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### Topic: What is the equilibrium constant for this reaction?  (Read 668 times)

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

$$K_c=\frac{0.00018}{(0.00027-x)*(0.00085-x)}=\frac{2x}{(0.00027-x)*(0.00085-x)}\Rightarrow 2x=0.00018,x=0.00009, K_c=\frac{0.00018}{(0.00018)*(0.00076)}=1315.789 liters/mol$$
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#### 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?
AWK

#### 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 $[Fe(SCN)]^{2+}$ is 2x because one x part we take from $Fe^{3+}$ and anther x part we take from $(SCN)^{-}$. Adding both, we get 2x.
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#### Borek

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##### Re: What is the equilibrium constant for this reaction?
« Reply #5 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.
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#### 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.
Any science consists of the following process.
1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

#### 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

$K_c=\frac{0.00018}{(0.00009)*(0.00067)}=2985.07463$ liters/mol.   Is this equilibrium constant correctly computed?
« Last Edit: November 23, 2020, 04:51:05 AM by Win,odd Dhamnekar »
Any science consists of the following process.
1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

#### 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.