Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: keme on April 14, 2008, 06:56:53 PM
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Hello chemists I am a grade 12 student hoping on going to Mac for chemical engineering and I need help with this lab.
We are given these equations:
ΔHf NH3(g) = -46.36 kJ
ΔHf HCl(g) = -92.65 kJ
Nh3(g) = NH3(aq) ΔH = -30.62 kJ
HCl(g) = HCl(aq) ΔH = -75.01 kJ
And these reagents
1.0 M aqeous ammonia solution
1.0 M hydrochloric acid solution
~ 2 g solid ammonium chloride
I need some help with the prelab questions:
Write the thermochemical equation for the formation of ammonium chloride from its elements in their standard states
Is it: 1/2 N2(g) + (3/2) H2(g) + 1/2 H2(g) + (1/2) Cl2(g) = NH4Cl(g) ΔH = -139.01 kJ
The next question asks why would it be impractical/unsafe to Determining the Enthalpy of Formation of Ammonium Chloride from its elements in their standard states
Is it because the gases formed (HCl and NH3) are toxic/poison?
The final prelab question asks to write a series of thermochemical equations to outline a safe/practical pathway for the Formation of solid Ammonium Chloride
1/2 N2(g) + (3/2) H2(g) = NH3(g) ΔHf = -46.36 kJ
1/2 H2(g) + (1/2) Cl2(g) = ΔHf HCl(g) ΔHf = -92.65 kJ
Nh3(g) = NH3(aq) ΔH = -30.62 kJ
HCl(g) = HCl(aq) ΔH = -75.01 kJ
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NH3(aq) + HCl(aq) = NH4Cl(now is it s or aq?)
Were asked to find solid but im not sure if it still aq or s?
ΔH = -46.36 kJ + -92.65 kJ + -30.62 kJ + -75.01 kJ = -244.64 (aq)
I hope eveything is right 8) Thanks for looking!
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Is it: 1/2 N2(g) + (3/2) H2(g) + 1/2 H2(g) + (1/2) Cl2(g) = NH4Cl(g) ΔH = -139.01 kJ
The next question asks why would it be impractical/unsafe to Determining the Enthalpy of Formation of Ammonium Chloride from its elements in their standard states
How easy it is to react nitrogen with hydrogen to produce ammonia?
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Is it: 1/2 N2(g) + (3/2) H2(g) + 1/2 H2(g) + (1/2) Cl2(g) = NH4Cl(g) ΔH = -139.01 kJ
The next question asks why would it be impractical/unsafe to Determining the Enthalpy of Formation of Ammonium Chloride from its elements in their standard states
How easy it is to react nitrogen with hydrogen to produce ammonia?
According to The Haber Process, not easy.
But we can assume a ΔHf of -46.36 kJ
hoepfully lol ::)
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The next question asks why would it be impractical/unsafe to Determining the Enthalpy of Formation of Ammonium Chloride from its elements in their standard states
How easy it is to react nitrogen with hydrogen to produce ammonia?
According to The Haber Process, not easy.
For me that translates to "impractical" :)