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### Topic: Gas equilibrium/Partial pressures  (Read 10288 times)

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#### chemicalsister

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##### Gas equilibrium/Partial pressures
« on: April 01, 2004, 04:06:14 PM »
Hey, Newbie here.  A computer guru friend of mine suggested this site to me and said to check it out.  I'm a college student and am having a heck of a time with this homework packet.  If anyone can help I would greatly appreciate it.  Well, here's my problem:
C2H6O2

CH4 + CO2 = CO + 2H2
a) what is the percent yield of H2 when an equimiolar mixture of CH4 and CO2 at 20 atm reaches equilibrium at 1200K, at which Kp=3.548x106?
b) What is the percent yield at 1300K at which Kp =2.626x107?

First off, I'm sorry about how this is all typed and hope that you can make sense of some of it.  If anyone would care to share how to properly type this information in I could really use it as you can tell.

Im a little familiar with percent yield but I haven't worked with it in awhile.  I'm just having dificulities figuring out how to use the necessary information.  Thanks again.
« Last Edit: April 24, 2004, 07:08:36 PM by hmx9123 »

#### gregpawin

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##### Re:Percent yield in equilibrium reactions
« Reply #1 on: April 01, 2004, 04:47:03 PM »
First of all, when you're going to use those sub or super position commands, you have to select the text and then press the appropriate button.  If you press the button before or after you've typed your stuff, it won't do anything useful.  The sup and sub tags have to be before and after the text you're targeting.

Percent yield is simply the experimental yield divided by the theoretical yield and multiplied by 100%.  So, you need two kinds of information, some kind of information about the reactants and an "actual" measurement of the final product to compare the two.
I've got nothin'

#### Mitch

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##### Re:Percent yield in equilibrium reactions
« Reply #2 on: April 01, 2004, 04:56:50 PM »
That seems very strange. You can't calculate a percent yield without having an experimental number too. Do you think they simply meant calculate the yield?

By the way we enjoy knowing how people got here, how did your friend find out about us?
« Last Edit: April 01, 2004, 05:35:21 PM by Mitch »
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
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#### chemicalsister

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##### Re:Percent yield in equilibrium reactions
« Reply #3 on: April 01, 2004, 06:53:27 PM »
Thanks for the help.  I'm glad to know that I'm not crazy for thinking that I was lacking necessary information.

As for finding this place, I was complaining about my difficulties with my homework to my friend. So he simply used the webferret search engine and typed in something to do with help on chemistry homework and this site came up as one. I must say that it is far greater than the other sites that showed up! I'm sure I'll be pestering you guys a lot.

#### Seymor-Omnis

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##### Re:Percent yield in equilibrium reactions
« Reply #4 on: April 01, 2004, 07:13:39 PM »
All I can say is the same thing happened to me, except I found it by accident  .  I can also say I actually found a new "hobby."

#### AWK

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##### Re:Percent yield in equilibrium reactions
« Reply #5 on: April 02, 2004, 05:46:01 AM »
This is rather complex and tedious problem, but can be solved without other data.
First of all, a correction to your equation:
CH4 + CO2 = 2CO + 2H2
Since mixture is equimolar, lets take 1 mole of methane and carbon dioxide.
In our case, theoretical yield of H2 in moles is two times more than
number of moles (x) of methane or carbon dioxide converted to products.
Then 2x moles of carbon monoxide or hydrogen will be obtained.

We use Kp so we should calculate partial pressures for all compounds
in the equilibrium.
pCH4 = pCO2 = 10(1-x)
pCO = pH2 = 20x/
This led us to the equation (attached), but solved in mole fraction because only in this case K may be unitless..

It seems to me values of Kp are far too high and reaction should result in
almost theoretical yield for these data.
« Last Edit: April 05, 2004, 01:09:17 AM by AWK »
AWK