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### Topic: Partial Pressure problem  (Read 8402 times)

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#### zmasterflex

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##### Partial Pressure problem
« on: March 27, 2014, 05:59:45 PM »
At a certain temperature the equilibrium constant Kp = .132 for the reaction:
PCl5(g) <--> PCl3(g) + Cl2(g)
At equilibrium, the partial pressures of both PCl5 and PCl3 are 100 mmHg. What is the total pressure of the equilibrium system in mmHg?

Here is what I got; [PCl3][Cl2]/[PCl5] = .132 and the partial pressures should be equal to their molar fractions. So [100][Cl2]/[100] = .132. this is only possible if Cl2 has a pressure of .132 mmHg which is not an answer. Something tells me I'm setting this up wrong. Thanks for any advice...

#### Borek

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##### Re: Partial Pressure problem
« Reply #1 on: March 27, 2014, 06:10:05 PM »
At first sight I see nothing wrong with your calculations, but you have not answered the question that is asked.
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#### sjb

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##### Re: Partial Pressure problem
« Reply #2 on: March 27, 2014, 06:16:13 PM »
At a certain temperature the equilibrium constant Kp = .132 for the reaction:
PCl5(g) <--> PCl3(g) + Cl2(g)
At equilibrium, the partial pressures of both PCl5 and PCl3 are 100 mmHg. What is the total pressure of the equilibrium system in mmHg?

Here is what I got; [PCl3][Cl2]/[PCl5] = .132 and the partial pressures should be equal to their molar fractions. So [100][Cl2]/[100] = .132. this is only possible if Cl2 has a pressure of .132 mmHg which is not an answer. Something tells me I'm setting this up wrong. Thanks for any advice...

Total pressure, not partial pressure of dichlorine?

#### zmasterflex

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##### Re: Partial Pressure problem
« Reply #3 on: March 27, 2014, 06:25:02 PM »
The answers given are 100,200,300,400 and 332 mmHg. The one that might seem correct based on my calculation is 100 + 100 + .132 = close to 200mmHg. The correct answer given is 300mmHg. Is 200 the correct answer? (thanks)

#### Borek

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##### Re: Partial Pressure problem
« Reply #4 on: March 27, 2014, 06:48:55 PM »
Are units of Kp given?

If not, it is most likely using standard pressure, so you have to convert mmHg to atm (or bar).
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#### zmasterflex

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##### Re: Partial Pressure problem
« Reply #5 on: March 27, 2014, 06:51:57 PM »
Shouldn't Kp be unit-less? It's just a ratio of products over reactants?

#### Shipwreck

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##### Re: Partial Pressure problem
« Reply #6 on: March 28, 2014, 10:07:00 AM »
you have to convert mmHg to atm (or bar).

This.

(Pp/R*T) = n/V, so you need to convert mmHg to atmospheres so that the ideal gas constant will be valid:

100mmHg / (760mmHg/1 atm) = 0.132 atm

Kp = .132 = (.132)(PCl2) / .132
PCl2 = 0.132

Total pressure = sum of all partial pressures:

.132 + .132 + .132 = 0.396 atm

.396atm * 760mmHg/1 atm = 301mmHg

« Last Edit: March 28, 2014, 10:45:31 AM by Shipwreck »

#### Borek

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##### Re: Partial Pressure problem
« Reply #7 on: March 28, 2014, 12:44:14 PM »
you need to convert mmHg to atmospheres so that the ideal gas constant will be valid:

You can express ideal gas constant in whatever units you want, including mmHg. Check wikipedia page for the ideal gas constant for a huge list of different values expressed using different units. They are all perfectly valid.

Kp can be expressed whatever pressure units you want as well - you just have to be clear about units used, or you have to be clear about the fact Kp is unitles. In the latter case the value depends on the reference (standard state) pressure assumed to be 1. This can be either 1 atm or 1 bar, as the standard state definition has changed, but some books still list values using atm (1 bar being the standard state as of today).
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#### Shipwreck

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##### Re: Partial Pressure problem
« Reply #8 on: March 28, 2014, 01:27:39 PM »
you need to convert mmHg to atmospheres so that the ideal gas constant will be valid:

You can express ideal gas constant in whatever units you want, including mmHg.

Yeah, you're right. I should have mentioned that this conversion is done assuming Kp was calculated using atm (standard state) pressure units. My bad.

#### Borek

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##### Re: Partial Pressure problem
« Reply #9 on: March 28, 2014, 03:29:04 PM »
atm (standard state) pressure units

http://goldbook.iupac.org/S05921.html

1 bar (105 Pa) is the standard state since 1982.
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##### Re: Partial Pressure problem
« Reply #10 on: March 29, 2014, 07:54:41 PM »
atm (standard state) pressure units

http://goldbook.iupac.org/S05921.html

1 bar (105 Pa) is the standard state since 1982.

Well I think the answer is pretty similar anyway (as you would expect, since 1 atm is so close to 1 bar). I got almost exactly 299 mmHg.