Question

using average bond enthalpies (linked above), estimate the enthalpy change for the following reaction:
4nh₃(g) + 5o₂(g) → 4no(g) + 6h₂o(g)
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Explanation:

To solve the enthalpy change using bond enthalpies, we use the formula:
$\Delta H = \sum (\text{Bond enthalpies of reactants broken}) - \sum (\text{Bond enthalpies of products formed})$

Step 1: Identify bonds in reactants

• $\ce{NH3}$: Each $\ce{NH3}$ has 3 $\ce{N-H}$ bonds. For $4\ce{NH3}$, total $\ce{N-H}$ bonds = $4 \times 3 = 12$.
• $\ce{O2}$: Each $\ce{O2}$ has 1 $\ce{O=O}$ bond. For $5\ce{O2}$, total $\ce{O=O}$ bonds = $5 \times 1 = 5$.

Step 2: Identify bonds in products

• $\ce{NO}$: Each $\ce{NO}$ has 1 $\ce{N=O}$ bond. For $4\ce{NO}$, total $\ce{N=O}$ bonds = $4 \times 1 = 4$.
• $\ce{H2O}$: Each $\ce{H2O}$ has 2 $\ce{O-H}$ bonds. For $6\ce{H2O}$, total $\ce{O-H}$ bonds = $6 \times 2 = 12$.

Step 3: Use standard bond enthalpies (values from references):

• $\ce{N-H}$: $391\ \text{kJ/mol}$
• $\ce{O=O}$: $495\ \text{kJ/mol}$
• $\ce{N=O}$: $607\ \text{kJ/mol}$ (for $\ce{NO}$)
• $\ce{O-H}$: $463\ \text{kJ/mol}$

Step 4: Calculate energy to break reactant bonds

$\sum (\text{Reactant bonds}) = (12 \times 391) + (5 \times 495)$
$= 4692 + 2475 = 7167\ \text{kJ}$

Step 5: Calculate energy released to form product bonds

$\sum (\text{Product bonds}) = (4 \times 607) + (12 \times 463)$
$= 2428 + 5556 = 7984\ \text{kJ}$

Step 6: Calculate $\Delta H$

$\Delta H = 7167 - 7984 = -817\ \text{kJ}$

Answer:

$\boxed{-817}$ (Note: The exact value may vary slightly based on bond enthalpy sources, but this is the typical result.)