anonymous
  • anonymous
Calculate the no. of moles of Cl_2 produced at equilibrium , when 1 mole of PCl_5 is heated at 250°C , in a vessel having capacity of 10dm^3 . Kc for the reaction = 0.041
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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aaronq
  • aaronq
write a balanced reaction
anonymous
  • anonymous
help plz!! its a matter of life and death !!! ...i need to solve it within today ..so plzz dont go plzzzzzzz
anonymous
  • anonymous
there is no given equation

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aaronq
  • aaronq
you have to write it yourself
anonymous
  • anonymous
okay i'll try
anonymous
  • anonymous
PCl_5 ---> PCl_3 + Cl_2
anonymous
  • anonymous
^is this correct??
aaronq
  • aaronq
yep, thats good, now write an equilibrium expression
anonymous
  • anonymous
how ?
aaronq
  • aaronq
A -> B + C K = [B][C]/[A]
aaronq
  • aaronq
products over reactants
anonymous
  • anonymous
yes i know that
anonymous
  • anonymous
i guess these expressions are for reversible reactions ??
anonymous
  • anonymous
K = [PCl_3][Cl_2] / [PCl_5]
aaronq
  • aaronq
yeah, if you have an equilibrium the reaction is reversible
aaronq
  • aaronq
okay, now with the reaction you wrote: PCl_5 ---> PCl_3 + Cl_2 make an I.C.E. table
anonymous
  • anonymous
now what to do ? ...i have solved this equilibrium expression problems ...but this one is out of my mind
anonymous
  • anonymous
what is I.C.E table ?
aaronq
  • aaronq
concentrations of species at initial, change, equilibrium
anonymous
  • anonymous
1/10 ?
aaronq
  • aaronq
to find the initial amount of PCl5 you have to use PV=nRT
aaronq
  • aaronq
oh wait no you don't
aaronq
  • aaronq
yeah 0.1 is right
anonymous
  • anonymous
i guess we have to suppose it as x and then we have to solve a quadratic
aaronq
  • aaronq
yes you do
anonymous
  • anonymous
so is it x or 0.1 ?
aaronq
  • aaronq
well the change is x, the initial conc is 0.1
anonymous
  • anonymous
0.1-x ?
aaronq
  • aaronq
yep, thats PCl5, what about the rest?
anonymous
  • anonymous
there was nothing mentioned to PCl_3 and Cl_2 ...how to find that ?
aaronq
  • aaronq
well if you subtract x from PCl5, you get x Cl2 and x PCl3, right?
anonymous
  • anonymous
yeah!!
aaronq
  • aaronq
PCl_5 ---> PCl_3 + Cl_2 I 0.1 0 0 C -x +x +x E 0.1 -x x x K = x^2/(0.1-x) right?
anonymous
  • anonymous
wait let me solvve now
anonymous
  • anonymous
this is giving complicated answer
anonymous
  • anonymous
my equation is x^2+0.041x-0.0041=0
aaronq
  • aaronq
looks good, try it, heres the answer http://www.wolframalpha.com/input/?i=0.041+%3D+x%5E2%2F%280.1-x%29
anonymous
  • anonymous
the answer is coming wrong
aaronq
  • aaronq
you're probably not solving the quadratic correctly
anonymous
  • anonymous
no i mean the answer in wolframalpha is wrong...it is a different anser for x
aaronq
  • aaronq
it's not 0.047 ?
anonymous
  • anonymous
no
anonymous
  • anonymous
x=0.35
anonymous
  • anonymous
^this should come
aaronq
  • aaronq
hm even if you use the initial 1 mole PCl5, Cl2 = 0.18 moles
anonymous
  • anonymous
(1-2x)/10 --- > x/10 + x/10
anonymous
  • anonymous
^is this true
anonymous
  • anonymous
this gives x=0.35
aaronq
  • aaronq
no, it would only be 1 x not 2x because 1 PCl5 breaks apart into Cl2 and PCl3
aaronq
  • aaronq
what book are you using?
anonymous
  • anonymous
i asked a friend of mine
abb0t
  • abb0t
|dw:1370231337357:dw|
anonymous
  • anonymous
^this is the moles of CL_2 ?
aaronq
  • aaronq
hahaha .. um well i can tell you that the equilibrium expression is K = x^2/(0.1-x)
aaronq
  • aaronq
number 1 http://darlenewall.ca/grade12/Answers_Equilibrium_Problems.pdf number 5 https://mymission.lamission.edu/userdata%5Cpaziras%5CChem102%5CReview_14ANS.pdf

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