To find the enthalpy change (ΔH) for the given reaction:
1. Start with the given reactions and their enthalpy changes:
- Two moles of ammonia gas (NH₃) plus three moles of nitrous oxide gas (N₂O) produce four moles of nitrogen gas (N₂) and three moles of water (H₂O) in liquid form, with ΔH = -1010 kJ.
- One mole of nitrous oxide gas (N₂O) plus three moles of hydrogen gas (H₂) yield one mole of hydrazine (N₂H₄) and one mole of water (H₂O) in liquid form, with ΔH = -317 kJ.
- One mole of hydrazine (N₂H₄) plus one mole of oxygen gas (O₂) form one mole of nitrogen gas (N₂) and two moles of water (H₂O) in liquid form, with ΔH = -623 kJ.
- Three moles of hydrogen gas (H₂) plus three moles of oxygen gas (O₂) produce three moles of water (H₂O) in liquid form, with ΔH = -286 kJ.
2. Rearrange and combine these reactions to match the target reaction:
- Multiply the third reaction by 2 to get two moles of hydrazine (N₂H₄) plus two moles of oxygen gas (O₂) forming two moles of nitrogen gas (N₂) and four moles of water (H₂O) in liquid form, with ΔH = -1246 kJ.
- Add the first, second, and modified third reactions to get:
Two moles of ammonia gas (NH₃) plus three moles of nitrous oxide gas (N₂O) plus one mole of nitrous oxide gas (N₂O) plus three moles of hydrogen gas (H₂) plus two moles of hydrazine (N₂H₄) plus two moles of oxygen gas (O₂) forming four moles of nitrogen gas (N₂) plus three moles of water (H₂O) plus one mole of water (H₂O) plus two moles of nitrogen gas (N₂) plus four moles of water (H₂O).
3. Simplify and cancel out common species on both sides of the reaction to get:
Two moles of ammonia gas (NH₃) plus four moles of nitrous oxide gas (N₂O) plus three moles of hydrogen gas (H₂) yield six moles of nitrogen gas (N₂) and six moles of water (H₂O) in liquid form.
4. Calculate the enthalpy change for the target reaction as the sum of the enthalpy changes for the individual reactions involved:
ΔH(target) = -1010 kJ + (-317 kJ) + (-1246 kJ)
ΔH(target) = -2573 kJ
Therefore, the enthalpy change for the reaction 2NH₃(g) + 9O₂(g) → N₂H₄(l) + H₂O(l) is -2573 kJ.
Answer:
The AH for the reaction 2NH3(g) + O2(g) → N2H4(l) + H2O(l) is -999 kJ.
Explanation:
To find the AH for the reaction 2NH3(g) + O2(g) → N2H4(l) + H2O(l), we can use the given data for the individual reactions:
1. 2NH3(g) → N2(g) + 3H2(g) ΔH = -1010 kJ
2. N2(g) + 3H2(g) → N2H4(l) + H2O(l) ΔH = -317 kJ
3. N2H4(l) + O2(g) → N2(g) + 2H2O(l) ΔH = -623 kJ
Now, we need to reverse the second reaction and multiply it by 2 in order to match the stoichiometry of the desired reaction. This will give us:
-2(N2(g) + 3H2(g) → N2H4(l) + H2O(l)) ΔH = 2(317) kJ
Adding the three reactions together, we get:
2NH3(g) + O2(g) → N2H4(l) + H2O(l) ΔH = -1010 + 2(317) - 623 kJ
ΔH = -1010 + 634 - 623 kJ
ΔH = -999 kJ
Therefore, the AH for the reaction 2NH3(g) + O2(g) → N2H4(l) + H2O(l) is -999 kJ.
