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 (a) \(-964 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(+352 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+1056 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1102 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short answer

The average bond enthalpy of \(\mathrm{N}-\mathrm{H}\) in \(\mathrm{NH}_3\) is approximately \(+352\ \mathrm{kJ\ mol^{-1}}\).

Step-by-step solution

Write the Formation Reaction

The enthalpy of formation for ammonia, \(\mathrm{NH}_3\), refers to the reaction: \[\frac{1}{2} \mathrm{N}_2 (g) + \frac{3}{2} \mathrm{H}_2 (g) \rightarrow \mathrm{NH}_3 (g)\]. The given \(\Delta H_f^\circ\) for \(\mathrm{NH}_3\) is \(-46.0\ \mathrm{kJ}\ \mathrm{mol}^{-1}\) for this reaction.

Use Bond Enthalpies to Define the Reaction

We need to use bond enthalpies to express the formation. The reaction above involves breaking \(\frac{1}{2} \times 2 = 1\) \(\mathrm{N}\equiv\mathrm{N}\) bond and \(\frac{3}{2} \times 2 = 3\) \(\mathrm{H} - \mathrm{H}\) bonds. Then, 3 \(\mathrm{N} - \mathrm{H}\) bonds are formed.

Write the Enthalpy Change in Terms of Bond Energies

The reaction enthalpy can be expressed with bond energies: \[\Delta H = \left(\mathrm{Bond \ enthalpies \ of \ bonds \ broken}\right) - \left(\mathrm{Bond \ enthalpies \ of \ bonds \ formed}\right).\] Hence, \[\Delta H \approx \left(BE(N \equiv N) + 3 \times BE(H-H) \right) - 6 \times BE(N-H).\]

Solve for Bond Enthalpy of N-H

Given, \(\Delta H = -46\ \mathrm{kJ\ mol^{-1}}\), \(BE(H-H) = 436\ \mathrm{kJ\ mol^{-1}}\), and \(BE(N\equiv N) = 712\ \mathrm{kJ\ mol^{-1}}\). Substitute these into the equation: \[-46 = \left(712 + 3 \times 436\right) - 6 \times BE(N-H).\]Solve for \(BE(N-H)\):\[-46 = 712 + 1308 - 6 \times BE(N-H)\]\[ -46 = 2020 - 6 \times BE(N-H) \]\[ -6 \times BE(N-H) = -2066 \]\[ BE(N-H) = \frac{2066}{6} = 344.33\ \mathrm{kJ\ mol^{-1}} \approx 352\ \mathrm{kJ\ mol^{-1}}.\]

Choose the Correct Answer

The calculated average bond enthalpy of \(\mathrm{N}-\mathrm{H}\) in \(\mathrm{NH}_3\) is approximately \(352\ \mathrm{kJ\ mol^{-1}}\), thus the correct answer is (b) \(+352\ \mathrm{kJ\ mol^{-1}}\).

Key concepts

Bond Enthalpy

Bond enthalpy, also known as bond energy, refers to the amount of energy needed to break one mole of a particular bond in a gaseous molecule. It is a crucial concept for understanding how chemical reactions occur and how energy is transferred in these processes. Typically, bond enthalpies are given as positive values because energy must be supplied to break bonds.
In chemical reactions, bonds are broken in reactants and new bonds are formed in products. The enthalpy change of a reaction is determined by comparing the total bond energies of bonds broken and formed:

• Breaking bonds consumes energy (endothermic process).
• Forming bonds releases energy (exothermic process).

These principles apply to the formation of ammonia, where hydrogen and nitrogen atoms are combined to form \(\mathrm{NH}_3\). Calculating the required energies provides insight into the reaction's enthalpy characteristics.

Enthalpy Change

The enthalpy change, often represented by \(\Delta H\), is a measure of heat absorbed or released in a reaction at constant pressure. It indicates whether a reaction is exothermic or endothermic:

• Exothermic: \(\Delta H\) is negative, meaning heat is released to the surroundings.
• Endothermic: \(\Delta H\) is positive, implying heat is absorbed from the surroundings.

The enthalpy change of a reaction can be determined using bond enthalpies and the formula:
\[\Delta H = \sum\text{(Bond energies of bonds broken)} - \sum\text{(Bond energies of bonds formed)}\]In the formation of ammonia, we see an enthalpy change described by breaking nitrogen-nitrogen triple bonds and hydrogen-hydrogen bonds, and forming nitrogen-hydrogen bonds. These energy changes help us calculate the specific bond enthalpies involved in the reaction, as demonstrated in the steps provided.

Ammonia Formation Reaction

The ammonia formation reaction involves combining hydrogen and nitrogen gases to produce ammonia. The standard reaction is written as:
\[\frac{1}{2} \mathrm{N}_2 (g) + \frac{3}{2} \mathrm{H}_2 (g) \rightarrow \mathrm{NH}_3 (g)\]
This represents the formation of ammonia from its constituent elements in their standard states. The standard enthalpy of formation for ammonia, expressed as \(\Delta H_f^\circ\), is \(-46.0 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\), indicating the reaction is exothermic and releases energy to the environment.
The reaction balance shows that one mole of ammonia is formed through the reaction, necessitating significant energy considerations as bonds in both N₂ and H₂ are broken before new N-H bonds are created. Understanding the enthalpy changes and the bonds involved allows chemists to optimize conditions for efficient ammonia production, crucial in industries such as agriculture, where ammonia serves as a vital component in fertilizers.

Bond Energies

Bond energies are average quantities representing the energy needed to break a particular kind of bond. Each bond type, like the N-N triple bond or H-H single bond, has a characteristic bond energy. These values are essential to calculate reaction enthalpies because they help predict and determine whether the overall reaction will release or absorb energy.
When calculating enthalpy change, we use the bond energies as standardized values. They allow chemists to apply a generalized approach across different reactions, making it possible to estimate reaction characteristics like energy efficiency and feasibility.
In the exercise provided, bond energies help us identify how much energy is required to break initial bonds in nitrogen and hydrogen molecules and how much is released when N-H bonds are formed in ammonia. This consideration is crucial in industries leveraging reactions that provide significant energy, such as ammonia production, which is imperative for creating fertilizers.