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Topic: Haber Process Explained in terms of Gas Laws  (Read 1142 times)

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Offline cstewart2231121988

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Haber Process Explained in terms of Gas Laws
« on: May 06, 2020, 03:16:47 PM »
So I missed a Gas Laws lab a while ago and I can't make it up due to quarantine. My teacher says I can instead explain the Haber Process in terms of Gas Laws. I understand the process but can't really explain it in terms of Gas Laws. So far I tried explaining it with Charle's law but it doesn't seem right. Any help is appreciated.

Offline NeeleshGupta23

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Re: Haber Process Explained in terms of Gas Laws
« Reply #1 on: May 06, 2020, 10:23:08 PM »
It would be helpful to explain it in terms of moles of coefficients. You can do this through using vapor pressure and number of moles. When the gasses combine, liquid ammonia is formed. Discussing the vapor pressure decreasing as more ammonia forms will likely be the answer your teacher is wanting. There is more pressure and more heat so the energy required to turn the gasses into a liquid is lowered thus creating liquid ammonia and incorporating the ideal gas law would help this proof. This is also a good example of lower IMFs creating a higher vapor pressure of the products. Putting the question in terms of the moles of coefficients would be a good way to utilize Avogadro's law. Since you are lessening the system by 2 moles, a lot more pressure and temperature is required.

P1V1n2T2=P2V2n1T1

I hope this helped. Have a great day. :]


Offline Borek

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Re: Haber Process Explained in terms of Gas Laws
« Reply #2 on: May 07, 2020, 03:38:10 AM »
When the gasses combine, liquid ammonia is formed.

No, ammonia is a gas as well, so large part of what you wrote later doesn't make much sense.

BP of ammonia is -33°C, Haber process takes place in temperatures around 500°C.
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