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### AuthorTopic: Equilibrium (ICE) Method of successive approximations!  (Read 862 times)

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#### spKelpDiver

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##### Equilibrium (ICE) Method of successive approximations!
« on: September 21, 2012, 04:12:51 PM »

Consider the following reaction:
2SO2(g) + O2(g) <-> 2SO3(g)
Kp=0.355 at 950 K
A 2.75 L reaction vessel at 950 K initially contains 0.100 mol of SO2 and 0.100 mol of O2.
Calculate the total pressure (in atmospheres) in the reaction vessel when equilibrium is reached.

Working with Kp so I solved for the partial pressure of each reactant and plugged it into my ICE table.

So now I make my ICE table.
2SO2(g) + O2(g) <-> 2SO3(g)

I         2.83atm    2.83atm      0

C        -2x            -x            +2x

E       2.83-2x     2.83-x         2x

.355 = (2x)2 / (2.83-2x)2(2.83-x)

.355 = 4x2 / (2.83-2x)2(2.83-x)

So I know I need to use "X is small method" before I move on to the "method of successive approximations."

.355 = 4x2 / (2.83)2(2.83) Did I do something wrong at this step?

solving for x.. x =1.42

Now I check the ratio to make sure its less than 5% deviation

1.42/2.83 = 50% ..WAY off. Did I do something wrong at this step?

So now I am supposed to use this ridiculous method of successive approximation.

So I plug the value of x (from the smaller amount method) back into the equation and solve until I get a number close the value of x.

.355 = 4x2 / (2.83-2(1.42))2(2.83-(1.42)

x=0.00355.... not close to 1.42 Did I do something wrong at this step?
so I plug this value (0.00355) back into the equation, and hope that x is close to 00355.

.355 = 4x2 / (2.83-2(.00355))2(2.83-(.00355))

x= 1.41 again!!!!! I must be doing something wrong.

So just a quick note about myself. When I get stuck on something like this....I literally have dreams about it. I can't put it down, I can't move passed it... I obsess over it. I know I need to work on this. I thought this method was supposed to be easier and quicker than manually solving for x!!!!
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#### Borek

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##### Re: Equilibrium (ICE) Method of successive approximations!
« Reply #1 on: September 22, 2012, 12:46:52 AM »

While I understand what you are doing it is the first time I see this approach pretending to be a well defined method.

So I know I need to use "X is small method" before I move on to the "method of successive approximations."

.355 = 4x2 / (2.83)2(2.83) Did I do something wrong at this step?

solving for x.. x =1.42

Now I check the ratio to make sure its less than 5% deviation

1.42/2.83 = 50% ..WAY off. Did I do something wrong at this step?

Assuming your math is OK you know "x is small" approximation is wrong in this case.

Quote
So now I am supposed to use this ridiculous method of successive approximation.

Nothing ridiculous about iterative methods - actually their variants are basis of many numerical methods.

Quote
So I plug the value of x (from the smaller amount method) back into the equation and solve until I get a number close the value of x.

No idea why you want to start with the x values calculated earlier - not that it is a bad starting point, I just fail to see why it is better than any other.

Quote
.355 = 4x2 / (2.83-2(1.42))2(2.83-(1.42)

You better rearrange your equation to the form xn+1=f(xn), then it is easy to either write a program or even use a spreadsheet for calculations.

#### spKelpDiver

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##### Re: Equilibrium (ICE) Method of successive approximations!
« Reply #2 on: September 22, 2012, 11:53:11 AM »

This is the method that is being described in my chemistry book. The book says first to use the "x is small" method and figure out the percent error. If it is greater than 5%, they say to move on to the method of successive approximations.

From the book...

"substitute the value obtained for x in the "x is small method" back into the original cubic equation, but only at the exact spot where x was assumed to be negligible and then solve the equation for x again. Continue this procedure until the value of x obtained from solving the equation is the same as the one that is substituted back into the equation."

Can somebody check my math and make sure I am not making a careless mistake? Am I wrong to drop both x values in the denominator? So far I can only find examples of this method where there is only one x variable in the denominator.
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#### Borek

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##### Re: Equilibrium (ICE) Method of successive approximations!
« Reply #3 on: September 22, 2012, 10:05:50 PM »

Am I wrong to drop both x values in the denominator? So far I can only find examples of this method where there is only one x variable in the denominator.

Assumption is A-x≈A for x << A. You can use it as many times as you need.
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