April 21, 2019, 04:31:28 AM
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Topic: increasing the stoichiometric coefficients  (Read 639 times)

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Offline magnus

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increasing the stoichiometric coefficients
« on: February 07, 2019, 02:56:53 AM »
For this reaction Kc is 55.2, increasing the stoichiometric coefficients the value is always the same ...

  H2 (g) + I2 (g) ⇄ 2 HI (g) risulta Kc=55.2 a 698 K.
2 H2 (g) +2 I2 (g) ⇄ 4 HI (g)

Kc= [HI]² / [H₂][I₂]
 ??? ??? ???

Offline chenbeier

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Re: increasing the stoichiometric coefficients
« Reply #1 on: February 07, 2019, 04:43:09 AM »
No it would be Kc =[HI]4/([H2]2*[I2]2 )

Offline magnus

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Re: increasing the stoichiometric coefficients
« Reply #2 on: February 08, 2019, 02:26:32 AM »
ok, but with the same concentration values, doubling the stechimetric values ​​Kc will always be 55.2, it is a constant ...

Offline mjc123

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Re: increasing the stoichiometric coefficients
« Reply #3 on: February 08, 2019, 04:39:38 AM »
No, it would be 55.22.
[HI]4/[H2]2[I2]2 = ([HI]2/[H2][I2])2

Offline magnus

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Re: increasing the stoichiometric coefficients
« Reply #4 on: February 10, 2019, 01:50:45 PM »
Ok

Thanks

Offline Vidya

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Re: increasing the stoichiometric coefficients
« Reply #5 on: February 11, 2019, 07:53:05 PM »
For this reaction Kc is 55.2, increasing the stoichiometric coefficients the value is always the same ...

  H2 (g) + I2 (g) ⇄ 2 HI (g) risulta Kc=55.2 a 698 K.
2 H2 (g) +2 I2 (g) ⇄ 4 HI (g)

Kc= [HI]² / [H₂][I₂]
 ??? ??? ???
If you multiply an equilibrium reaction by a coefficient n then new K  is n raised to power of previous K
Mathematics on K expression
Every textbook of chemistry has great explanation of it.

Offline mjc123

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Re: increasing the stoichiometric coefficients
« Reply #6 on: February 12, 2019, 08:10:14 AM »
I think you mean "new K is previous K raised to power of n"

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