To calculate the heat of reaction [tex]\(\Delta H\)[/tex] for the reaction
[tex]\[
2 \text{NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g),
\][/tex]
we can use the bond energies given in the question. The given values for bond energies (in kJ/mol) are:
- N-H bond energy = 391 kJ/mol
- N≡N triple bond energy = 941 kJ/mol
- H-H bond energy = 432 kJ/mol
### Step-by-Step Solution:
1. Determine the bonds broken in the reactants:
In the reactants, we have [tex]\(2 \text{NH}_3\)[/tex] molecules.
Each NH[tex]\(_3\)[/tex] molecule has 3 N-H bonds.
Therefore, the total number of N-H bonds broken is:
[tex]\[
2 \text{ NH}_3 \times 3 \text{ N-H bonds/NH}_3 = 6 \text{ N-H bonds}
\][/tex]
Energy required to break these bonds:
[tex]\[
6 \text{ N-H bonds} \times 391 \text{ kJ/mol} = 2346 \text{ kJ}
\][/tex]
2. Determine the bonds formed in the products:
In the products, we have [tex]\(\text{N}_2\)[/tex] and [tex]\(3 \text{H}_2\)[/tex].
Each [tex]\(\text{N}_2\)[/tex] molecule involves a N≡N triple bond.
Each [tex]\(\text{H}_2\)[/tex] molecule involves a H-H bond.
Therefore, the total number of bonds formed is:
[tex]\[
1 \text{ N≡N bond} + 3 \text{ H-H bonds}
\][/tex]
Energy released from these bonds:
[tex]\[
(1 \text{ N≡N bond} \times 941 \text{ kJ/mol}) + (3 \text{ H-H bonds} \times 432 \text{ kJ/mol})
\][/tex]
Simplifying:
[tex]\[
941 \text{ kJ} + 1296 \text{ kJ} = 2237 \text{ kJ}
\][/tex]
3. Calculate the enthalpy change [tex]\(\Delta H\)[/tex]:
[tex]\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\][/tex]
Substituting the obtained values:
[tex]\[
\Delta H = 2346 \text{ kJ} - 2237 \text{ kJ} = 109 \text{ kJ}
\][/tex]
Thus, the heat of reaction [tex]\(\Delta H\)[/tex] for the reaction [tex]\(2 \text{NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g)\)[/tex] is:
[tex]\[
\Delta H = 109 \text{ kJ/mol}
\][/tex]
