The image contains a chemical reaction and thermodynamic calculations involving enthalpy changes ($\Delta H$) for bond dissociation energies. The second part seems to refer to the subsequent reaction below the first calculation.

Second Reaction Analysis

The reaction is: $\text{NH}_\gamma(g) + \text{NO}_\gamma(g) + \text{H}_\gamma O(g) \rightarrow \text{N}_2(g) + \text{H}_2 O(g)$

1. The bond dissociation energies for $N=O$ and $N-H$ are mentioned at the bottom:

   • $N=O = 601 \, \text{kJ/mol}$
   • $N-H = 407 \, \text{kJ/mol}$

2. Enthalpy Calculation ($\Delta H$): The enthalpy change is calculated using bond energies: [ \Delta H = \left[\text{Bonds broken}\right] - \left[\text{Bonds formed}\right] ]

   • Bonds broken:

     • $1 \times N-H = 407 \, \text{kJ/mol}$
     • $1 \times N=O = 601 \, \text{kJ/mol}$

   • Total: $407 + 601 = 1008 \, \text{kJ/mol}$

   • Bonds formed:

     • $1 \times N\equiv N$ (triple bond) = $944 \, \text{kJ/mol}$
     • $1 \times O-H = 463 \, \text{kJ/mol}$

   • Total: $944 + 463 = 1407 \, \text{kJ/mol}$

   $\Delta H = 1008 - 1407 = -399 \, \text{kJ/mol}$

Thus, the enthalpy change for the second reaction is $-399 \, \text{kJ/mol}$.

Do you want further details or clarifications?

Here are five related questions to deepen your understanding:

1. What is the significance of negative $\Delta H$ in chemical reactions?
2. How are bond dissociation energies experimentally determined?
3. Why is the triple bond $N\equiv N$ more stable than single or double bonds?
4. Can you explain how bond energy values are used to predict reaction spontaneity?
5. What are the applications of enthalpy calculations in industrial chemistry?

Tip: Always check the bond types and their corresponding energies carefully to avoid calculation errors.