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Topic: Determining the Enthalpy of Formation of Ammonium Chloride?  (Read 30359 times)

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Offline keme

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Determining the Enthalpy of Formation of Ammonium Chloride?
« on: April 14, 2008, 06:56:53 PM »
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
----------------------------------------------------------------------
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!


Offline Borek

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Re: Determining the Enthalpy of Formation of Ammonium Chloride?
« Reply #1 on: April 14, 2008, 07:04:59 PM »
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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Offline keme

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Re: Determining the Enthalpy of Formation of Ammonium Chloride?
« Reply #2 on: April 14, 2008, 08:49:04 PM »
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  ::)

Offline Borek

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Re: Determining the Enthalpy of Formation of Ammonium Chloride?
« Reply #3 on: April 15, 2008, 03:31:18 AM »
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" :)
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