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Equilibrim constant after equilibrium?

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Scholar
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Respect (80%):
Date Posted: 4/25/2008 10:43:06 PM  Status: Live
Equilibrim constant after equilibrium?
Course Textbook Chapter Problem
General Chemistry Chemistry (9th) by Chang N/A N/A
Question Details:
2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400oC. After equilibrium was established, it was found that 28 percent of the NOCl had dissociated according to the equation:

2NOCl(g) 2NO(g) + Cl2(g)

Calculate the equilibrium constant, Kc, for the reaction.
  1. 0.021
  2. 0.039
  3. 0.169
  4. 26
  5. 47
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Answers:

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Oracle
Karma Points: 9,935
Date Posted: 4/25/2008 11:22:00 PM  Status: Live
Asker's Rating: Lifesaver   
Response:
2NOCl(g) 2NO(g) + Cl2(g)
Set up ICE table:
 
2NOCl(g)
2NO(g)
Cl2(g)
Initial
2.50/2.50 = 1.00 M
 
 
Change   
-0.28 M
+0.28 M
+ 0.14 M
Equilibrium
0.72 M
0.28 M   
0.14 M
2NOCl(g) 2NO(g) + Cl2(g)
K = [NO]2 [ Cl2] / [NOCl]2 = (0.28) 2 (0.14) / (0.72)2 = 0.021 (answer a)
Bertalsteinein's Comment:
Right! Thank you!
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Guru
Karma Points: 1,798
Date Posted: 4/25/2008 11:33:39 PM  Status: Live
Asker's Rating: Helpful   
Response:
Degree of disociation(x)= 0.28
             2NOCl(g) 2NO(g) + Cl2(g)
   let       a(1-x)                 ax         ax/2
 
    a=2.5  , x=0.28 and V=2.5 Lit
at equilibrium,
              [NOCl]= a(1-x) =2.5(1-0.28)/2.5=0.72mol/lit
              [NO] = ax= 2.5 x 0.28/2.5=0.28mol/lit
              [cl2] = ax/2=2.5 x 0.28/2 x 2.5=0.14mol/lit
  Kc = [NO] 2[Cl2 ]/[NOCl]2
        =0.28 x 0.28 x 0.14 /0.72 x 0.72
        =0.021
 
Option a)
 
 
Bertalsteinein's Comment:
Thank you for the confirmation!



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