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Step 1: Identify the Relevant Data
• Standard enthalpy of formation of NH3: –46.0 kJ mol–1 (per mole of NH3)
• Enthalpy of formation of H2 from its atoms: –436 kJ mol–1
• Enthalpy of formation of N2 from its atoms: –712 kJ mol–1
• We need to find the average bond enthalpy of the N–H bond in NH3.
Step 2: Write the Balanced Reaction for Formation of Ammonia
Ammonia is formed by the reaction:
$ N_2 + 3 H_2 \rightarrow 2 NH_3 $
According to the data given, the standard enthalpy change for forming 1 mole of NH3 is –46.0 kJ mol–1. Therefore, for forming 2 moles of NH3, the enthalpy change is:
$ \Delta H = 2 \times ( -46.0 )\,\text{kJ mol}^{-1} = -92 \,\text{kJ mol}^{-1} $
Step 3: Express the Bond Enthalpy Balance
When calculating bond enthalpies, bond breaking is considered endothermic (positive enthalpy), and bond formation is considered exothermic (negative enthalpy):
Bond breaking in reactants: N≡N and H–H bonds
Bond formation in products: N–H bonds
However, we often use tabulated values (formation enthalpies) and bond energies carefully. Let $ x $ be the average bond enthalpy of the N–H bond.
From the given data (changed to appropriate signs for bond breaking):
Enthalpy required to break 1 mole of N2 into atoms: $ +712 \,\text{kJ mol}^{-1} $
Enthalpy required to break 1 mole of H2 into atoms: $ +436 \,\text{kJ mol}^{-1} $
For the formation of 2 moles of NH3, 6 N–H bonds are formed. So the total enthalpy for forming 6 N–H bonds is $ 6 \times ( -x ) = -6x $.
Step 4: Set Up the Enthalpy Equation
Using the standard enthalpy change for the overall reaction ($ \Delta H = -92 \,\text{kJ mol}^{-1} $), we can write:
$ \Delta H = \bigl[ \text{(energy to break 1\,N}_2) + 3 \times \text{(energy to break 1\,H}_2) \bigr] \;-\; \bigl[ \text{energy released in forming 6\,N–H bonds} \bigr] $
In symbols:
$ -92 = ( +712 ) + 3 \times ( +436 ) - ( 6x ) $
Step 5: Solve for the N–H Bond Enthalpy
Combine the numeric terms on the right side:
$ -92 = 712 + (3 \times 436) - 6x $
$ -92 = 712 + 1308 - 6x $
$ -92 = 2020 - 6x $
Add 92 to both sides:
$ 0 = 2020 + 92 - 6x $
$ 6x = 2020 + 92 $
$ 6x = 2112 $
Hence:
$ x = \frac{2112}{6} = 352 \,\text{kJ mol}^{-1} $
Step 6: State the Final Answer
The average bond enthalpy of the N–H bond in NH3 is +352 kJ mol–1.
