1. Hydrogen (H 2 ) and iodine (I 2 ) react to produce hydroiodic acid (HI).Hydrogen (H 2 ) and iodine (I 2 ) react to produce hydroiodic acid (HI). Forward Reaction: H 2(g) + I 2(g) → 2HI (g) However, as soon as some HI is produced it begins to dissociate back to H 2 and I 2.However, as soon as some HI is produced it begins to dissociate back to H 2 and I 2. Reverse Reaction: 2HI (g) → H 2(g) + I 2(g) These 2 opposing reactions, 1 forward and 1 reverse, go on simultaneously. Ultimately, the rate of the forward reaction will equal the rate of the reverse reaction and the system is in dynamic equilibrium.These 2 opposing reactions, 1 forward and 1 reverse, go on simultaneously. Ultimately, the rate of the forward reaction will equal the rate of the reverse reaction and the system is in dynamic equilibrium. H 2(g) + I 2(g) = 2HI (g) CH 104: DETERMINATION OF AN EQUILIBRIUM CONSTANT EQUILIBRIUM

2. How many moles of H 2(g), I 2(g), and HI (g) are initially used in experiment 1?How many moles of H 2(g), I 2(g), and HI (g) are initially used in experiment 1? H 2(g) = 0.0015 moles, I 2(g) = 0.0015 moles, and HI (g) = 0 moles.H 2(g) = 0.0015 moles, I 2(g) = 0.0015 moles, and HI (g) = 0 moles. How many moles of H 2(g), I 2(g), and HI (g) are initially used in experiment 2?How many moles of H 2(g), I 2(g), and HI (g) are initially used in experiment 2? H 2(g) = 0 moles, I 2(g) = 0 moles, and HI (g) = 0.0015 moles.H 2(g) = 0 moles, I 2(g) = 0 moles, and HI (g) = 0.0015 moles. How many moles of H 2(g), I 2(g), and HI (g) are initially used in experiment 3?How many moles of H 2(g), I 2(g), and HI (g) are initially used in experiment 3? H 2(g) = 0.0015 moles, I 2(g) = 0.0015 moles, and HI (g) = 0.0015 moles.H 2(g) = 0.0015 moles, I 2(g) = 0.0015 moles, and HI (g) = 0.0015 moles. In summary, all 3 experiments reach a dynamic equilibrium regardless of the initial concentrations of H 2(g), I 2(g), and HI (g).In summary, all 3 experiments reach a dynamic equilibrium regardless of the initial concentrations of H 2(g), I 2(g), and HI (g).EQUILIBRIUM

3. H 2(g) + I 2(g) = 2HI (g) At equilibrium the rate of the forward reaction equals the rate of the reverse reaction. H 2(g) and I 2(g) are still producing HI (g). And HI (g) is still producing H 2(g) and I 2(g). However, the final concentrations of H 2(g), I 2(g), and HI (g) are constant.At equilibrium the rate of the forward reaction equals the rate of the reverse reaction. H 2(g) and I 2(g) are still producing HI (g). And HI (g) is still producing H 2(g) and I 2(g). However, the final concentrations of H 2(g), I 2(g), and HI (g) are constant.EQUILIBRIUM

4. If the rate of people going to the first floor equals the rate of people going to the second floor, the number of people on each floor remains constant and the 2 groups are at dynamic equilibrium.If the rate of people going to the first floor equals the rate of people going to the second floor, the number of people on each floor remains constant and the 2 groups are at dynamic equilibrium.EQUILIBRIUM

5. For the general reaction:For the general reaction: aA + bB = cC + dD Equilibrium Constant = K c =Equilibrium Constant = K c = WhereWhere [A], [B], [C], and [D] are the concentrations of reactants and products in moles per liter.[A], [B], [C], and [D] are the concentrations of reactants and products in moles per liter. a, b, c, and d are the stoichiometric coefficients from the balance reaction.a, b, c, and d are the stoichiometric coefficients from the balance reaction. K c is the equilibrium constant.K c is the equilibrium constant. The value of K c depends on the particular reaction and on the temperature.The value of K c depends on the particular reaction and on the temperature.EQUILIBRIUM [C] c [D] d Products [A] a [B] b Reactants

6. In today’s experiment you will measure the equilibrium constant (K c ) for the following reaction.In today’s experiment you will measure the equilibrium constant (K c ) for the following reaction. Fe 3+ + SCN – = FeSCN 2+ K c = K c = Solutions of known iron(III) nitrate (Fe(NO 3 ) 3 ) and potassium thiocyanate (KSCN) will be mixed in prescribed proportions. The concentration of FeSCN 2+ at equilibrium will be measured spectrophotometrically. These data will be used to calculate K c.Solutions of known iron(III) nitrate (Fe(NO 3 ) 3 ) and potassium thiocyanate (KSCN) will be mixed in prescribed proportions. The concentration of FeSCN 2+ at equilibrium will be measured spectrophotometrically. These data will be used to calculate K c. MEASURING AN EQUILIBRIUM CONSTANT [FeSCN 2+ ] [FeSCN 2+ ] [Fe 3+ ][SCN – ]

7. If 10.00 mL of 0.00200 M Fe(NO 3 ) 3, 5.00 mL of 0.00200 M KSCN, and 5.00 mL of distilled water are mixed together, what is the initial concentration of Fe 3+ ?If 10.00 mL of 0.00200 M Fe(NO 3 ) 3, 5.00 mL of 0.00200 M KSCN, and 5.00 mL of distilled water are mixed together, what is the initial concentration of Fe 3+ ? [Fe 3+ ] initial = 0.00100 M[Fe 3+ ] initial = 0.00100 M What is the initial concentration of SCN – ?What is the initial concentration of SCN – ? [SCN – ] initial = 0.000500 M[SCN – ] initial = 0.000500 M What is the initial concentration of FeSCN 2+ ?What is the initial concentration of FeSCN 2+ ? [FeSCN 2+ ] initial = 0 M[FeSCN 2+ ] initial = 0 M If the concentration of FeSCN 2+ at equilibrium is 0.000120 M, what is the equilibrium concentration of Fe 3+ ?If the concentration of FeSCN 2+ at equilibrium is 0.000120 M, what is the equilibrium concentration of Fe 3+ ? Fe 3+ + SCN – = FeSCN 2+ [Fe 3+ ] equilibrium = 0.00100 M – 0.000120 M = 0.00088 M[Fe 3+ ] equilibrium = 0.00100 M – 0.000120 M = 0.00088 M What is the equilibrium concentration of SCN – ?What is the equilibrium concentration of SCN – ? [SCN – ] equilibrium = 0.000500 M – 0.000120 M = 0.000380 M[SCN – ] equilibrium = 0.000500 M – 0.000120 M = 0.000380 M MEASURING AN EQUILIBRIUM CONSTANT

8. An ICE (Initial, Change, Equilibrium) table is used to solve these types of problems.An ICE (Initial, Change, Equilibrium) table is used to solve these types of problems. What is the equilibrium constant?What is the equilibrium constant? K c = K c = K c = = 3.6x10 2 MEASURING AN EQUILIBRIUM CONSTANT Reaction: Fe 3+ + SCN – = FeSCN 2+ Initialconcentrations: 0.00100 M 0.000500 M 0 M Change: – 0.000120 M + 0.000120 M Equilibriumconcentrations: 0.00088 M 0.000380 M 0.000120 M [FeSCN 2+ ] [FeSCN 2+ ] [Fe 3+ ][SCN – ] [0.000120] [0.000120][0.00088][0.000380]

9. SCHEMATIC OF A SPECTROPHOTOMETER The most common light source for the visible region spectrophotometry is a tungsten filament incandescent lamp. A tungsten lamp emits useful light from approximately 325 nm to 3,000 nm.The most common light source for the visible region spectrophotometry is a tungsten filament incandescent lamp. A tungsten lamp emits useful light from approximately 325 nm to 3,000 nm. A monochromator uses a prism or a diffraction grating to separate polychromatic (many wavelengths) light into monochromatic (single wavelength) light.A monochromator uses a prism or a diffraction grating to separate polychromatic (many wavelengths) light into monochromatic (single wavelength) light. A cell or cuvette is used to hold the sample during analysis.A cell or cuvette is used to hold the sample during analysis. The detector uses a phototube or a photomultiplier tube to convert light into an electrical signal that is sent to a recorder or computer.The detector uses a phototube or a photomultiplier tube to convert light into an electrical signal that is sent to a recorder or computer.

10. THE SPECTRONIC 20D SPECTROPHOTOMETER The controls. Loading a sample. THE HACH DR2010 SPECTROPHOTOMETER

11. In today’s experiment you will use the Beer-Bouguer-Lambert law, more commonly called Beer’s law, to measure the equilibrium concentration of FeSCN 2+.In today’s experiment you will use the Beer-Bouguer-Lambert law, more commonly called Beer’s law, to measure the equilibrium concentration of FeSCN 2+. A sample in a cell or cuvette during spectrophotometric analysis. P o = the power of monochromaticlight entering the sample.P o = the power of monochromatic light entering the sample. P = the power of monochromaticlight leaving the sample.P = the power of monochromatic light leaving the sample. a = the absorptivity constant, which depends on the wavelength and the nature of the absorbing compound.a = the absorptivity constant, which depends on the wavelength and the nature of the absorbing compound. b = the path length through the absorbing compound.b = the path length through the absorbing compound. c = the concentration of absorbing compound in the cuvette.c = the concentration of absorbing compound in the cuvette. ABSORBANCE AND CONCENTRATION

12. SOURCES Beck, J. 2006. Unit 3 Spectrophotometry. Available: http://iws.ccccd.edu/jbeck/Spectrophotometryweb/Page.ht ml [accessed 2 October 2006].Beck, J. 2006. Unit 3 Spectrophotometry. Available: http://iws.ccccd.edu/jbeck/Spectrophotometryweb/Page.ht ml [accessed 2 October 2006]. http://iws.ccccd.edu/jbeck/Spectrophotometryweb/Page.ht ml http://iws.ccccd.edu/jbeck/Spectrophotometryweb/Page.ht ml Christian, G.D. 1986. Analytical Chemistry, 3rd ed. New York, NY: John Wiley & Sons, Inc.Christian, G.D. 1986. Analytical Chemistry, 3rd ed. New York, NY: John Wiley & Sons, Inc. Harris, D.C. 1999. Quantitative Chemical Analysis, 5th ed. New York, NY: W.H. Freeman Company.Harris, D.C. 1999. Quantitative Chemical Analysis, 5th ed. New York, NY: W.H. Freeman Company. McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall.McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall. Petrucci, R.H. 1985. General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.Petrucci, R.H. 1985. General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.