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Nitrogen dioxide is produced from ammonia and air as shown in these equations 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) ΔH = -909 kJ mol⁻¹ 2NO(g) + O₂(g) → 2NO₂(g) ΔH = -115 kJ mol⁻¹ What is the enthalpy change (in kJ mol⁻¹) for the following reaction? 4NH₃(g) + 7O₂(g) → 4NO(g) + 6H₂O(g) - AQA - A-Level Chemistry - Question 8 - 2019 - Paper 3
Question 8
Nitrogen dioxide is produced from ammonia and air as shown in these equations 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) ΔH = -909 kJ mol⁻¹ 2NO(g) + O₂(g) → 2NO₂(g) ... show full transcript
Worked Solution & Example Answer:Nitrogen dioxide is produced from ammonia and air as shown in these equations 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) ΔH = -909 kJ mol⁻¹ 2NO(g) + O₂(g) → 2NO₂(g) ΔH = -115 kJ mol⁻¹ What is the enthalpy change (in kJ mol⁻¹) for the following reaction? 4NH₃(g) + 7O₂(g) → 4NO(g) + 6H₂O(g) - AQA - A-Level Chemistry - Question 8 - 2019 - Paper 3
Step 1
Calculate the enthalpy change for 4NH₃(g) + 7O₂(g) → 4NO(g) + 6H₂O(g)
Answer
To find the enthalpy change for the reaction 4NH₃(g) + 7O₂(g) → 4NO(g) + 6H₂O(g), we can break it down using Hess's Law.
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From the first equation:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
ΔH = -909 kJ -
The second equation gives us the enthalpy for the formation of NO₂:
2NO(g) + O₂(g) → 2NO₂(g)
ΔH = -115 kJ
Thus, to get our desired equation we need to:- Use the first equation as is.
- Use the second equation but in the reverse direction and multiplied by 2 (to get 4NO). This means we need to add the enthalpy.
- Reverse the second reaction:
4NO(g) → 2NO(g) + O₂(g)
This means we take the positive value: ΔH = +230 kJ
Now we combine:
Final ΔH = -909 kJ + 230 kJ
= -679 kJ
The overall enthalpy change for the reaction 4NH₃(g) + 7O₂(g) → 4NO(g) + 6H₂O(g) is -679 kJ.