Question

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: (g) (g) (g) In the second step, ammonia and oxygen react to form nitric oxide and water: (g) (g) (g) (g) Calculate 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 .

Answer 1

Answer:
`\Delta H = -272.25kJ` for 1 mole of NO.

Step-by-step explanation: Hess' Law of Constant Summation or Hess' Law states that the total enthalpy change of a reaction with multiple stages is the sum of the enthalpies of all the changes.

For this question:

1)
`N_(2)_((g)) + 3H_(2)_((g))` =>
`2NH_(3)_((g))`
`\Delta H=-92kJ`

2)
`4NH_(3)_((g))+5O_(2)_((g))` =>
`4NO_((g))+6H_(2)O_((g))`
`\Delta H=-905kJ`

Amonia (
`NH_(3)_((g))`) appeares as product in the first equation and as reagent in the 2 reaction, so when adding both, there is no need to inverse reactions. However, in the 2nd, there are 4 moles of that molecule, so to cancel it, you have to multiply by 2 the first chemical equation and enthalpy:

`2N_(2)_((g)) + 6H_(2)_((g))` =>
`4NH_(3)_((g))`
`\Delta H=-184kJ`

Now, adding them:

`2N_(2)_((g)) + 6H_(2)_((g))` =>
`4NH_(3)_((g))`
`\Delta H=-184kJ`

`4NH_(3)_((g))+5O_(2)_((g))` =>
`4NO_((g))+6H_(2)O_((g))`
`\Delta H=-905kJ`

`2N_(2)_((g))+6H_(2)_((g))+5O_(2)_((g))=>4NO_((g))+6H_(2)O_((g))`
`\Delta H = -185-905`

`2N_(2)_((g))+6H_(2)_((g))+5O_(2)_((g))=>4NO_((g))+6H_(2)O_((g))`
`\Delta H = -1089kJ`

Note net enthalpy is for the formation of 4 moles of nitric oxide.

For 1 mole:

`\Delta H = (-1089)/(4)`

`\Delta H=-272.25kJ`

To form 1 mol of nitric oxide from nitrogen, oxygen and hydrogen, net change in enthalpy is
`\Delta H=-272.25kJ`.