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

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.

1. From the first equation:
   4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
   ΔH = -909 kJ

2. 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.