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Topic: What is the equilibrium constant for this reaction?  (Read 1794 times)

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Offline Win,odd Dhamnekar

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What is the equilibrium constant for this reaction?
« on: November 21, 2020, 03:18:20 AM »
 How to compute equilibrium constant and equilibrium concentrations of reactants?
« Last Edit: November 21, 2020, 03:29:00 AM by Win,odd Dhamnekar »
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Offline Borek

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Re: What is the equilibrium constant for this reaction?
« Reply #1 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.
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Offline Win,odd Dhamnekar

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Re: What is the equilibrium constant for this reaction?
« Reply #2 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]
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Offline AWK

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Re: What is the equilibrium constant for this reaction?
« Reply #3 on: November 22, 2020, 12:58:51 AM »
Why 2x?
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Offline Win,odd Dhamnekar

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Re: What is the equilibrium constant for this reaction?
« Reply #4 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.
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Offline Borek

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Re: What is the equilibrium constant for this reaction?
« Reply #5 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.
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Offline Win,odd Dhamnekar

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Re: What is the equilibrium constant for this reaction?
« Reply #6 on: November 22, 2020, 08:40:11 AM »
 Thanks for advice.
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Offline Win,odd Dhamnekar

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Re: What is the equilibrium constant for this reaction?
« Reply #7 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?   
« Last Edit: November 23, 2020, 04:51:05 AM by Win,odd Dhamnekar »
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Any science consists of the following process.
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Offline mjc123

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Re: What is the equilibrium constant for this reaction?
« Reply #8 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.
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