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Kc

Finding the Equilibrium Constant, Kc Tulsa Community College

Kc

PURPOSE: To determine the equilibrium constant for Fe3+ + SCN- ⇌ FeSCN2+ utilizing Beer’s law. PROCEDURE: Calibrate the spectrometer. Mix 18 mL of 0.2 M Fe(NO 3)3 with 2 ml of 0.002 M KSCN and let it sit for 10 minutes. Fill the cuvette with the standard solution and collect the wavelength versus absorbance data. Record the wavelength of maximum absorbance. Select Absorbance versus Concentration and calculate the concentration of FeSCN 2+. Record the absorbance and rinse out the cuvette. [FeSCN2+] in standard solution Absorbance of standard solution 2mL * 0.002 M= 20 mL * x = 0.0004/20 = 0.76 2*10-4 Set up these solutions and let them each sit for 10 minutes. Measure the absorbance of each. Rinse out the test tubes and neutralize the waste beaker. Trial Number 1 2 3 4 1. 2. 3. 4.

0.0020 M 0.0020 M High-Purity Absorbance Fe(NO3)3, mL KSCN, mL Water, mL 5 (0.0001 M) 5 (0.0001 M) 0 0.5 5 (0.0001 M) 4 1 0.4 5 (0.0001 M) 3 2 0.3 5 (0.0001 M) 2 3 0.2 (2*10-4)/(0.8) = (x)/(0.5) = 1.25*10-4 = x (2*10-4)/(0.8) = (x)/(0.4) = 1*10-4 = x (2*10-4)/(0.8) = (x)/(0.3) = 7.5*10-5 = x (2*10-4)/(0.8) = (x)/(0.2) = 5*10-5 = x Calculate the Equilibrium Constants for your trials and record them in the chart below.

Trial 1 0.000125/0.00089 2 = 0.140

Average KC 0.140 + 0.112 + 0.084 + 0.056 = 0.392/4 = 0.098 Use a R.I.C.E. chart to calculate the equilibrium concentrations of FeSCN 2+, Fe3+ and SCN-. Calculate the percent of SCN- that reacted to form FeSCN 2+. Reaction Initial Change Equilibrium

Trial 2 0.0001/0.00089 2 = 0.112

Fe3+ 0.18 M -x 0.18-x

Trial 3 0.000075/0.00089 2 = 0.084

+ SCN 2*10-4 M -x (2*10-4)-x

Trial 4 0.00005/0.00089 2 = 0.056

⇌ FeSCN2+ 0M +x x

Kc

K = 0.14 = x / ((0.18-x)((2*10-4)-x) 0.14 = x / (3.6*10-5 – 0.1802x + x2) 5.04*10-6 – 0.025228x + x2 = x X2 – 1.025228x + x2 = 0 (1.025228 ± sqrt (-1.025228)2 – 4(1)(5.04*10-6))/2 (1.025228 ± sqrt 1.05 – 2.06*10-5)/2 (1.025228 ± sqrt 1.05107)/2 (1.025228 + 1.025218)/2

(1.025228 - 1.025218)/2

2.05/2

9.823*10-6/2

1.025 ≠ x

4.916*10-6 = x

% of SCN- that reacted in the standard solution = 4.916*10-6/0.002 * 100 = 2.458%

2.458%

CHEMICALS USED: •

Fe3+

•

SCN-

•

FESCN2+

•

HNO3

•

KSCN

CONCLUSIONS: Our assumption was that the % of SCN- that reacted in the standard solution would be 100%. It was actually 2.458%....