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srinimag
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Hydrazine sulfate (N2H5+HSO4-) and sodium nitrite (NaNO2) possible reaction.
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ChenBeier
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What about of a conproportion to nitrogen.
Try to balance it by your self.
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srinimag
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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


Last edited by srinimag on Tue Dec 31, 2019 4:36 pm; edited 1 time in total
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ChenBeier
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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
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srinimag
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Thanks.
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srinimag
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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
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ChenBeier
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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.
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srinimag
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Reactant
N2H5+ HSO4- (S) -894.495 (ΔH˚f (kJ/mol)) 282.742 (S° (J/K·mol))
NaNO2 (S) -886.900 (ΔH˚f (kJ/mol)) 131.700 (S° (J/K·mol))

Product

N2 (g) 0 ΔH˚f (kJ/mol) 191.6 S° (J/K·mol)
H20 -285.8 ΔH˚f (kJ/mol) 70.0 S° (J/K·mol)
Na2SO4 -1387.1 ΔH˚f (kJ/mol) 149.6 S° (J/K·mol)
H2SO4 -814 ΔH˚f (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.Cool + 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.Cool
=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?
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ChenBeier
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Looks ok
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srinimag
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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.
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ChenBeier
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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.
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