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posted by  Crusoe on 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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posted by io2chip on 5/11/2008 10:50:31 AM  |  status: Live
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Response Details:

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.
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posted by vsiddhardha on 5/11/2008 10:58:34 AM  |  status: Live
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Crusoe's comment:
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Response Details:
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

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posted by gopackgo on 5/11/2008 11:18:45 AM  |  status: Live
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Response Details:
The equation would be
 
[P of NCl3]2
over
][P of N2][P of Cl2]3
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