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Author Topic: Equilibrium Concentrations  (Read 14909 times)

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nikita

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Equilibrium Concentrations
« on: March 03, 2009, 11:59:23 AM »

I really need help here guys!!!!

Here is the question:

For the following reaction
Kc = 255 at 1000 K
CO(g) + Cl2(g) ::equil:: COCl2 (g)
 
A reaction mixture initially contains a CO concentration of 0.1490 M and a Cl2 concentration of 0.174 M  at 1000 K.

What is the equilibrium concentration of CO  at 1000 ?
What is the equilibrium concentration of Clat 1000 ?
What is the equilibrium concentration of COCl2 at 1000 ?

ok.  ive tried this prob a number of times and even asked some people in my class and everyone comes up with the same number.  and its wrong.  i know you guys here can figure out what i am doing wrong....

      CO(g) + Cl2(g) ::equil:: COCl2 (g)
I:  .1490M     .174M                              0
C:    -x          -x                                 +x
E:  .1490 - x   .174-x                            x
---------------------------------------------------------------
x/(.1490-x)(.174-x) = 255

x/255(.0259 -.323x+x2)
6.60-82.4x+255x2 = x
6.60-83.4x+255x2 = 0

through the quadratic equation, i get
x= .193 or x= .134
since x cannot be greater than the original concentration, i picked .134

Substituting back in for x,

.1490-.134 = .0150

sig figs aside, i put the following answers in and they were all wrong:

   5.00×10−3   
   1.50×10−2   
   1.50×10−2   
   1.40×10−2   Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.
   1.50×10−2   

so, could it be .013?  note the message that came up when i tried .014.   i tried the equation in about 5 or 6 different quadratic equation solvers and got .134 on all of them except for one, where i got .133.  when i substitute back in with .133, i get .016 as an answer. i havent attempted .016 as an answer, but the message leads me to believe its closer to .014 than .016.   and i only have one attempt left, so i, er, have to be right.  Oh, and when i did the quadratic by hand, i came out with .015.  can anyone tell me what i am doing wrong?  i would really hate to get this wrong due to a rounding error, but i am certain that i have the concept correct.  any info at all is much appreciated.  esp if i am messing up my sig figs. 


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Borek

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Re: Equilibrium Concentrations
« Reply #1 on: March 03, 2009, 12:11:01 PM »

Why do you round down intermediate results? You should do your calculation with full precision and round down only the final result.
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nikita

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Re: Equilibrium Concentrations
« Reply #2 on: March 03, 2009, 12:21:52 PM »

because im not that smart, i guess?

here is the full calculation:

x/(.1490-x)(.174-x) = 255

x/.025926-.323x+x2 = 255

255(.025926-.323x+x2) = x

6.61113-82.365+255x2 = x
6.61113-83.365+255x2

x= 0.19162863848223255
x= 0.13529293014521843

.1490 - 0.13529293014521843 = .0137070699
= .0137?  (because .1490 has 3 sig figs?)

(edited to apply 255 for x2 coefficient)
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nikita

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Re: Equilibrium Concentrations
« Reply #3 on: March 03, 2009, 12:23:49 PM »

actually, im subtracting, so should it be .0137 because of the decimal places?
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nikita

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Re: Equilibrium Concentrations
« Reply #4 on: March 03, 2009, 07:37:06 PM »

yes, the answer was .0137.  thanks for reminding me about rounding.  ive learned my lesson.
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