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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: magnus on February 07, 2019, 02:56:53 AM

Title: increasing the stoichiometric coefficients
Post by: magnus 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]
 ??? ??? ???
Title: Re: increasing the stoichiometric coefficients
Post by: chenbeier on February 07, 2019, 04:43:09 AM
No it would be Kc =[HI]4/([H2]2*[I2]2 )
Title: Re: increasing the stoichiometric coefficients
Post by: magnus 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 ...
Title: Re: increasing the stoichiometric coefficients
Post by: mjc123 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
Title: Re: increasing the stoichiometric coefficients
Post by: magnus on February 10, 2019, 01:50:45 PM
Ok

Thanks
Title: Re: increasing the stoichiometric coefficients
Post by: Vidya 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.
Title: Re: increasing the stoichiometric coefficients
Post by: mjc123 on February 12, 2019, 08:10:14 AM
I think you mean "new K is previous K raised to power of n"