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posted by  ss83 on 7/9/2008 9:04:19 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{SO}}_2 {\rm{Cl}}_2 \left( g \right)\; \rightleftharpoons \;{\rm{SO}}_2 \left( g \right)\; + \;{\rm{Cl}}_2 \left( g \right)
K_{\rm{c}} = 2.99\; \times \;10^{ - 7} at 227^\circ \rm C

If a reaction mixture initially contains 0.177 M {\rm{SO}}_2 {\rm{Cl}}_2, what is the equilibrium concentration of {\rm{Cl}}_2 at 227 ^\circ {\rm C}?
  [{\rm Cl}_2]  =
 ? \rm M

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posted by james jagan on 7/10/2008 12:30:52 AM  |  status: Live
Asker's Rating: Helpful   
ss83's comment:
"not quite the right answer, but thank you for helping me"
Response Details:
 We know that
      Kp = Kc(RT)Δn
   Here Δn is the difference in stoichometric cofficients for the reaction
  ∴ Δn = moles of products - moles of reactants
            = 2-1
            = 1
  Given                                     Need
  K_{\rm{c}} = 2.99\; \times \;10^{ - 7} 
  R = 8.314 J/Kmol  
  T = 227+273
    = 500K                               Kp
   Substitute these values in above equvation we get
  
 Kp = (2.99*10-7)(8.314)(500)
       = 0.0012429
   
 Kp for above reaction
  Kp = [SO2][Cl2]/[SO2Cl2]
  at equilibrium concentration
   Kp = [Cl2]2/[SO2Cl2]
 
  Obtained      
  Kp = 0.0012429
 Given
 [SO2Cl2] = 0.177M
  
Need
 [Cl2]
 
  substitute these values we get
 [Cl2] = √Kp*[SO2Cl2]
          = √(0.0012429)(0.177)
          = 0.0148M
 
   
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posted by AJ <3 on 7/10/2008 1:07:16 AM  |  status: Live
Asker's Rating: Lifesaver   
ss83's comment:
"wow, thank you!"
Response Details:
 You first start with the "ICE box" strategy:     

                      {\rm{SO}}_2 {\rm{Cl}}_2 \left( g \right)\; \rightleftharpoons \;{\rm{SO}}_2 \left( g \right)\; + \;{\rm{Cl}}_2 \left( g \right) 
initial         :      0.177 M {\rm{SO}}_2 {\rm{Cl}}_2                         0                              0
change      :               - X                                  + X                           + X                     
equilibrium:      0.177 M - X                          0 + X                        0 + X

We can set up an equation for this reaction such that:
 


And this is the following dta we have right now:

K_{\rm{c}} = 2.99\; \times \;10^{ - 7}
[SO2Cl2] = 0.177 M - X 
[SO2] =  0 + X  
[Cl2] = 0 + X
(Temperature is not important because it is constant throughout the whole reaction, so don't worry about it)

Now we plug everything into our equation and it should look like:



Now let's solve for X with some math:

         Muiltiply each side by 0.177M - X
                Distribute  
          Subtract X2 to each side.

That last step takes the form of   aX2 + bX - C = 0. This is where we need to use the quadratic formula:

= X

So lets go ahead and plug everything in and solve for X:

=
=                            8.9x10-14 is very small so adding it is insignificant
=                                Because of the + and - , lets split these two up.
=
=               We want the Positive value, so 2.3x10-4

Now we have solved for X and we can go ahead and find the equilibrium. So lets go back to the "ICE box" strategy and plug back our X's.
                      {\rm{SO}}_2 {\rm{Cl}}_2 \left( g \right)\; \rightleftharpoons \;{\rm{SO}}_2 \left( g \right)\; + \;{\rm{Cl}}_2 \left( g \right) 
initial         :      0.177 M {\rm{SO}}_2 {\rm{Cl}}_2                         0                              0
change      :               - X                                  + X                           + X                     
equilibrium:      0.177 M - 2.3x10-4                          0 + 2.3x10-4                        0 + 2.3x10-4

And so, we have our equilibrium concentrations:

[SO2Cl2] = 0.177 M - 2.3x10-4    =  0.176 M
[SO2] =  0 + 2.3x10-4                   =  2.3x10-4 M 
[Cl2] = 0 + 2.3x10-4                      =  2.3x10-4 M           This is the answer you asked for.
Please ask for further assistance. AJ
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