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posted by  ss83 on 7/9/2008 9:01:25 PM  |  status: Live  

Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant

Course Textbook Chapter Problem
General Chemistry N/A 14 N/A
Question Details:
Consider the following reaction:
{\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right)
K_{\rm{p}} = 0.0611 at 2000 \rm K
A reaction mixture initially contains a {\rm CO} partial pressure of 1322 torr and a {\rm{H}}_2 {\rm{O}} partial pressure of 1772 torr at 2000 {\rm K}.

Calculate the equilibrium partial pressure of {\rm CO}_2.
  [{\rm CO}_2]  =
 ? \rm torr

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posted by AJ <3 on 7/10/2008 1:33:34 AM  |  status: Live
Asker's Rating: Lifesaver   
ss83's comment:
"i appreciate your helping me, but the answer was wrong. I'm not sure how to get to the correct answer so if you resubmit with a correct answer I will give you another 10points. Thanks."
Response Details:
I answered another question like this one, so look it up too =]
Well here we go, lets do the "ICE box" method again.
                      {\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right) 
initial            :  1322 torr             1772 torr                   0 torr                0 torr
change         :       - X                    - X                           + X                + X 
equilibrium   :  1322 - X              1772 - X                     X                      X

Let's set up the equation using the following formula:



Here we can use the quadratic formula, another method is to approximate. You can't always approximate so don't always use it =D. If you prefer to see work with the quadratic formula then just ask again and I'll do it over. But anyways, by approximating, we will assume that the value subtracting from 1322 and 1772 will be very small and insignificant.
So we assume (1322 - X) ˜ 1322 and (1772 - X) ˜ 1772. Then we get:
 
Let's continue with the math now:



No we have our X so plug it back in and solve for our new pressures.
                       {\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right) 
initial            :  1322 torr             1772 torr                   0 torr                0 torr
change         :       - X                    - X                           + X                + X 
equilibrium   :1322 - 378.3277  1772 - 378.3277     378.3277          378.3277
 
PCO = 944 torr
PH2O = 944 torr
PCO2 = 378 torr      This is your answer.
PH2 = 378 torr
 
Please ask for further assistance.
 
Please ask for further assistance. AJ
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posted by AJ <3 on 7/10/2008 9:22:19 PM  |  status: Live
Asker's Rating: Helpful   
ss83's comment:
"thanks for helping"
Response Details:
Woops, sorry. We need to use the quadratic formula instead of using an estimating =/ So lets try this again.

    {\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right) 
initial            :  1322 torr             1772 torr                   0 torr                0 torr
change         :       - X                    - X                           + X                + X 
equilibrium   :  1322 - X              1772 - X                     X                      X

Let's set up the equation using the following formula:

aX2 + bX + c = 0
  


X can't be negative so we choose the positive value.
X = 303 torr
[CO] = 1322 - X = 1019
[H2O] = 1772 - X = 1469
[CO2] = X = 303 Now this is your answer =]
[H2] = X = 303
 
You can also check you work for solving X. If you want to, just plug back X into your equation:
   It's very close to the actual answer, and that's just do to rounding off somewhere. This is how you would check your work just to make sure the X value you solved for doesn't give you a bogus answer, which I should I have done when I sent the first answer. Sorry about that.

Please ask for further assistance.
Please ask for further assistance. AJ
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posted by Ainala8055 on 7/10/2008 9:43:57 PM  |  status: Live
Asker's Rating: Not Helpful   
ss83's comment:
"sorry this is still wrong. If you would like to see how it was done above to see your mistakes it's available :) Thanks anyway"
Response Details:
 
{\rm{CO}}\left( g \right)\; + \;{\rm{H}}_2 {\rm{O}}\left( g \right)\; \rightleftharpoons \;{\rm{CO}}_2 \left( g \right)\; + \;{\rm{H}}_2 \left( g \right) 
Kp is calculated only for gases
Kp=PP.CO2 x PP.H2 / PP.CO x PP.H2O
0.0611=X2 / 1772 x 1322
X2=140555
X=378
equilibrium partial pressure of {\rm CO}_2=378
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