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Determining Eq Constant Essay

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Below is an essay on "Determining Eq Constant" from Anti Essays, your source for research papers, essays, and term paper examples.

October, 2011

TITLE: The Equilibrium Constant

PURPOSE: In this experiment I calculated the equilibrium constant for the reaction shown below under different conditions to determine if the equilibrium constant is really constant.
Fe3+ (aq) + SCN– (aq)      FeSCN2+   (aq)  

PROCEDURE:
This lab, The Equilibrium Constant,   was provided by the teacher, who got it from Flinn ChemTopic Labs Volume 15: Equilibrium, Irene Cesa, Senior Editor, Flinn Scientific, Inc., Batavia, IL, © 2003, pp. 29-44.

In Part A of the lab, I prepared six solutions – including a reference solution (known concentration) and a series of five test solutions (varied in concentration of SCN– (aq)). In Part B of the lab, I measured the absorbance of the solutions using a Spec 20. After collecting the data, I calculated the unknown equilibrium concentrations of the FeSCN2+   (aq)   which I then used to determine the Keq.

DATA COLLECTION   combined with
DATA PROCESSING & PRESENTATION:


Table 1:   Preparing the Solutions

Reagents (Solutions)
0.0020 M   Fe3+ 0.0020 M SCN– Distilled   H2O
+/-– .0005 M*
(25% error ) +/-– .0005 M*
(25% error )
* uncertainty due to solution prep
Volume of Solution Used
+/-– 0.005 mL**
(0.1% error) +/-– 0.005 mL**
(0.1% error) +/-– 0.005 mL**
(0.1% error)
Test Solution # 1 5.00 1.00 4.00
Test Solution # 2 5.00 2.00 3.00
Test Solution # 3 5.00 3.00 2.00
Test Solution # 4 5.00 4.00 1.00
Test Solution # 5 5.00 5.00 0.00
**uncertainty of graduated pipette / bulb system

NOTE: It is clear that the uncertainty of the solution prep is very high… and will far exceed
any uncertainty due to measuring with the pipette.



Table 2:   Initial Concentration and Measured Absorbance of Solutions

Reagents
Temperature of Solutions
(assumed to be room temperature):   Absorbance

+/- 0.01
Solution  [ Fe3+ ]
in M [ SCN– ]
in M
Uncertainty  +/- 25 %         +/- 25 % +/- 2.5 %   (avg)
Test Solution # 1 1.0 x 10–3...

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