# Stahl, Equilibrium COnstant of Reaction Post Lab - Ashlynn...

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Ashlynn Stahl 11/22/2019 Chem Lab I Equilibrium Constant of a Reaction Post Lab 1) Looking at you data, what was your equilibrium constant (Kc)? Our equilibrium constant for test tube #1 as 84.125, #2 was 73.917, #3 was 66.825, and #4 was 64.33. Calculations: #1. Kc = 3.365 x 10 -5 / (.001)(.0004) = 84.125 #2. Kc = 4.435 xx 10 -5 / (.001)(.0006) = 73.917 #3. Kc = 5.346 x 10 -5 / (.001)(.0008) = 66.825 #4. Kc = 6.433 x 10 -5 / (.001)(.001) = 64.33 2)Explain the concept of calibration using a graph and description (Can be in terms of Beer’s Law or in general). A calibration curve or standard curve on a graph is used to determine the concentration of a substance in a sample by comparing the unknown sample to a constant set of samples with known concentrations. 4)For the reaction N H 3 + H + + N H + 4 +at 20.0∘ C, the equilibrium concentrations were as follows: NH 3 =2.0 x 10 -4 M, H + =2.0 x 10 -4 M, and NH 4 + =18.0M. Calculate the equilibrium constant for the reaction. What does this mean in terms of which 'side" of the equation

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Unformatted text preview: is favored? Kc= [ 18.0 ] [ 2.0 × 10 − 4 ] × [ 2.0 × 10 − 4 ] ¿ [ 18.0 ] [ 4 × 10 − 8 ] = 4.5 × 10 8 This means that the right side of the equation is favored because the equilibrium constant is much larger than one. 5)What does a very small equilibrium constant mean? (2.5X10-10 , as an example) A very small equilibrium constant means that the reactants or, left side of the equation is being favored. Data Table Test Tube #1 #2 #3 #4 Standard #5 Absorbance 0.192 0.253 0.305 0.367 1.141 Test Tube #1 #2 #3 #4 [Fe3+]i .001 .001 .001 .001 [SCN-]i 0.0004 0.0006 0.0008 0.001 [FeSCN2+]eq 3.365 x 10-5 4.435 x 10-5 5.346 x 10-5 6.433 x 10-5 [Fe3+]eq 9.664 x 10-4 9.557 x 10-4 9.466 x 10-4 9.357 x 10-4 [SCN-]eq 3.664 x 10-4 5.557 x 10-4 2.654 x 10-5 9.357 x 10-4 Kc 84.125 73.917 88.825 64.33 Average Kc value 72.299...
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