Question

Use the following standard enthalpies of formation to estimate the \(\mathrm{N}-\mathrm{H}\) bond energy in ammonia: \(\mathrm{N}(\mathrm{g}), 472.7 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}(\mathrm{g})\), \(216.0 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{NH}_{3}(g),-46.1 \mathrm{~kJ} / \mathrm{mol}\). Compare your value to the one in Table \(8.4\).

Short answer

In this exercise, we estimate the N-H bond energy in ammonia using the standard enthalpies of formation for N(g), H(g), and NH3(g). Following the steps and applying Hess's Law, we first write the balanced formation reactions for the given substances and then calculate the enthalpy change for the reaction that forms one N-H bond: ΔH = -ΔH_f(NH3) + ΔH_f(N) + 3 * ΔH_f(H) = 46.1 + 472.7 + 3 * 216.0 Next, we calculate the N-H bond energy by dividing the obtained enthalpy change by the number of N-H bonds broken in the reaction: N-H bond energy = ΔH / 3 Finally, we compare our calculated N-H bond energy with the value provided in Table 8.4 to check the accuracy of our estimation.

Step-by-step solution

Write the balanced formation reactions for the given substances

First, we need to know the balanced formation equations of N2, H2, and NH3. Recall that a balanced formation equation is the reaction where one mole of a substance is formed from its constituent elements in their standard states. For N2 (g): \[\frac{1}{2}N_2(g) \rightarrow N (g)\] For H2 (g): \[\frac{1}{2}H_2(g) \rightarrow H (g)\] For NH3 (g): \[N (g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g)\]

Calculate the enthalpy change for the formation of N-H bond

Now, we need to calculate the enthalpy change for the reaction that forms one N-H bond. To do this, we will construct a reaction that breaks one mole of ammonia into its component elements and breaks one nitrogen and three hydrogen molecules in the process. This looks like: \[NH3 (g) \rightarrow N (g) + \frac{3}{2} H_2(g)\] Using Hess's Law, we can calculate the enthalpy change for this reaction by summing the enthalpy changes of the reverse of the ammonia formation reaction and the formation reactions of nitrogen and hydrogen. The enthalpy change for the reverse reaction is the negative of the standard enthalpy of formation of ammonia, i.e., ΔH = 46.1 kJ/mol. Enthalpy change for the reaction: ΔH = -ΔH_f(NH3) + ΔH_f(N) + 3 * ΔH_f(H) = 46.1 + 472.7 + 3 * 216.0

Calculate the N-H bond energy

Now, divide the obtained enthalpy change by the number of N-H bonds broken in the reaction (3) to get the N-H bond energy. N-H bond energy = ΔH / 3

Compare with the value in Table 8.4

Finally, we will compare our calculated N-H bond energy with the value provided in Table 8.4 to check the accuracy of our estimation.

Key concepts

Enthalpy Change

Enthalpy change refers to the heat content change in a chemical reaction at constant pressure.
It is represented by the symbol \( \Delta H \). When a reaction releases heat to the surroundings, it is exothermic and \( \Delta H \) is negative.
Conversely, if the reaction absorbs heat, it is endothermic and \( \Delta H \) is positive.
For the formation of a bond, such as the N-H bond in ammonia, the reaction often releases energy.
This release signifies an exothermic process where the resultant compound has lower energy than its reactants.
The enthalpy change is crucial for understanding how energy is involved in creating or breaking bonds.
By calculating these changes, we can determine factors like bond strength and the stability of molecules.

Hess's Law

Hess's Law is a vital principle in chemistry that states the total enthalpy change in a chemical process is the same, regardless of the steps the process takes.
This law is helpful because it allows the calculation of enthalpy for reactions that are difficult to measure directly.
In our exercise, Hess's Law is used to estimate the N-H bond energy in ammonia.
We begin with a known enthalpy value for the formation of ammonia and use Hess’s Law to calculate the energy required to revert ammonia back to its gaseous elements.
Essentially, this approach means that even if we need to break reactions into multiple steps, the accumulated enthalpy changes will give us a clear picture of the overall energy change.
This technique is invaluable when calculating bond energies in complex molecules.

Enthalpies of Formation

Enthalpy of formation is the change in enthalpy when one mole of a compound forms from its elements in their standard states.
It is symbolized as \( \Delta H_f \) and can be thought of as the energy milestone for creating a specific compound.
For ammonia, the enthalpy of formation helps us understand the energy implications of forming this molecule from nitrogen and hydrogen gas.
The values needed in our exercise are given as: \( \Delta H_f \) for nitrogen (472.7 kJ/mol), for hydrogen (216.0 kJ/mol), and for ammonia (-46.1 kJ/mol).
These values are essential for the calculation of the N-H bond energy using Hess’s Law.
The information allows us to see where and how much energy is released or needed during the formation process.

Ammonia Formation Reaction

Ammonia is a common compound with various applications, and its formation reaction is fundamental in understanding its energy properties.
The standard formation reaction of ammonia involves nitrogen and hydrogen gases forming \( NH_3 \):\[ N (g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g) \]In this process, bonds between nitrogen and hydrogen create ammonia, releasing energy which is indicative of the formation being exothermic.
This reaction is fundamental in reducing free gaseous nitrogen and hydrogen into a stable compound such as ammonia.
By understanding this conversion, chemists can gauge the bond energies within ammonia, aiding in applications from fertilizers to industrial processes.
This reaction’s enthalpy element is considered when calculating the N-H bond energy, proving essential in multiple practical chemistry areas.