To determine the change in enthalpy (ΔH) for the reaction [tex]\( \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \)[/tex], we will use the given bond energies:
- Energy for breaking one mole of [tex]\( \text{N} \equiv \text{N} \)[/tex] is 942 kJ.
- Energy for breaking one mole of [tex]\( \text{H-H} \)[/tex] is 432 kJ.
- Energy for forming one mole of [tex]\( \text{N-H} \)[/tex] is 386 kJ.
### Step-by-Step Solution:
1. Energy Required to Break Bonds:
- Breaking 1 mole of [tex]\( \text{N} \equiv \text{N} \)[/tex] bonds:
[tex]\[ 1 \times 942 \text{ kJ} = 942 \text{ kJ} \][/tex]
- Breaking 3 moles of [tex]\( \text{H-H} \)[/tex] bonds:
[tex]\[ 3 \times 432 \text{ kJ} = 1296 \text{ kJ} \][/tex]
- Total energy for breaking these bonds:
[tex]\[ 942 \text{ kJ} + 1296 \text{ kJ} = 2238 \text{ kJ} \][/tex]
2. Energy Released in Forming Bonds:
- Forming 2 moles of NH[tex]\(_3\)[/tex] involves forming 6 N-H bonds (since each NH[tex]\(_3\)[/tex] molecule has 3 N-H bonds):
[tex]\[ 6 \times 386 \text{ kJ} = 2316 \text{ kJ} \][/tex]
3. Calculate ΔH (Change in Enthalpy):
- The total energy required to break the bonds is 2238 kJ.
- The total energy released in forming the bonds is 2316 kJ.
- The change in enthalpy for the reaction (ΔH) is the difference between the energy required to break the bonds and the energy released in forming the bonds:
[tex]\[ \Delta H = 2238 \text{ kJ} - 2316 \text{ kJ} = -78 \text{ kJ} \][/tex]
### Final Answer:
The enthalpy change for the reaction, expressed to two significant figures, is [tex]\(-78\)[/tex] kilojoules.
Thus, the enthalpy change for the reaction is [tex]\(-78\)[/tex] kilojoules.
