Problem 171 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard enthalpy of formation of \(\mathrm{NH}_{3}\) is \(-46.0\) \(\mathrm{kJ} \mathrm{mol}^{-1}\). If the enthalpy of formation of \(\mathrm{H}_{2}\) from its atoms is \(-436 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and that of \(\mathrm{N}_{2}\) is \(-712 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), the average bond enthalpy of \(\mathrm{N}-\mathrm{H}\) bond in \(\mathrm{NH}_{3}\) is \([\mathbf{2 0 1 0}]\) (a) \(-964 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(+352 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+1056 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1102 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The average N-H bond enthalpy in NH₃ is +352 kJ/mol (option b).

Step 1: Write the Formation Reaction

Write the balanced chemical equation for the formation of ammonia from its elements in their standard states: \[\frac{1}{2} \mathrm{N}_2(g) + \frac{3}{2} \mathrm{H}_2(g) \rightarrow \mathrm{NH}_3(g)\] The enthalpy change for this reaction is the standard enthalpy of formation of ammonia, given as \(-46.0\ \mathrm{kJ}\ \mathrm{mol}^{-1}\).

Step 2: Determine the Bond Enthalpy for H-H and N≡N Bonds

The enthalpy change to form \(\mathrm{H}_2\) from its atoms is \(-436 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Hence, breaking the bonds in \(\mathrm{H}_2\) would be \(+436 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Similarly, breaking the \(\mathrm{N}_2\) molecule requires \(+712 \mathrm{~kJ}\ \mathrm{mol}^{-1}\).

Step 3: Calculate Total Bond Energy of Reactants

Calculate the total bond energy required to break the bonds in the reactants: \[\text{Total } = \frac{1}{2}(712) + \frac{3}{2}(436) \]\[ = 356 + 654 = 1010 \ \mathrm{kJ}\ \text{per mole of } \mathrm{NH}_3\]

Step 4: Use Hess's Law to Find Bond Enthalpy of N-H

By Hess's Law,\[\text{Total energy to break bonds} - \text{Bond enthalpy of new bonds formed} = \Delta H_f \text{ (enthalpy of formation)}\]\[1010 - 3x = -46\]where \(x\) is the bond enthalpy of one \(\mathrm{N-H}\) bond. Solve for \(x\).

Step 5: Solve for N-H Bond Enthalpy

Rearrange the equation to find the average \(\mathrm{N-H}\) bond enthalpy:\[3x = 1010 + 46\]\[3x = 1056\]\[x = \frac{1056}{3}\]\[x = 352\ \mathrm{kJ}\ \mathrm{mol}^{-1}\]

Step 6: Confirm and Select the Answer

The average bond enthalpy of the \(\mathrm{N-H}\) bond is \(352\ \mathrm{kJ} \mathrm{~mol}^{-1}\). Thus, the correct answer is option (b) \(+352\) \(\mathrm{kJ} \mathrm{~mol}^{-1}\).

Key Concepts

Bond Enthalpy

Bond enthalpy is a measure of the strength of a chemical bond. It represents the energy required to break one mole of bonds in a substance, usually in its gaseous state. The concept is central to understanding chemical reactions, as breaking and forming bonds are key steps. For instance, in the exercise, we calculate the bond enthalpies for \( ext{H-H}\) and \( ext{N≡N}\) bonds. Using bond enthalpies helps us to estimate the energy changes in a chemical reaction. To break hydrogen molecules \( (\text{H}_2)\), the bond enthalpy is \( +436 \, \text{kJ} \, \text{mol}^{-1}\). Meanwhile, breaking the triple bond in nitrogen \( (\text{N}_2)\) requires \( +712 \text{ kJ} \text{ mol}^{-1}\). By evaluating these energies, we deduce the energy involved in breaking bonds in the reactants. This information is crucial for calculating the average bond enthalpy for \( \text{N-H}\) bonds in ammonia.

Hess's Law

Hess's Law is a principle that states the total enthalpy change during a chemical reaction is the same, regardless of the pathway or steps taken. This means you can add up the enthalpy changes for individual steps to find the overall enthalpy change for the reaction. It's like taking different routes to a destination but still covering the same distance in total.In the exercise, we use Hess's Law to determine the enthalpy required to form ammonia (\( \text{NH}_3 \)). We add up the energies needed to break \( \frac{1}{2} \, \text{N}_2 \) and \( \frac{3}{2} \text{H}_2\) to qualify for the formation of new \( \text{N-H}\) bonds. This total bond-breaking energy, which is \( 1010 \, \text{kJ} \, \text{mol}^{-1} \,\), is used along with the given standard enthalpy of formation of ammonia \( (–46 \, \text{kJ} \, \text{mol}^{-1})\) to calculate the average bond enthalpy of the \( \text{N-H}\) bonds.

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to transform reactants into products. This process typically involves breaking existing bonds and forming new ones, which involves changes in energy, known as enthalpy. Each chemical reaction is unique, having specific energy requirements to move from reactants to products. In the context of ammonia formation, the reaction converts nitrogen and hydrogen gases into ammonia gas. We calculate the energy needed to do this. The exercise demonstrates that by breaking reactant bonds and forming stronger ones in ammonia, we evaluate the average bond enthalpy of new bonds, which gives insights into the energetics of the reaction. Understanding these principles helps predict the spontaneity and energy efficiency of reactions, critical for industrial applications like fertilizer production.

Ammonia Formation

Ammonia formation is a chemical process that combines nitrogen and hydrogen gases to create ammonia \( (\text{NH}_3)\). It's an exothermic reaction, releasing energy as heat. The standard enthalpy of formation is \( -46.0 \text{ kJ} \text{ mol}^{-1}\), suggesting that forming ammonia from its elemental gases releases energy.The balanced chemical equation \( \frac{1}{2}\, \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g)\) showcases this formation process, elucidating the stoichiometry. Generally, ammonia is pivotal in agriculture, especially in fertilizers, due to its high nitrogen content, essential for plant growth.In the exercise, we calculate average bond enthalpies in newly formed \( \text{N-H}\) bonds by considering the energies involved in breaking and forming bonds. This gives more profound insight into the efficiency and dynamics of the ammonia synthesis process, illuminating its importance in both chemistry and industry.