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Answer:
ΔH = - 272.255 kJ
Explanation:
Given that
In first step ,reaction is given as
N2(g) + 3H2(g) →2NH3(g) ΔH=−92 kJ
In second step ,reaction is given as
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905 kJ
Now multiple by in the first equation ,we get
2N2(g) + 6H2(g) →4NH3(g) ΔH=− 184 kJ
Now by adding the above equation
2N2(g) + 6H2(g) →4NH3(g) ΔH=− 184 kJ
<u>4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905 kJ</u>
2 N2+6 H2 + 5 O2 → 4 NO + 6 H2O ΔH= - 1089 k J
For one mole of nitric oxide (NO)
ΔH = - 272.255 kJ
Therefore change in the enthalpy will be - 272.255 kJ.