Partial pressure question- this makes no sense!

[virtualmaster999]

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Hey everyone! Could someone explain why, apparently, the answer is b? I really think it's D, 1.8. If it starts at 1.2 and goes down one fourth, then instead of producing 2.4, wouldn't it be 1.8??




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[DAT Destroyer]

Hey everyone! Could someone explain why, apparently, the answer is b? I really think it's D, 1.8. If it starts at 1.2 and goes down one fourth, then instead of producing 2.4, wouldn't it be 1.8??




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Set up your table.....Start... Change... Equilibrium........at equilibrium N2 is 1.2 -x. We are told that N2 left is 0.9...thus x equals 0.3 OK,,,,,,now.... NH3 at equilibrium is 2x......or 2 (.3) or 0.6.

Hope this helps......

Dr. Jim Romano

 

[goutamk]

Just think of it in terms of moles. You know that the change in moles of N2 was 1.2-0.9 = 0.3 moles lost.
The stochiometric ratio of N2:NH3 is 1:2 so for every mole of N2 lost, you get 2 moles of NH3 made. 0.3x2 = 0.6

 

[virtualmaster999]

Just think of it in terms of moles. You know that the change in moles of N2 was 1.2-0.9 = 0.3 moles lost.
The stochiometric ratio of N2:NH3 is 1:2 so for every mole of N2 lost, you get 2 moles of NH3 made. 0.3x2 = 0.6

But why are you doing the change in miles rather than the current amount? Is it because it's an equilibrium problem?


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[FeralisExtremum]

But why are you doing the change in miles rather than the current amount? Is it because it's an equilibrium problem?


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Since these are all gases, you can keep it in atmospheres if you want.

 

[100%permwinner]

This is more of a stoichiometry exercise than equilibrium. Partial pressure of each gas is proportional to the moles according to Dalton's rule. When the pres of N2 is 0.9 atm at the end of the reaction, there must be 0.3 atm of N2 consumed during the process. At the same time, there must be 0.9 atm of H2 consumed because 3 moles of H2 reacts with 1 mole of N2. On the product side, there must be 0.6 atm of NH3 produced during the reaction because the mole ratio of N2 to NH3 is 1:2. B is the correct answer.

Hey everyone! Could someone explain why, apparently, the answer is b? I really think it's D, 1.8. If it starts at 1.2 and goes down one fourth, then instead of producing 2.4, wouldn't it be 1.8??




Sent from my iPhone using Tapatalk[/QUOTE]

 

[virtualmaster999]

This is more of a stoichiometry exercise than equilibrium. Partial pressure of each gas is proportional to the moles according to Dalton's rule. When the pres of N2 is 0.9 atm at the end of the reaction, there must be 0.3 atm of N2 consumed during the process. At the same time, there must be 0.9 atm of H2 consumed because 3 moles of H2 reacts with 1 mole of N2. On the product side, there must be 0.6 atm of NH3 produced during the reaction because the mole ratio of N2 to NH3 is 1:2. B is the correct answer.

Hey everyone! Could someone explain why, apparently, the answer is b? I really think it's D, 1.8. If it starts at 1.2 and goes down one fourth, then instead of producing 2.4, wouldn't it be 1.8??




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[/QUOTE]
Ohhh ok so you're worrying about the pressure that is actually used, not what it drains down to right? Because then it would be like a regular stoichiometry (ex: "how many moles of y are produced if 2 miles of x is used)? Is that right?


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[100%permwinner]

Ohhh ok so you're worrying about the pressure that is actually used, not what it drains down to right? Because then it would be like a regular stoichiometry (ex: "how many moles of y are produced if 2 miles of x is used)? Is that right?


Sent from my iPhone using Tapatalk[/QUOTE]

In a way yes. You have zero atm of NH3 in the system at the start of the reaction. When 0.3 atm of N2 disappears during the reaction, the pressure of NH3 increases by 0.6 atm at the same time.