Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia:N2(g) + 3H2(g) → 2NH3(g) =ΔH−92.kJIn the second step, ammonia and oxygen react to form nitric oxide and water:4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) =ΔH−905.kJCalculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions.Round your answer to the nearest kJ.

To calculate the net change in enthalpy for the formation of one mole of nitric oxide, we need to consider both reactions.

1. For the first reaction, the enthalpy change (ΔH) is -92 kJ for the formation of 2 moles of ammonia. Therefore, the enthalpy change for the formation of 1 mole of ammonia would be -92 kJ / 2 = -46 kJ.

2. For the second reaction, the enthalpy change (ΔH) is -905 kJ for the formation of 4 moles of nitric oxide. Therefore, the enthalpy change for the formation of 1 mole of nitric oxide would be -905 kJ / 4 = -226.25 kJ.

3. To find the total enthalpy change for the formation of 1 mole of nitric oxide from nitrogen, hydrogen, and oxygen, we need to add the enthalpy changes for the two reactions. This is because the first reaction is necessary to produce the ammonia that is used in the second reaction.

So, the total enthalpy change is -46 kJ (from the first reaction) + -226.25 kJ (from the second reaction) = -272.25 kJ.

Therefore, the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen, and oxygen is approximately -272 kJ (rounded to the nearest kJ).