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Topic: Kc Equilibrium Problem - Need to know Method of Successive Approximations  (Read 9882 times)

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Offline edoble

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Here is the problem, any help would be really helpful, thank you.


At T=350°, the value for Kc for the reaction N2 (g) +3H2 (g) = 2NH3 (g) is 8. If 2 moles of N2, 1 mole of H2, and 2 moles of NH3 are put into an empty 2L flask, what will be the concentration of H2 at equilibrium?

Offline Hunter2

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #1 on: November 06, 2012, 07:59:52 AM »
Write down the law of mass action for this reaction.

Then you have to think that some of the chemicals will convert and disappear and new product is formed, for this you use x.  Some x will go away and some in a ratio will be produced.

Offline edoble

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #2 on: November 06, 2012, 09:52:03 AM »
I made the ICE Chart but after that I am confused because I need to use the Method of Successive Approximations to get my X. This is what I have so far:

        N2     +     3H2      =     2NH3

I       1              0.5                1

C      -x            -3x               +2x
 
E     1-x          0.5-3x            1+2x


Kc =     (1+2X)^2      = 8
       (1-x)*(0.5-3x)^3

At this point I need to use Method of Successive Approximations to solve for X. I do not know how to do it since my Kc formula is complicated.

Is my ICE chart correct?

Any help would be much appreciated, thank you.

Offline XGen

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #3 on: November 06, 2012, 10:01:22 AM »
Can you assume that x will be negligible for any of those concentrations?

Offline Hunter2

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #4 on: November 06, 2012, 10:06:27 AM »
I made the ICE Chart but after that I am confused because I need to use the Method of Successive Approximations to get my X. This is what I have so far:

        N2     +     3H2      =     2NH3

I       1              0.5                1

C      -x            -3x               +2x
 
E     1-x          0.5-3x            1+2x


Kc =     (1+2X)^2      = 8
       (1-x)*(0.5-3x)^3

At this point I need to use Method of Successive Approximations to solve for X. I do not know how to do it since my Kc formula is complicated.

Is my ICE chart correct?

Any help would be much appreciated, thank you.

I think this should be correct. Of course you need a math program to solve for x. But you can try to simplify the formula by using the binomial formulas.

Offline Hunter2

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #5 on: November 06, 2012, 10:09:48 AM »
Can you assume that x will be negligible for any of those concentrations?

I dont think so.


Offline edoble

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #7 on: November 06, 2012, 11:31:19 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

When I use that X value to find Concentration of H2 at equilibrium I get:

From ICE chart: H2=0.5-3x
                      H2=0.5-[3(1.0628)]
                      H2= - 2.6884

I get a negative solution...I believe something is wrong...please help I am confused, thank you.

Offline Hunter2

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #8 on: November 06, 2012, 12:33:20 PM »
You have to choose another x. 1.0628 is not the solution we looking for.

I have to change my answer to XGen. He was right.

Offline Borek

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #9 on: November 06, 2012, 12:39:48 PM »
First things first - what is the method of successive approximations?

Edit: something is wrong, I got the same 1.0628 with a different approach. Are you sure about the Kc value? I seem to remember it is much lower.

Edit2: ROFL, everything is OK. Nice question :) 1.0628 is not the only real root.
« Last Edit: November 06, 2012, 12:53:39 PM by Borek »
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Offline Hunter2

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #10 on: November 06, 2012, 12:55:39 PM »
If all values true I think 1 *10-16 is the right x. What means not much change.

Offline XGen

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #11 on: November 06, 2012, 07:14:37 PM »
Perhaps looking at Q prior to doing any calculations might be helpful.

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #12 on: November 07, 2012, 01:00:06 AM »
What is in this case the Q. Because I cannot see without calculation whats going on.

Offline curiouscat

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #13 on: November 07, 2012, 08:09:28 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?

Offline curiouscat

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Re: Kc Equilibrium Problem - Need to know Method of Successive Approximations
« Reply #14 on: November 07, 2012, 08:27:10 AM »

At this point I need to use Method of Successive Approximations to solve for X. I do not know how to do it since my Kc formula is complicated.


Rearrange your equality  to get:

[tex]
x=1-\frac{(1+2*x)^2}{8*(0.5-3*x)^3}
[/tex]


Now choose any reasonable guess for x. Already 0≤x≤1  seems the domain from your ICE chart.
Use this guess to calculate a new x. You'll find this converges to a solution (often, not always).

In less than 4 iterations I have your Wolfram solution.

e.g. Starting from x=0.1
0.10
-21.5000
0.9992
1.0721
1.0617
1.0630
1.0628

Starting from x=0.9

0.90
1.0920
1.0592
1.0633
1.0628

etc.

Note this only gives one solution. Which is the key in this problem. :)

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