Gen Chem Partial Pressure Question help

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pretty sure its a simple problem, but for some reason im lost on it...
i dont understand Qvault's explanation, so if anyone can help with their own explanation itd be great

6 moles of nitrogen and an unknown amount of hydrogen are held at 40atm. The partial pressure exerted by the nitrogen is 24atm. How many moles of hydrogen are present?



also, has anyone heard of this - · For a gas, the equilibrium expression is written in terms of pressure, not concentration.
.N.2(.g..).+.3..H2(.g..)...↽.−.−.⇀.2..NH3(.g..).. --- Keq =.(..P..NH3.)2 ./ ..(.(.P.H2)3(.P.N2).).
I had selected [Nh3]^2 / [H2]^3[N2] like ive always seen and i think theres an exact same problem with the Haber Process in the Destroyer too, but i never saw the "P" in the Keq before
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Just do Pn2= mole fraction x Ptot
mole fraction is 6/6+x, Pn2, the partial pressure of N2 is 24 atm, and Ptot is 40. Solve for x.
 
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duke2121 said:
Just do Pn2= mole fraction x Ptot
mole fraction is 6/6+x, Pn2, the partial pressure of N2 is 24 atm, and Ptot is 40. Solve for x.
Click to expand...

whats Pn2?....pressure of N2?
so then 24 = (6/6+x)(40) which leads to x = 4 right? nice bc thats the answer...so that was just the partial pressure formula right? partial pressure = mole fraction/mole total)(total pressure)?


idk whats with these mole fraction/partial pressure problems thats giving me trouble when they used to be so easy...heres another one if you or anyone can please help:

The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?
 
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Member 8586 said:
pretty sure its a simple problem, but for some reason im lost on it...
i dont understand Qvault's explanation, so if anyone can help with their own explanation itd be great

6 moles of nitrogen and an unknown amount of hydrogen are held at 40atm. The partial pressure exerted by the nitrogen is 24atm. How many moles of hydrogen are present?



also, has anyone heard of this - · For a gas, the equilibrium expression is written in terms of pressure, not concentration.
.N.2(.g..).+.3..H2(.g..)...↽.−.−.⇀.2..NH3(.g..).. --- Keq =.(..P..NH3.)2 ./ ..(.(.P.H2)3(.P.N2).).
I had selected [Nh3]^2 / [H2]^3[N2] like ive always seen and i think theres an exact same problem with the Haber Process in the Destroyer too, but i never saw the "P" in the Keq before
Click to expand...

Yes equilibrium constant for a gas can also be written in terms of their partial pressure. If you watch chad's videos you will see this.
PN2= partial pressure of N2 and so on.
 
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bump..help with this question please?

The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?

SN1 said:
Yes equilibrium constant for a gas can also be written in terms of their partial pressure. If you watch chad's videos you will see this.
PN2= partial pressure of N2 and so on.
Click to expand...

thanks..simple enough to remember
 
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Member 8586 said:
bump..help with this question please?

The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?



thanks..simple enough to remember
Click to expand...

I was confused when I saw this problem. I've never seen it in destroyer nor Chad.
What I did was just simple cross multiply to get the solubility.

(0.21/1.25) = (1/X)
And if you solve for the X, I think I got this right. But not sure why I got right :(
Anyone?:rolleyes:
 
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dental13 said:
I was confused when I saw this problem. I've never seen it in destroyer nor Chad.
What I did was just simple cross multiply to get the solubility.

(0.21/1.25) = (1/X)
And if you solve for the X, I think I got this right. But not sure why I got right :(
Anyone?:rolleyes:
Click to expand...

That seems too easy LOL. Do you know the correct answer?
 
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you can use the equation pv =nrt
and since v, r, and t are constant
p1/n1 = p2/n2
your answer should come out to 4
 
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i gotta look into the previous problem still, but thanks for the responses... any help with this one..idk why im having trouble with these when partial pressure used to be the easiest calculation lol:

When it is bottled, the partial pressure of [FONT=MathJax_Main]CO.[FONT=MathJax_Main]2. over a 1 L bottle of soda is 4.0 atm. The concentration of dissolved [FONT=MathJax_Main]CO.[FONT=MathJax_Main]2. is 0.12M. The soda is opened at left at 1.0 atm. What will be the new concentration of dissolved [FONT=MathJax_Main]CO.[FONT=MathJax_Main]2. ?
 
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Member 8586 said:
i gotta look into the previous problem still, but thanks for the responses... any help with this one..idk why im having trouble with these when partial pressure used to be the easiest calculation lol:

When it is bottled, the partial pressure of [FONT=MathJax_Main]CO.[FONT=MathJax_Main]2. over a 1 L bottle of soda is 4.0 atm. The concentration of dissolved [FONT=MathJax_Main]CO.[FONT=MathJax_Main]2. is 0.12M. The soda is opened at left at 1.0 atm. What will be the new concentration of dissolved [FONT=MathJax_Main]CO.[FONT=MathJax_Main]2. ?
Click to expand...

That's all they give you? Is the final 1.0 atm the pressure of the the vapor pressure of CO2, or is it the total vapor pressure of the solution? Do they give you a molecular formula for what "soda" is?

If they gave the molecular formula for soda in this problem, it would be easy since you can figure out the mole fraction and go from there. Since they don't give it to you, I don't know how to convert 1L = 1000g = ? moles. Otherwise, I'm also unsure of how to do this problem with the information you gave. All I know is, solubility of CO2 in soda decreases because the pressure above the solution drops.

BUMPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Also, can you please give us an answer to the first problem so we know what's going on?
 
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Felix24 said:
That's all they give you? Is the final 1.0 atm the pressure of the the vapor pressure of CO2, or is it the total vapor pressure of the solution? Do they give you a molecular formula for what "soda" is?

If they gave the molecular formula for soda in this problem, it would be easy since you can figure out the mole fraction and go from there. Since they don't give it to you, I don't know how to convert 1L = 1000g = ? moles. Otherwise, I'm also unsure of how to do this problem with the information you gave. All I know is, solubility of CO2 in soda decreases because the pressure above the solution drops.

BUMPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Also, can you please give us an answer to the first problem so we know what's going on?
Click to expand...

sorry...the very first problem? yeah the answer was 4...i havent got a chance to look at the others yet...im going to look back at them when i review them
 
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Could you post the answer for this asap?

When it is bottled, the partial pressure of CO2 over a 1 L bottle of soda is 4.0 atm. The concentration of dissolved CO2 is 0.12M. The soda is opened at left at 1.0 atm. What will be the new concentration of dissolved CO2 ?

I'm taking my DAT soon and I would like to know what the answer is and how to do it, lol.
 
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Felix24 said:
Could you post the answer for this asap?

When it is bottled, the partial pressure of CO2 over a 1 L bottle of soda is 4.0 atm. The concentration of dissolved CO2 is 0.12M. The soda is opened at left at 1.0 atm. What will be the new concentration of dissolved CO2 ?

I'm taking my DAT soon and I would like to know what the answer is and how to do it, lol.
Click to expand...

i just logged into qvault to check back at these problems but apparently their server has been down since yesterday!! wtf! i have a schedule to keep to...ill let u know once i can get in
 
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premac said:
Is 0.03 M the answer?
Seems like a simple proportion problem.
Click to expand...


wow i think youre rite... (4atm)(.12M)=(1atm)(M2) ? but then wouldnt M2 be 0.48M which doesnt seem to make sense that theres more CO2 in it if cap is left open (even though P down = M up)..? or am i setting it up incorrectly? lol

are maybe the numbers used just poor choices bc they dont make sense in that gas solubility would go UP with INCREASED pressure and vice versa??
 
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Member 8586 said:
wow i think youre rite... (4atm)(.12M)=(1atm)(M2) ? but then wouldnt M2 be 0.48M which doesnt seem to make sense that theres more CO2 in it if cap is left open (even though P down = M up)..? or am i setting it up incorrectly? lol

are maybe the numbers used just poor choices bc they dont make sense in that gas solubility would go UP with INCREASED pressure and vice versa??
Click to expand...

It would be (4atm / 1 atm) = (0.12M / M2).
You are comparing the RATIO of partial pressure with the RATIO of concentration.
The ratio between the two should be equal.
 
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premac said:
It would be (4atm / 1 atm) = (0.12M / M2).
You are comparing the RATIO of partial pressure with the RATIO of concentration.
The ratio between the two should be equal.
Click to expand...

o yup youre right...and just verified by finding out on qvault..im not sure if i ever saw a Molarity to Pressure setup before...where have you seen these problems?

The partial pressure over a dissolved gas is proportional to the concentration of that gas in solution. This is Henry's Law.
[FONT=MathJax_Main]Concentration.[FONT=MathJax_Main]=.[FONT=MathJax_Main](.[FONT=MathJax_Main]0.12.[FONT=MathJax_Main] M.[FONT=MathJax_Main]).[FONT=MathJax_Main](.[FONT=MathJax_Main]1.0.[FONT=MathJax_Main] atm.[FONT=MathJax_Main]).[FONT=MathJax_Main]4.0.[FONT=MathJax_Main] atm.[FONT=MathJax_Main]=.[FONT=MathJax_Main]0.03.[FONT=MathJax_Main] M

.
[FONT=MathJax_Main] anything for this one?.
The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?

I just figured 100/21 = 4.76 and multiplied that by (1.25 x 10^-4) = 6 x 10^-4 but i dont know if i had the correct thinking on this? They have the answer in setup form as (1.25 x 10^-4) / 0.21 ...and both equal approximately 6x10^-4

I guess its similar to the above problem? ratio would be Mole fraction 1 / Mole fraction 2 = Concentration 1 / Concentration 2? so 0.21/1 = (1.25x10^-4)/C2
​
 
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Member 8586 said:
o yup youre right...and just verified by finding out on qvault..im not sure if i ever saw a Molarity to Pressure setup before...where have you seen these problems?

The partial pressure over a dissolved gas is proportional to the concentration of that gas in solution. This is Henry's Law.
[FONT=MathJax_Main]Concentration.[FONT=MathJax_Main]=.[FONT=MathJax_Main](.[FONT=MathJax_Main]0.12.[FONT=MathJax_Main] M.[FONT=MathJax_Main]).[FONT=MathJax_Main](.[FONT=MathJax_Main]1.0.[FONT=MathJax_Main] atm.[FONT=MathJax_Main]).[FONT=MathJax_Main]4.0.[FONT=MathJax_Main] atm.[FONT=MathJax_Main]=.[FONT=MathJax_Main]0.03.[FONT=MathJax_Main] M

.
[FONT=MathJax_Main] anything for this one?.
The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?

I just figured 100/21 = 4.76 and multiplied that by (1.25 x 10^-4) = 6 x 10^-4 but i dont know if i had the correct thinking on this? They have the answer in setup form as (1.25 x 10^-4) / 0.21 ...and both equal approximately 6x10^-4

I guess its similar to the above problem? ratio would be Mole fraction 1 / Mole fraction 2 = Concentration 1 / Concentration 2? so 0.21/1 = (1.25x10^-4)/C2
​
Click to expand...

Yeah, it should be the same concept.
I've seen the setup in Chad's notes somewhere. Either that, or DAT destroyer. Can't recall the exact source...
 
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Member 8586 said:
o yup youre right...and just verified by finding out on qvault..im not sure if i ever saw a Molarity to Pressure setup before...where have you seen these problems?

The partial pressure over a dissolved gas is proportional to the concentration of that gas in solution. This is Henry's Law.
[FONT=MathJax_Main]Concentration.[FONT=MathJax_Main]=.[FONT=MathJax_Main](.[FONT=MathJax_Main]0.12.[FONT=MathJax_Main] M.[FONT=MathJax_Main]).[FONT=MathJax_Main](.[FONT=MathJax_Main]1.0.[FONT=MathJax_Main] atm.[FONT=MathJax_Main]).[FONT=MathJax_Main]4.0.[FONT=MathJax_Main] atm.[FONT=MathJax_Main]=.[FONT=MathJax_Main]0.03.[FONT=MathJax_Main] M

.
[FONT=MathJax_Main] anything for this one?.
The solubility of dissolved oxygen in water is 1.25 x 10-4 M at 25°C, where the mole fraction of oxygen is 0.21 and atmospheric pressure is 1.0 atm. In a pure oxygen atmosphere at the same pressure, what would the solubility be?

I just figured 100/21 = 4.76 and multiplied that by (1.25 x 10^-4) = 6 x 10^-4 but i dont know if i had the correct thinking on this? They have the answer in setup form as (1.25 x 10^-4) / 0.21 ...and both equal approximately 6x10^-4

I guess its similar to the above problem? ratio would be Mole fraction 1 / Mole fraction 2 = Concentration 1 / Concentration 2? so 0.21/1 = (1.25x10^-4)/C2
​
Click to expand...


I asked chad about that question and he said it is very unlikely that we will get one of those on the dat.
 
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Member 8586 said:
wow i think youre rite... (4atm)(.12M)=(1atm)(M2) ? but then wouldnt M2 be 0.48M which doesnt seem to make sense that theres more CO2 in it if cap is left open (even though P down = M up)..? or am i setting it up incorrectly? lol

are maybe the numbers used just poor choices bc they dont make sense in that gas solubility would go UP with INCREASED pressure and vice versa??
Click to expand...


When I first read the question I also thought of the problem as a simple proportion problem and set up my proportion exactly like you did but I realize that

Pv=nRT

and since p and v is on the same side then you must keep both volume and pressure on the same side.

in contrast to M1V1=M2V2

molarity(moles/L) is on the opposite side of volume in Pv=nrt so that is why it is written like that

I think I am right? lol
 
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