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CHEMICAL EQUILIBRIUM

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Pupil
Karma Points: 53
(University of South Africa)
Respect (90%):
Date Posted: 5/11/2008 3:29:29 AM  Status: Live
CHEMICAL EQUILIBRIUM
Course Textbook Chapter Problem
General Chemistry module che102p 5 1
Question Details:
Consider the following reaction;
Use this reaction to drive the relation between
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Answers:

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Mentor
Karma Points: 415
Date Posted: 5/11/2008 10:50:31 AM  Status: Live
Asker's Rating: Lifesaver   
Response:

kp= Keq=
kc=kp/(rt)2 w/ 2 is n which is calculated by n=2-(1+3)= -2
 kc is known by using concentration (aqueous soln)
  kp is known by using partial pressure (gaseous soln)
in this case we have gaseous solution--> use kp to calculate the constant equilibrium of this problem.
Crusoe's Comment:
thankx

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Expert
Karma Points: 1,031
Date Posted: 5/11/2008 10:58:34 AM  Status: Live
Asker's Rating: Helpful   
Response:
the rel. b/w kp and kc is
   kp=kc(rt)n

where n= diff. in the no. of moles of products -- no.of moles of reactants
           n=2-(1+3)= -2

kp=kc(rt)-2

kp=kc/(rt)2
Crusoe's Comment:
thankx


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Oracle
Karma Points: 7,091
(University of South Florida)
Date Posted: 5/11/2008 11:18:45 AM  Status: Live
Asker's Rating: Helpful   
Response:
The equation would be
 
[P of NCl3]2
over
][P of N2][P of Cl2]3
Crusoe's Comment:
thankx
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