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Jr. Member

Joined: 29 Dec 2019
Posts: 6

# Reaction

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

Distinguished Member

Joined: 27 Sep 2017
Posts: 332
Location: Berlin, Germany
 What about of a conproportion to nitrogen. Try to balance it by your self.

Jr. Member

Joined: 29 Dec 2019
Posts: 6
 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 commentLast edited by srinimag on Tue Dec 31, 2019 4:36 pm; edited 1 time in total

Distinguished Member

Joined: 27 Sep 2017
Posts: 332
Location: Berlin, Germany
 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

Jr. Member

Joined: 29 Dec 2019
Posts: 6
 Thanks.

Jr. Member

Joined: 29 Dec 2019
Posts: 6
 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

Distinguished Member

Joined: 27 Sep 2017
Posts: 332
Location: Berlin, Germany
 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.

Jr. Member

Joined: 29 Dec 2019
Posts: 6
 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. + 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?

Distinguished Member

Joined: 27 Sep 2017
Posts: 332
Location: Berlin, Germany
 Looks ok

Jr. Member

Joined: 29 Dec 2019
Posts: 6
 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.

Distinguished Member

Joined: 27 Sep 2017
Posts: 332
Location: Berlin, Germany
 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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