# this question is from edexcel chemistry unit 4 january 2018

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#1
https://imgur.com/cPz5wXf

i dont get it why its C
A and D are not correct i get that but whats wrong with B (this is the one i think should have been the correct answer)
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1 month ago
#2
Assuming you already know how to calculate the equilibrium constant,
kp(1) = ([HI]^2)/([H][I])
kp(2) = ([HI])/(([H]^(1/2)([I]^(1/2))
If you square both sides of the second equation you get:
(kp(2))^2 = ([HI]^2)/([H][I])
which if the equivalent of kp(1)
Hence kp(1)=(kp(2))^2
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#3
(Original post by lege-lego)
Assuming you already know how to calculate the equilibrium constant,
kp(1) = ([HI]^2)/([H][I])
kp(2) = ([HI])/(([H]^(1/2)([I]^(1/2))
If you square both sides of the second equation you get:
(kp(2))^2 = ([HI]^2)/([H][I])
which if the equivalent of kp(1)
Hence kp(1)=(kp(2))^2
okay,got it but one last thing what happens to the denomenator after squaring?
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#4
(Original post by riamu)
okay,got it but one last thing what happens to the denomenator after squaring?
(1/2H2)^2 and (1/2I2)^2????
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1 month ago
#5
(Original post by riamu)
(1/2H2)^2 and (1/2I2)^2????
Could you write that out in a different way?
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1 month ago
#6
(Original post by riamu)
okay,got it but one last thing what happens to the denomenator after squaring?
The same thing that happens to the numerator.
(a/b)^2 = (a^2)/(b^2)
Is that what you wanted to know?
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#7
(Original post by lege-lego)
Could you write that out in a different way?
https://imgur.com/IzNg9FR
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#8
(Original post by lege-lego)
The same thing that happens to the numerator.
(a/b)^2 = (a^2)/(b^2)
Is that what you wanted to know?
https://imgur.com/IzNg9FR ----correct me on this
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#9
(Original post by lege-lego)
The same thing that happens to the numerator.
(a/b)^2 = (a^2)/(b^2)
Is that what you wanted to know?
will this affect their coeffecients as well?
and also why does the kp changes i mean they are only affected by temperature which i assume would be same for both so why it gets squared?
Last edited by riamu; 1 month ago
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#10
(Original post by riamu)
will this affect their coeffecients as well?
and also why does the kp changes i mean they are only affected by temperature which i assume would be same for both so why it gets squared?
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1 month ago
#11
I haven't really read the question yet, but yes kp is only affected by temperature I think, so if both the numerator and denominator were squared, the equilibrium would have to shift so that kp is the same even with squaring
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