Question: Calculate the heat of reaction $\Delta H$ for the following reaction:
$$2NH_3 (g) \rightarrow N_2 (g) + 3H_2 (g)$$
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Round your answer to the nearest kJ/mol.
kJ
āā
mol
1. **State the problem:** Calculate the heat of reaction $\Delta H$ for the reaction:
$$2NH_3 (g) \rightarrow N_2 (g) + 3H_2 (g)$$
2. **Formula used:** The heat of reaction can be estimated using bond energies:
$$\Delta H = \sum \text{Bond energies of bonds broken} - \sum \text{Bond energies of bonds formed}$$
3. **Identify bonds broken and formed:**
- Reactants: 2 moles of $NH_3$ each have 3 N-H bonds, so total bonds broken = $2 \times 3 = 6$ N-H bonds.
- Products:
- $N_2$ has one Nā”N triple bond formed.
- $3H_2$ molecules have $3$ H-H bonds formed.
4. **Bond energies (approximate values in kJ/mol):**
- N-H bond energy = 391 kJ/mol
- Nā”N triple bond energy = 945 kJ/mol
- H-H bond energy = 436 kJ/mol
5. **Calculate total bond energies broken:**
$$6 \times 391 = 2346 \text{ kJ}$$
6. **Calculate total bond energies formed:**
$$945 + 3 \times 436 = 945 + 1308 = 2253 \text{ kJ}$$
7. **Calculate $\Delta H$:**
$$\Delta H = 2346 - 2253 = 93 \text{ kJ}$$
8. **Interpretation:** The positive $\Delta H$ means the reaction is endothermic, absorbing 93 kJ per mole of reaction.
**Final answer:**
$$\boxed{93 \text{ kJ/mol}}$$