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### AuthorTopic: PCl5 dissociation+mixture density)  (Read 1410 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### Rutherford

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##### PCl5 dissociation+mixture density)
« on: December 06, 2012, 04:01:26 AM »

I asked few months ago a similar question, but I wouldn't ask this again if I understood it completely then. I've done these problems by memorizing rather than understanding them.

PCl5 dissociates on PCl3 and Cl2. Under 200°C and p=101.325Pa, the density of the gas mixture towards hydrogen is 70.2. How many percents of PCl5 dissociated on this temperature?
-I was solving it this way:
PCl5 PCl3 + Cl2
x-y        y        y

(1)208.5(x-y)+137.5y+71y=70.2*2
(2)x+y=1
How do I get those equations? Why has x+y to be 1?
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#### curiouscat

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##### Re: PCl5 dissociation+mixture density)
« Reply #1 on: December 06, 2012, 04:15:10 AM »

Why has x+y to be 1?

Because MW is the mass of 1 mole.

PS. Is the answer 47.848.5%? (My eqns. were set a wee bit differently; its easier to start with a basis of 1 gmol PCl5; then you only need one variable x)

Quote
How do I get those equations?

$$\rho=\frac{PM}{RT} \\ \frac{M_{mix}}{M_{H_2}}=70.2 \\ MW_{mix}=\displaystyle\sum_{i} MW_i \cdot x_i \\$$
« Last Edit: December 06, 2012, 04:33:00 AM by curiouscat »
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#### Rutherford

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##### Re: PCl5 dissociation+mixture density)
« Reply #2 on: December 06, 2012, 04:23:04 AM »

Okay, then I want it to be 2 moles.
x+y=2
417(x-y)+417y=70.2*2
When solved y is bigger than x, so more PCl5 dissociated than there was at the beginning. Why is it so?
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#### curiouscat

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##### Re: PCl5 dissociation+mixture density)
« Reply #3 on: December 06, 2012, 04:41:04 AM »

Okay, then I want it to be 2 moles.

Ok, fine. Have it be 2 moles.

Quote
x+y=2
417(x-y)+417y=70.2*2
When solved y is bigger than x, so more PCl5 dissociated than there was at the beginning. Why is it so?

Because you set up a wrong equation? Justify writing your second equation.

GIGO.
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#### Rutherford

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##### Re: PCl5 dissociation+mixture density)
« Reply #4 on: December 06, 2012, 04:53:31 AM »

I want to start from the last equation you wrote in your 1st post. x is the molar share of the components of the mixture. The sum of the molar shares must be equal to the number of moles of the gas mixture (1mol has a mass of 140.4g). Now back to the equations I wrote:
x+y=2
417(x-y)+417y=70.2*2*2
Why is again y bigger than x?
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#### curiouscat

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##### Re: PCl5 dissociation+mixture density)
« Reply #5 on: December 06, 2012, 05:00:29 AM »

417(x-y)+417y=70.2*2*2
Why is again y bigger than x?

Still the wrong eq.

Why 417?
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#### Rutherford

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##### Re: PCl5 dissociation+mixture density)
« Reply #6 on: December 06, 2012, 06:27:43 AM »

Right. It should be 208.5 so it is:
x+y=2
208.5(x-y)+208.5y=70.2*2*2
Just to conclude: when I have density towards a gas of a mixture given, using the ideal gas law I get that the molar mass of the mixture divided by the molar of the referent gas is equal to the density. The molar mass is the mass of 1 mole, so I can assume that there was 1 mole of the mixture. Then, using the formula you gave, I can calculate the unknowns.

Thanks again for the help. It is much clearer now. If I maybe get a doubt about this I will revive this topic, but I hope that it won't happen  .
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