Question

Given the bond energies of \(\mathrm{N} \equiv \mathrm{N}, \mathrm{H}-\mathrm{H}\) and \(\mathrm{N}-\mathrm{H}\) bonds as 945,436 and \(391 \mathrm{~kJ}\) mol \(^{1}\), respectively, the enthalpy of the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) is (1) \(-93 \mathrm{~kJ}\) (2) \(102 \mathrm{~kJ}\) (3) \(90 \mathrm{~kJ}\) (4) \(105 \mathrm{~kJ}\)

Short answer

-93 kJ/mol. The correct answer is (1) -93 kJ.

Step-by-step solution

Write the given bond energies

Given bond energies:- Bond energy of \(\text{N} \equiv \text{N}\) is 945 kJ/mol- Bond energy of \(\text{H}-\text{H}\) is 436 kJ/mol- Bond energy of \(\text{N}-\text{H}\) is 391 kJ/mol

Identify the bonds broken

In the reaction \( \text{N}_{2}(\text{g}) + 3 \text{H}_{2}(\text{g}) \rightarrow 2 \text{NH}_{3}(\text{g}) \), the bonds broken are:- 1 \(\text{N} \equiv \text{N}\) bond in \( \text{N}_{2} \)- 3 \(\text{H}-\text{H}\) bonds in \( \text{H}_{2} \)

Calculate the total energy required to break the bonds

\[ \text{Total energy required} = 1 \times 945 \text{kJ/mol} + 3 \times 436 \text{kJ/mol} \]\[ = 945 + 1308 \]\[ = 2253 \text{kJ/mol} \]

Identify the bonds formed

In the reaction, the bonds formed are:- 6 \(\text{N}-\text{H}\) bonds in 2 \(\text{NH}_{3}\)

Calculate the total energy released upon bond formation

\[ \text{Total energy released} = 6 \times 391 \text{kJ/mol} \]\[ = 2346 \text{kJ/mol} \]

Calculate the enthalpy change of the reaction

\[ \text{Enthalpy change} = \text{Total energy required} - \text{Total energy released} \]\[ = 2253 \text{kJ/mol} - 2346 \text{kJ/mol} \]\[ = -93 \text{kJ/mol} \]

Determine the answer

The enthalpy of the reaction is \(-93 \text{kJ/mol}\). Therefore, the correct answer is (1) \(-93 \text{kJ}\).

Key concepts

Bond Energy

Before diving into the calculations, it's important to understand what bond energy is. Bond energy is the amount of energy required to break one mole of a particular bond in a molecule in the gas phase.
Different bonds have different bond energies. For instance, the bond energy for a triple bond like \(\text{N} \equiv \text{N}\) in nitrogen gas (\text{N}_2) is 945 kJ/mol, which is quite high. Double and single bonds typically have lower bond energies.
Understanding bond energies helps us figure out how much energy is needed to break all the bonds in the reactants and how much energy is released when new bonds form to make the products.

Chemical Reactions

Chemical reactions involve breaking old bonds and forming new ones. In our exercise, for instance, we start with nitrogen gas (\text{N}_2) and three hydrogen gas (\text{H}_2) molecules.
To form ammonia (\text{NH}_3), we need to break the bonds in nitrogen gas and hydrogen gas first. We then form new bonds to create ammonia. The equation for this reaction is:
\(\text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \rightarrow 2 \text{NH}_3(\text{g})\).
This reaction shows the transformation from the reactants (\text{N}_2 and \text{H}_2) into the product (\text{NH}_3). Each step in this process requires specific amounts of energy input and output, which we can calculate using bond energies.

Enthalpy Change

Enthalpy change (\text{∆H}) of a reaction is the difference between the total energy required to break the bonds in the reactants and the total energy released when new bonds form in the products.
In our example, we calculated the total energy required to break the \(\text{N} \equiv \text{N}\) bond and three \(\text{H}-\text{H}\) bonds, which totaled 2,253 kJ/mol.
Next, we calculated the total energy released when forming six \(\text{N}-\text{H}\) bonds in the products (\text{NH}_3), which was 2,346 kJ/mol.
The enthalpy change helps us determine whether the reaction is exothermic (releasing heat) or endothermic (absorbing heat). In our case, the enthalpy change was \(\text{∆H} = -93 \text{kJ}/\text{mol}\), indicating that the reaction releases energy, making it exothermic.