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### Topic: Kc Equilibrium Problem - Need to know Method of Successive Approximations  (Read 9754 times)

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#### AWK

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« Reply #15 on: November 07, 2012, 11:02:20 AM »
Why do you think that ammonia is formed. May be it decomposes and negative root of the equation should be used.
AWK

#### curiouscat

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« Reply #16 on: November 07, 2012, 11:05:49 AM »
Why do you think that ammonia is formed. May be it decomposes and negative root of the equation should be used.

Does it have a negative root? Thought the only roots are 0 and 1.0628.

#### sjb

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« Reply #17 on: November 07, 2012, 11:11:33 AM »
What is in this case the Q. Because I cannot see without calculation whats going on.

See for instance http://dwb4.unl.edu/Chem/CHEM869V/CHEM869VLinks/learn.chem.vt.edu/tutorials/equilibrium/rxnquotient.html or similar

#### edoble

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« Reply #18 on: November 07, 2012, 11:18:45 AM »
I used www.wolframalpha.com to solve that above equation and this is what I get :

216x4-324x3+122x2-23x=0

The real root is X= 1.0628

I used Wolfram too and got two roots; not one: Why do you only see one root?

Using Wolfram Alpha to solve for X : 216x4-324x3+122x2-23x=0

The real solutions are:
X=0
X=1.06284

So thats why I quoted only X=1.06284

Is there something wrong with this problem; and YES I copied the question correctly from my chemistry book...word for word. Could be there is an error in the book..not sure.

#### Hunter2

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« Reply #19 on: November 07, 2012, 11:28:15 AM »
But 1.0628 is not the solution what gives the right hydrogen amount.

#### Borek ##### Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #20 on: November 07, 2012, 11:42:22 AM »
What is x?

What does x=1.0628 mean?

What does x=0 mean?
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#### Hunter2

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« Reply #21 on: November 07, 2012, 11:46:56 AM »
What is in this case the Q. Because I cannot see without calculation whats going on.

See for instance http://dwb4.unl.edu/Chem/CHEM869V/CHEM869VLinks/learn.chem.vt.edu/tutorials/equilibrium/rxnquotient.html or similar

We used Kp for that. p = pressure. I learned Q is used for heat capacity. But in different countries different letters for variables in use.

#### sjb

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« Reply #22 on: November 07, 2012, 12:04:38 PM »
We used Kp for that. p = pressure. I learned Q is used for heat capacity. But in different countries different letters for variables in use.

Kp to me suggests the conditions at equilibrium, not necessarily the conditions at the start of the reaction. which is what Q is used for in this case.

#### Hunter2

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« Reply #23 on: November 07, 2012, 12:08:28 PM »
x = 0 means we are already at equilibrium.  So far so good, but if I want to calculate the hydrogen (0,5-3x)3 = c(H2) we get 0.125 mol/l but we put 0.5 mol/l in? ?

For ammonia (1+2x)2 = c(NH3) give 1 mol/l and also for nitrogen  (1-x) give 1 mol/l. Why it doesnt fit for hydrogen.

the whole term give 1/0.125 = 8 = K

#### sjb

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« Reply #24 on: November 07, 2012, 12:11:03 PM »
x = 0 means we are already at equilibrium.  So far so good, but if I want to calculate the hydrogen (0,5-3x)3 = c(H2) we get 0.125 mol/l but we put 0.5 mol/l in??

For ammonia (1+2x)2 = c(NH3) give 1 mol/l and also for nitrogen  (1-x) give 1 mol/l. Why it doesnt fit for hydrogen.

the whole term give 1/0.125 = 8 = K

#### Hunter2

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« Reply #25 on: November 07, 2012, 12:15:11 PM »
Wow right. Its for hydrogen 0.125 (mol/l)3 so need 3rd. root what gives 0.5 mol/l Also for ammonia I need square root. (1+x)2 gives (mol/l)2.

#### curiouscat

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« Reply #26 on: November 07, 2012, 12:26:40 PM »

Using Wolfram Alpha to solve for X : 216x4-324x3+122x2-23x=0

The real solutions are:
X=0
X=1.06284

So thats why I quoted only X=1.06284

Is there something wrong with this problem; and YES I copied the question correctly from my chemistry book...word for word. Could be there is an error in the book..not sure.

Nothing wrong in the problem. But ignoring a zero root is the problem. That is the relevant root in your case.

They tricked you by giving you  a system already at equilibrium. #### Borek ##### Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #27 on: November 07, 2012, 12:26:50 PM »
Q is usually used for the reaction quotient.
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#### AWK

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« Reply #28 on: November 08, 2012, 05:18:44 AM »
Why do you think that ammonia is formed. May be it decomposes and negative root of the equation should be used.

Does it have a negative root? Thought the only roots are 0 and 1.0628.
If one of the solutions is exactly zero then we are  at the equilibrium point.
AWK

#### curiouscat

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« Reply #29 on: November 08, 2012, 08:09:29 AM »
Why do you think that ammonia is formed. May be it decomposes and negative root of the equation should be used.

Does it have a negative root? Thought the only roots are 0 and 1.0628.
If one of the solutions is exactly zero then we are  at the equilibrium point.

Yep. That's exactly the case here.