Use bond energies to calculate the enthalpy change in kJ for the reaction: 4NH3 (g) + 3O2 (g) = 2N2 (g) + 6H2O (g)
Added by Chad F.
Step 1
- For the formation of 6H2O (g), the enthalpy change is -286 kJ/mol per mole of water. Therefore, the total enthalpy change for the formation of products is: \[ \Delta H_{\text{products}} = 0 + 6(-286) = -1716 \text{ kJ} \] Show more…
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