Chapter 8: Problem 80
From the following data, calculate the average bond enthalpy for the NH bond: $$ \begin{array}{cl} \mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{2}(g)+\mathrm{H}(g) & \Delta H^{\circ}=435 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{NH}_{2}(g) \longrightarrow \mathrm{NH}(g)+\mathrm{H}(g) & \Delta H^{\circ}=381 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{NH}(g) \longrightarrow \mathrm{N}(g)+\mathrm{H}(g) & \Delta H^{\circ}=360 \mathrm{~kJ} / \mathrm{mol} \end{array} $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
NH Bond
In chemical reactions, these NH bonds may need to be broken, which requires energy. Understanding the strength of the NH bond is crucial for predicting how ammonia will behave in different chemical processes. This is why calculating the average bond enthalpy is important.
Enthalpy Change
In our exercise, each step of breaking the NH bonds has an associated enthalpy change:
- The first reaction breaks an NH bond in \( \text{NH}_3 \), contributing 435 kJ/mol.
- The second step involves NH splitting further, requiring 381 kJ/mol.
- The final step breaks another NH bond, needing 360 kJ/mol.
Bond Breaking
In our case, breaking the NH bonds in ammonia is demonstrated in three successive steps. Each step represents the breaking of one NH bond, and the enthalpy changes reflect the energy required for each step. Calculating the total enthalpy change for the entire process helps us determine the strength of these bonds in a measurable way.
Ammonia
This structure results in the presence of three NH bonds, making ammonia a key compound for studying bond enthalpies. Ammonia is not only a common product in fertilizers but also a starting material in many chemical syntheses. Understanding the bond energies in ammonia helps us grasp its stability and reactivity, allowing us to predict how it behaves under various conditions.