, while forming new bonds releases energy.

In the context of the exercise, we look at the energies associated with breaking one \( ext{N} ext{N}\) and three \( ext{H} ext{H}\) bonds versus forming six \( ext{N} ext{H}\) bonds. High bond energy indicates a strong bond which requires more energy to break, thereby contributing to the stability of the molecule. Calculating bond energies helps chemists understand and predict reaction outcomes and design processes accordingly.

Enthalpy Change

Enthalpy change, denoted as \( riangle H\), is another key concept in chemical reactions. It measures the total energy change within a system. Specifically, it accounts for both the energy absorbed and released during the breaking and forming of bonds.
During a reaction:

• Energy absorbed when bonds break is counted as positive.
• Energy released when new bonds form is negative.

The enthalpy change of a reaction can be calculated using the formula:\[\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}\]In the given reaction, we calculate the enthalpy change using bond energies \(Q_1\), \(Q_2\), and \(Q_3\). Applying the bond energy values into the formula helps determine if a reaction is endothermic (absorbs heat) or exothermic (releases heat), providing insights into the reaction’s thermodynamics.

Chemical Bonding

Chemical bonding underpins the formation and breakage of bonds in a reaction. It involves the forces that hold atoms together to form molecules. There are several types of chemical bonds, including ionic, covalent, and metallic. In the context of this reaction, covalent bonding plays a crucial role.
Covalent bonds occur when atoms share electrons to achieve a full outer electron shell, contributing to molecular stability. In this reaction:

• The \( ext{N} ext{N}\) triple bond is very strong due to high electron sharing.
• \( ext{H} ext{H}\) bonds are simpler and involve single electrons from hydrogen atoms.
• \( ext{N} ext{H}\) bonds are the result of nitrogen and hydrogen sharing electrons, forming ammonia.

Understanding chemical bonding is essential for predicting the behavior of molecules and the energy changes in reactions. The formation and breaking of these bonds cause the energy changes observed in the reaction, aligning with the fundamental principles of chemical thermodynamics.

Haber Process

The Haber Process is a pivotal industrial method for ammonia synthesis, vital for fertilizers and several other chemical products. It involves the reaction of nitrogen gas with hydrogen gas under high temperatures and pressures to form ammonia. This process is directly related to the concept of enthalpy change and bond energy.
Steps involved in the Haber Process:

• The nitrogen \(( ext{N}_2)\) and hydrogen gases \(( ext{H}_2)\) are combined in a 1:3 ratio.
• Conditions are controlled (pressures around 200 atmospheres and temperatures of about 450°C) to optimize ammonia production.
• A catalyst, typically iron, is used to accelerate the reaction rate.

During the production, the strong \( ext{N} ext{N}\) bonds are broken, and new \( ext{N} ext{H}\) bonds form to produce ammonia \(( ext{NH}_3)\). The energy dynamics involved in this process highlight the critical role of bond energies and enthalpy changes in achieving commercially viable yields.