Chemistry Forum

Hydrazine sulfate (N₂H₅+HSO₄-) and sodium nitrite (NaNO₂) possible reaction.

What about of a conproportion to nitrogen.
Try to balance it by your self.

I have never used a comproportionate equation and calculated free energy. Can you help me to derive this equation?
hydrazine sulfate (NH₅+HSO₄-) + sodium nitrite (NaNO₂) reaction.

I am hypothesizing the reaction

NaNO₂ + N₂H₆SO₄ → H₂SO₄ + NaN₃ + 2H₂O
H₂SO₄ + NaN₃ + 2H₂O → NaHSO₄ + HN₃
Also
NaN₃ + H₂O = HN₃ + NaOH

Please comment

I told nitrogen will be developed.

First you have N₂H₅ HSO₄ + 2 NaNO₂ => N₂H₄ + 2HNO₂ + Na₂SO₄

Redox reaction

N₂H₄ => N₂ + 4 H+ + 4e- Oxidation

2 HNO₂ + 6 H+ + 6 e- => N₂ + 4 H₂O Reduction

Balance Electrons

3 N₂H₄ => 3 N₂ + 12 H+ + 12 e-

4 HNO₂ + 12 H+ + 12 e- => 2 N₂ + 8 H₂O

Addition

3 N₂H₄ + 4 HNO₂ => 5 N₂ + 8 H₂O

Substitution
3 N₂H₅ HSO₄ + 4 NaNO₂ => 5 N₂ + 8 H₂O + 2 Na₂SO₄ + H₂SO₄

Finished

Thanks.

I am curious the above reaction is spontaneous. These two can react without a catalyst or at an atmospheric temperature/pressure. Can you help me in getting its free energy Delta G. Thanks

It will go spontaneous. Its common way to destroy ammonia, hydoxylamine and hydrazine.
I think some homework you should do by yourself, to look up the enthalpies.

Reactant
N₂H₅+ HSO₄- (S) -894.495 (ΔHËšf (kJ/mol)) 282.742 (S° (J/K·mol))
NaNO₂ (S) -886.900 (ΔHËšf (kJ/mol)) 131.700 (S° (J/K·mol))

Product

N₂ (g) 0 ΔHËšf (kJ/mol) 191.6 S° (J/K·mol)
H₂₀ -285.8 ΔHËšf (kJ/mol) 70.0 S° (J/K·mol)
Na₂SO₄ -1387.1 ΔHËšf (kJ/mol) 149.6 S° (J/K·mol)
H₂SO₄ -814 ΔHËšf (kJ/mol) 156.9 S° (J/K·mol)

3 N₂H₅ HSO₄ + 4 NaNO₂ => 5 N₂ + 8 H₂O + 2 Na₂SO₄ + H₂SO₄

Δ G = Δ H − T Δ S° calculation
Δ H = Δ H product - Δ H reactant
= (5 (0) + 8(-285. + 2(-1387.1) + (-814)) – (3(-894.495) + 4(-886.90+))
= (0 + (-2286.4) + (-2774.2) + (-814)) – ((-2695.485) + (-3467.6))
= -4246.6 + 6163.085
Δ H =1916.485 kJ/mol

Δ S° = Δ S° product - Δ S° reactant
= (5 (191.6) + 8(70) + 2(149.6) + (156.9)) – (3(282.742) + 4(131.7))
= (958+560+299.2+156.9) – (848.226+526.
=1974.1-1375.026
=599.074 J/K·mol
=0.599074 kJ/k.mol

Tk=25oC + 273.15 = 298.15 K

Δ G = 1916.485 kJ/mol - 273.15 K x 0.599074 J/K·mol
=1916.485-163.637 kJ/mol
=1752.848 kJ/mol
My calculation looks not spontaneous. Can you correct me?

Looks ok

All N on the product side is in the 0 oxidation state. Is that going to affect the enthalpy?
Since the reaction is spontaneous, and delta G is >o having difficulties in getting this reaction.

The resulting Enthalpy is calculated by the difference of Product and educt. For what you need it, airbag in car, rocket fuel , or only to destroy hydrazine.