Define the term mean bond enthalpy.
Hydroxylamine (NH2OH\text{NH}_2\text{OH}NH2OH) decomposes in the gas phase according to the equation:
2NH2OH(g)→N2(g)+2H2O(g)+H2(g) 2\text{NH}_2\text{OH}(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(g) + \text{H}_2(g) 2NH2OH(g)→N2(g)+2H2O(g)+H2(g)The enthalpy change for this reaction is ΔH=−336 kJ mol−1\Delta H = -336\text{ kJ mol}^{-1}ΔH=−336 kJ mol−1. Each molecule of hydroxylamine contains two N−H\text{N}-\text{H}N−H bonds, one N−O\text{N}-\text{O}N−O bond, and one O−H\text{O}-\text{H}O−H bond.
Table 1 shows some mean bond enthalpy values:
| Bond | Mean bond enthalpy / kJ mol−1\text{kJ mol}^{-1}kJ mol−1 |
|---|---|
| N−H\text{N}-\text{H}N−H | 388 |
| O−H\text{O}-\text{H}O−H | 464 |
| H−H\text{H}-\text{H}H−H | 436 |
| N≡N\text{N}\equiv\text{N}N≡N | 944 |
Use the data in Table 1 to calculate a value for the N−O\text{N}-\text{O}N−O bond enthalpy in hydroxylamine.
