# Chemical Forums

## 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 $Fe^{3+}= 0.00027 M, (SCN)^{-}=0.00085 M$. Initial concentration of product $Fe(SCN)^{2+}=0$. At the equilibrium, concentration of  $Fe(SCN)^{2+}=0.00018=2x$

$$K_c=\frac{0.00018}{(0.00027-x)*(0.00085-x)}=\frac{2x}{(0.00027-x)*(0.00085-x)}\Rightarrow 2x=0.00018,x=0.00009, K_c=\frac{0.00018}{(0.00018)*(0.00076)}=1315.789 liters/mol$$
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 $[Fe(SCN)]^{2+}$ is 2x because one x part we take from $Fe^{3+}$ and anther x part we take from $(SCN)^{-}$. 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
Equilibrium concentration of product $[Fe(SCN)]^{2+}$ is 2x because one x part we take from $Fe^{3+}$ and anther x part we take from $(SCN)^{-}$. 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
$K_c=\frac{0.00018}{(0.00009)*(0.00067)}=2985.07463$ liters/mol.   Is this equilibrium constant correctly computed?