Class Notes (837,539)
Canada (510,303)
Chemistry (543)
CHEM 1AA3 (67)

Determination of an Equilibrium Constant for a Chemical Reaction Lab 4.docx

4 Pages
Unlock Document

Adam Hitchcock

Determination of an Equilibrium Constant for a Chemical Reaction Purpose: To find the equilibrium constant of the (Kc) for the reaction: Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+ (aq) through several different concentrations of reactants, using spectrophotometric analysis. Procedure: Please refer to the CHEM 1A03/1E03/1AA3 Lab Manual 2013, pages 43-47, for a detailed procedure. Part A: Observations: Table 1: Values of Absorbance and Concentration for Different Volumes of Potassium Thiocyanate and 0.200 mol/LFerric Nitrate Test Tube Volume of Absorbance (A) Concentration Number KSCN (mL) of FeSCN2+ (mol/L) 1 1 0.182 4.00 x 10-6 2 2 0.368 8.00 x 10-6 3 3 0.669 1.20 x 10-5 Qualitative Observations: FeSCN2+(aq) and KSCN (aq) were both clear, colourless liquids initially. Solution turned to an orange and brown colour when the ferric nitrate was added to the potassium thiocyanate. Sample Calculations: Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+ (aq) n(SCN-) = cv n(SCN-) = (.002 Mol/L) (.001(L)) n(SCN-) = .000002 Mol Due to 1:1 Mol ratio, n(FeSCN2+) = .000002Mol [FeSCN2+] = n/v [FeSCN2+] = (.000002(Mol)/.05(L)) [FeSCN2+] = .00004Mol/L Graph 1: Absorbance vs. Concentration of FeSCN2+ Constant The above graph shows the proportional relationship between the concentration and the absrbance of FeSCN2+. Using a line of best fit, a proportional constant of 5223 is determined to be the slope. Part B: Sample Calculations: Initial [F3+(aq] in Test Tube #1: C V 1= C 2V2 (0.00200 mol/L)(0.0050L) = (C 2)(0.010L) C = 0.00100 mol/L Initial [SCN-(aq] in Test Tube #1: C1V 1= C2V 2 (0.00200 mol/L)(0.0010L) = (C 2)(0.010L) C =0.000200 mol/L Equilibrium [FeSCN 2+(aq] in Test Tube #1: A = Absorbance = 0.243A y = proportionality constant = 5223 A/mol/L A = y[FeSCN 2+] [FeSCN 2]= A/y 0.243A/5223 A/mol/L = 4.65x10 -5mol/L Observations: ICE Table for Test Tube #1: Reaction: Fe 3+(aq+ SCN -(aq⇌ FeSCN 2+(aq) Initial 1.00x10^-3mol/L 2.00x10^-3mol/L ------- Concentration (I) Change in -x -x +x Concentration (C) Equilibrium 1.00x10^-3 – x 2.00x10^-4 – x 4.65x10^-5 Concentration (E) x=4.65x10^-5 mol/L Equilibrium concentration for Fe^3+ (aq) = (1.00x10^-3 mol/L) - (4.65x10^-5 mol/L) = 9.54x10^-4 mol/L Equilibrium concentration for SCN^-1 (aq) = (2.00x10^-4 mol/L) – (4.65x10^-5 mol/L) = 1.54x10^- 4 mol/L Equilibrium concentration for FeSCN^2+(aq) = 4.65x10^-5 mol/L Equilibrium Constant for Test Tube #1: Kc=Equilibrium Constant Kc=[Products]/[Reactants] Kc=[FeSCN^2+ (aq)]/[[Fe^3+ (aq)]*[SCN^-1 (aq)]] Kc=[4.65x10^-5]/[[1.54x10^-4]*[9.54x10^-4]] Kc=317 Observations: Table 3: Initial and Equilibrium Concentrations for Fe 3+(aq)and SCN -(aq, Equilibrium Concentrations for FeSCN 2+(aq, and Values for Equilibrium Constant For Each Different Solution Test Tube Initial [F3] Initial Equilibrium Equilibrium Equilibrium Kc Value Number (mol/L) [SCN -(a]) [FeSCN 2+(a]) [Fe3+(a] [SCN -(a]) (mol/L) (mol/L) (mol/L) (mol/L) 1 1.00x10-3 2.00x10-
More Less

Related notes for CHEM 1AA3

Log In


Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.