How Does Le Chatelier's Principle Apply to Fe3+ and SCN- Concentrations?

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Discussion Overview

The discussion centers on the application of Le Châtelier's Principle to the concentrations of Fe3+ and SCN- in the context of the formation of the FeSCN2+ complex. Participants explore the implications of concentration ratios and equilibrium constants in a chemical reaction involving these ions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asserts that a standard solution of FeSCN2+ can be prepared by using a low concentration of SCN- and a high concentration of Fe3+, suggesting that this drives the reaction to completion rather than equilibrium.
  • Another participant argues that while the equilibrium still applies, the excess Fe3+ leads to the consumption of nearly all SCN-, resulting in a final concentration of FeSCN2+ that closely matches the initial SCN- concentration.
  • Several participants express difficulty in visualizing the calculations related to the equilibrium constant Q and the concentrations involved, particularly in how to handle the large disparity between the concentrations of SCN- and Fe3+.
  • A participant provides a mathematical approach to derive the relationship between the concentrations, indicating that if the initial concentration of Fe3+ is significantly greater than that of SCN-, certain terms can be neglected in the calculations.
  • Another participant seeks clarification on the algebraic steps taken in the mathematical derivation, indicating a struggle with the underlying math but later expresses confidence in finding the answers independently.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and disagreement regarding the application of Le Châtelier's Principle and the interpretation of the equilibrium constant Q. Some participants support the idea that the reaction approaches completion, while others emphasize the relevance of equilibrium considerations.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the concentrations of SCN- and Fe3+, as well as the implications of neglecting certain terms in the calculations. The discussion reflects varying levels of comfort with the mathematical aspects involved.

Who May Find This Useful

This discussion may be useful for students or individuals interested in chemical equilibrium, particularly in the context of complex ion formation and the mathematical modeling of concentration relationships in reactions.

alingy1
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"The Beer-Lambert relationship must be solved using a FeSCN2+ standard solution. This standard solution can be prepared by mixing a SCN- solution that has a very low concentration with a Fe3+ solution that has avery high concentration. In this situation, the reaction is driven to completion instead of equilibrium, according to Le Châtelier [...]. "

I do not see how Le Châtelier applies here. Q=[FeSCN2+]/([SCN-][Fe3+]) How does having a low and a very high concentration help...
 
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The equilibrium still applies, but, since the SCN is mixed with an overabundance of Fe3, virtually all the SCN gets consumed. So the number of moles of FeSCN2 in the final solution is virtually identical to the number of moles of SCN that was present in the original SCN solution.

Chet
 
I am not able to picture myself this with calculations. In my head, when we calculate Q, the bottom part becomes a normal value (product of big value with small value)...
 
alingy1 said:
I am not able to picture myself this with calculations. In my head, when we calculate Q, the bottom part becomes a normal value (product of big value with small value)...
Q is not calculated. Q is the equilibrium constant.

Let SCN0 be the initial concentration of SCN in the mixture and Fe30 be the initial concentration of Fe3 in the mixture. Let x be the number of moles per liter of SCN that react with Fe3 to produce x moles /liter of FeSCN2. Then:

\frac{x}{(SCN_0-x)(Fe3_0-x)}=Q
Since Fe30 is going to be much greater than SCN_0, we can neglect x compared to Fe30. So the above equation becomes:
\frac{x}{(SCN_0-x)}=Q(Fe3_0)
So,
\frac{x}{(SCN_0)}=\frac{Q(Fe3_0)}{Q(Fe3_0)+1}
So, if Fe3_0 is high enough such that Q(Fe30)>>1,

x -->SCN0

This is the final concentration of FeSCN2 in the solution.
 
Wow, this is really clear! Thank you. One small question though. What is the algebra behind your third latex code and your second latex code. I do not understand how you got rid of the -x and divided the left side by Q(Fe30)+1.
*gulp* Sorry, my math is really weak. I admit it. EDIT: IT'S FINE. I COULD FIND IT! :)
 

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