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Topic: Chemical Equilibrium (Read 3602 times)
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Chocotaco
Regular Member
Posts: 13
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Chemical Equilibrium
«
on:
November 21, 2010, 06:24:40 PM »
Just have a few questions on a couple of problems, hopefully if I do this one right the other one would follow ok
After being introduced into a container at an initial pressure of 3.42 atm A
3
B
5
decomposes into A
2
B
4
and AB. At equilibrium the partial pressure of AB is 0.28 atm.
What is K
p
for this equilibrium?
So I'll do a table chart for it
A
3
B
5
<--> A
2
B
4
+ AB
Initial 3.42 | 0 | 0
Change -x | +x | +x
Equi 3.42-x| x | .28+x
K
p
= (x)(.28+x)/3.42-x
Am I going in the right direction or is it completely wrong? I'm not too sure if it is. Any help would be appreciated.
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rabolisk
Chemist
Full Member
Posts: 494
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Re: Chemical Equilibrium
«
Reply #1 on:
November 21, 2010, 10:17:13 PM »
You're going in the right direction.
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Chocotaco
Regular Member
Posts: 13
Mole Snacks: +0/-0
Re: Chemical Equilibrium
«
Reply #2 on:
November 21, 2010, 11:08:48 PM »
Ok alright that's great to hear
So...
Kp= (x)(.28+x)/3.42-x=
x
2
+1.28x-3.42=0
Using the quadratic formula, I would get x= 1.32, -2.6. The first value 1.32 would make sense...so K
p
would be 1.32? Would that be correct or missing a part?
Thanks
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rabolisk
Chemist
Full Member
Posts: 494
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Re: Chemical Equilibrium
«
Reply #3 on:
November 22, 2010, 01:06:58 PM »
Sorry I misread the question. At equilibrium the partial pressure of AB is 0.28 atm. What you have tabled is if the initial partial pressure is 0.28 atm, which is unsolvable.
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Chocotaco
Regular Member
Posts: 13
Mole Snacks: +0/-0
Re: Chemical Equilibrium
«
Reply #4 on:
November 22, 2010, 02:43:25 PM »
Although I'm not quite sure what you mean exactly, but given the problem has no answer?
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rabolisk
Chemist
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Posts: 494
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Re: Chemical Equilibrium
«
Reply #5 on:
November 22, 2010, 03:23:30 PM »
The problem is solvable. Note that K
p
is the ratio of equilibrium partial pressures. You are given the equilibrium partial pressure of one of the products.
You interpreted the problem wrong. Your calculations would be correct if you were given the initial partial pressure of AB. But this would not be solvable. Basically I mistakenly said you were doing the right thing when you were not. My apologies.
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Chocotaco
Regular Member
Posts: 13
Mole Snacks: +0/-0
Re: Chemical Equilibrium
«
Reply #6 on:
November 22, 2010, 04:37:36 PM »
I appreciate for helping out but I don't know what else to do. Maybe I'm over thinking but I tried something else and doubt it was even right. Basically did this:
A3B5 <--> A2B4 + AB
Initial 3.42 | 0 | 0
Change -x | +x | +x
Equi 3.42-x| x | x
x2/3.42-x= Kp
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rabolisk
Chemist
Full Member
Posts: 494
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Re: Chemical Equilibrium
«
Reply #7 on:
November 27, 2010, 10:31:25 AM »
That is correct, except you're given one more information.
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