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General : CHemical Equilibrium Question
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 Message 7 of 8 in Discussion 
From: MSN Nickname·Steve·  in response to Message 6Sent: 4/13/2008 5:53 PM
One more way I forgot to mention:  the method of successive approximations.  Here's how that works:
 
1.  Start with the original equilibrium constant expression and solve for x using the approximation that the 1.00 �?2x term is 1.00:
 
0.288   =          4x3              x  =  0.4160
                 (1.00 �?2(0))2  
 
2.  Now repeat the calculation, plugging x = 0.416 into the denominator term, and solve for x again:
 
0.288   =             4x3                x  =  0.1267
                (1.00 �?2(0.416))2  
 
3.  Keep repeating this process, plugging the previous value of x into the denominator and solving for x again.  The next calculation is,
 
0.288   =             4x3                x  =  0.3424
                (1.00 �?2(0.1267))2  
 
What will happen, if your initial value of x is not too far off, is convergence of x to a constant value.  Although I have written rounded values of x here, I did not round x in the actual calculations.  I just stored each value on my calculator and used the RCL button to recall it in the next iteration.  I continued calculating x this way and got the following values:
 
0.1927
0.3007
0.2253
0.2790
0.2414
0.2681
0.2493
0.2626
0.2532
0.2598
0.2552
0.2585
0.2561
0.2578
0.2566
0.2574
0.2569
0.2573
 
So we can tell that x is about 0.257 M.  When you plug this value of x back into the equilibrium constant expression, you get,
 
       4(0.257)3           =    0.287
 (1.00 �?2(0.257))2  
 
This is almost the correct value of Kc.  With a little trial and error, you can get another digit.  Trying x = 0.2571 as a guess, I get Kc = 0.28804, closer to the correct value of Kc of 0.288, so this is a better value of x.  Or, you could continue the above procedure and gradually improve you value of x that way.
 
 It is probably faster to use the cubic formula, but if you don't have that, this is an alternate method that should work.  A possible problem is, if your initial value of x is too far off, you might get convergence to one of the other roots of the equation instead of the physically valid one.  But most of the time, this method will give the right answer.
 
Steve