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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on November 21, 2020, 03:18:20 AM

Title: What is the equilibrium constant for this reaction?
Post by: Win,odd Dhamnekar on November 21, 2020, 03:18:20 AM

 How to compute equilibrium constant and equilibrium concentrations of reactants?

Title: Re: What is the equilibrium constant for this reaction?
Post by: Borek on November 21, 2020, 04:30:45 AM

This is not much different from other problems you did in the past. Just follow the stoichiometry to find concentrations.

And please note: you are here long enough to know the rules. Next post where you don't follow them will be deleted and you will receive a formal warning.

Title: Re: What is the equilibrium constant for this reaction?
Post by: Win,odd Dhamnekar on November 21, 2020, 11:40:37 PM

Initial concentration of Reactants [itex]Fe^{3+}= 0.00027 M, (SCN)^{-}=0.00085 M[/itex]. Initial concentration of product [itex]Fe(SCN)^{2+}=0[/itex]. At the equilibrium, concentration of  [itex] Fe(SCN)^{2+}=0.00018=2x[/itex]

[tex]K_c=\frac{0.00018}{(0.00027-x)*(0.00085-x)}=\frac{2x}{(0.00027-x)*(0.00085-x)}\Rightarrow 2x=0.00018,x=0.00009, K_c=\frac{0.00018}{(0.00018)*(0.00076)}=1315.789   liters/mol[/tex]

Title: Re: What is the equilibrium constant for this reaction?
Post by: AWK on November 22, 2020, 12:58:51 AM

Why 2x?

Title: Re: What is the equilibrium constant for this reaction?
Post by: Win,odd Dhamnekar on November 22, 2020, 03:49:01 AM

Equilibrium concentration of product [itex][Fe(SCN)]^{2+}[/itex] is 2x because one x part we take from [itex]Fe^{3+}[/itex] and anther x part we take from [itex] (SCN)^{-}[/itex]. Adding both, we get 2x.

Title: Re: What is the equilibrium constant for this reaction?
Post by: Borek on November 22, 2020, 04:26:40 AM

Quote from: Win,odd Dhamnekar on November 22, 2020, 03:49:01 AM

Equilibrium concentration of product [itex][Fe(SCN)]^{2+}[/itex] is 2x because one x part we take from [itex]Fe^{3+}[/itex] and anther x part we take from [itex] (SCN)^{-}[/itex]. Adding both, we get 2x.

No, you don't add them, that's not how the stoichiometry works.

Please google ICE table and learn what I, C and E stand for and how they are determined.

Title: Re: What is the equilibrium constant for this reaction?
Post by: Win,odd Dhamnekar on November 22, 2020, 08:40:11 AM

 Thanks for advice.

Title: Re: What is the equilibrium constant for this reaction?
Post by: Win,odd Dhamnekar on November 23, 2020, 03:53:09 AM

 I prepared ICE table as follows:

Reaction                        Fe³⁺           (SCN)⁻             [Fe(SCN)]²⁺
Initial amounts           0.00027       0.00085                0
Change in amounts    -0.00018      -0.00018           0.00018
Equilibrium amounts   0.00009       0.00067           0.00018

[itex]K_c=\frac{0.00018}{(0.00009)*(0.00067)}=2985.07463 [/itex] liters/mol.   Is this equilibrium constant correctly computed?   

Title: Re: What is the equilibrium constant for this reaction?
Post by: mjc123 on November 23, 2020, 08:13:07 AM

Yes, but to an absurd degree of precision, considering that the concentrations are only given to 2 significant figures.