E17: Chemical Equilibrium

Transcription

1 E17: Chemical Equilibrium 77 SCN" is converted to FeSCN++. Under these conditions, you can assume that the final concentration of FeSCN++ is equal to the initial concentration of SCN-. Chemical Equilibrium CHEMICAL REACTIONS occuf SO as to approach a state of chemical equilibrium. The equilibrium state can be characterized by specifying its equimbrium constant, i.e., by indicating the numerical value of the mass-action expression (text, Sec. 13.2). In this experiment you will determine the value of the equilibrium constant for the reaction Fe+3 + SCN-" ^ FeSCN+ + PROCEDURE Thoroughly clean your six small test tubes, rinse with distilled water, and let drain. To each of these test tubes add 5 ml. of M NaSCN. To the first test tube add 5 ml. of 0.20 M Fe(N08)3. This tube will serve as standard. For the other test tubes proceed as follows: Add 10 ml. of 0.20 M Fe(N03)8 to your graduated cylinder, fill to 25 ml. with distilled water, and stir thoroughly to mix. Pour 5 ml. of the resulting diluted solution for which the equilibrium condition is [FeSCN++] [Fe«][SCN- K In order to find the value of K, it is necessary to determine the concentration of each of the species Fe+^ SCN^, and FeSCN++ in the system at equihbrium. This will be done colorimetrically, taking advantage of the fact that FeSCN++ is the only highly colored species in the solution. The color intensity of a solution depends on the concentration of the colored species and on the depth of solution viewed. Thus, e.g., 2 cm. of a 0.1 M solution of a colored species appears to have the same color intensity as 1 cm. of a 0.2 M solution of a colored species. Consequently, if the depths of two solutions of unequal concentrations are chosen so that the solutions appear equally colored, then the ratio of concentrations is simply the inverse of the ratio of the two depths. It should be noted that this procedure permits only a comparison between concentrations. It does not give an absolute value of either one of the concentrations. To know absolute values, it is necessary to compare with a standard of known concentration. For color determination of FeSCN + + concentration, you need to have a standard solution in which the concentration of FeSCN++ is known. Such a solution can be prepared by starting with a small known concentration of SCN-^ and adding such a large excess of Fe+s that essentially all the 76 Fig. E20.1 (0.080 M Fe+^) into test tube 2. Discard all but 10 ml. of the diluted solution that is in the graduated cylinder, refill with distilled water to 25 ml, and stir thoroughly. Add 5 ml. of the resulting solution (0.032 M Fe+^) to test tube 3. Discard all but 10 ml. of the solution in the cyhnder, and agam dilute to 25 ml. Continue this procedure until you have added to each successive test tube 5 ml. of progressively more dilute Fe+^ solution. Now the problem is to determine the concentration of FeSCN++ m each test tube relative to the standard in test tube 1. Compare the color intensity in test tube 1 with that in each of the other test tubes. Take the two tubes to be compared, hold them side by side, and wrap a strip of paper around both. Look down through the solutions towards a white paper on your desk, as shown in Fig. E20.L If color intensities app<«ar identical, record this fact. If not, take test lube I and pour out, some of the standard

2 Experim^ents into a clean, dry beaker (you may need to pour some back!) until the color intensities appear identical. Measure the heights of solutions in the two tubes being compared. Do this comparison for all five tubes. DATA Test tube 2 Test tube 3 Test tube 4 Test tube 5 Test tube 6 Height of liquid Comparison height of liquid standard E17: Chemical Equilibrium QUESTIONS 20.1 Using the best value of K, calculate the concentration of SCN m test tube 1 at equilibrium Why are the values of K determined by test tubes 2 and 6 not so reuable as the others? 20.3 As you have noticed, a small change in height does not appreciably affect the color intensity. Assume that, in test tube 4, your height measured for the standard was 5% too great. How far off would this make K? 79 RESULTS Test tube 1 Init. cones. IFe+3] [SCN-] Equilib. cones. [FeSCN++l [Fe+«] [SCNi K Test tube 2 Test tube 3 Test tube 4 Test tube 5 Test tube 6 In calculating initial concentrations, assume that Fe(N03)3 and NaSCN are each completely dissociated. Remember also that mixing two solutions dilutes both of them. In calculating equifibrium concentrations, assume in test tube 1 that all the initial SCN" has been converted to FeSCN++. For the other test tubes, calculate FeSCN++ from the ratio of heights in the color comparison. Equilibrium concentrations of Fe+» and SCN" are obtained by subtracting FeSCN++ formed from initial Fe+8 and SCN". For each of test tubes 2 to 6 calculate the value of K. these values is most reliable. Decide which of

3 Chemistrv 40 PGL FeSCN^^ Equilibrium Lab Prelab Calculations Record your answers to all questions in the table below. Test Tube Initial Concentrations [Fen [SCN"] (mol/l) (mol/l) Equilibrium Concentrations [FeSCN^ (mol/l) [Fe^ (mol/l) [SCN"] (mol/l) Kc Write the balanced dissolving ionic equations (DIEs) for both reactants as well as the balanced overall ionic (OIE) and net ionic equations (HIE). 2. Calculate all appropriate initial and equilibrium concentrations for the reactants and products of the standard solution in test tube 1. Use an ICE table to calculate the concentration of FeSCN^^. 3. Calculate the concentration of iron (III) ions in test tube 2 after the first dilution. 4. Calculate the equilibrium concentrations of all chemicals in test tube 2 and the value of the equilibriimi constant given the following data: Test Tube 2 Height of Liquid (cm) 6.25 Comparison Height of Liquid Standard (cm) 5.94 /lo

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5 KP & Kc November 26, 2007 that each gas molecule exerts on the walls of its An equilibrium constant inwolwing the^partlai ' p i c S d P i S ^ Of y3s S', C3II Q "-pj is OfteH ijsso In The same rylas for solwing aqyiiibriuin problems with concentrations apply for gases when pressures are Kp

6 Kp & Kc November 26, 2007 Write the equilibriym constant empression'using nitrosyl chloride giwen the following equiiibrlum reaction. ' ' 2 NO(g)-f Cl2(g) 2N0CI (g) ^^^ ^^m Kp Expression

7 Kp & Kc November 26, 2007 R all th3t w%c r pres nts th quilibriuin constant when molar eoncentrations are inwoiwed. Als0 recall the ideal gas sqyatfon PV = nrt A usefyl equation can be deweloped between Kp andke Rearrange the ideal gas law equation to solve for molar concentration. This equation, can be further deweloped ysing a concrete example. ': Kp&Kc

8 Kp & Kc November 26, 2007 Conwert Kp to Kc for the synthesis reaction of gaseous nitrosyl chloride from nitrogen monoxide and chlorine gases.. 2NO(g) + Cl2(g) 2NOCy Nov 25-10:31 PM

9 Kp & Kc November 26, 2007 Ill ^^il^^'ioij An Kp Kc (RT) where An is the difference between the sum, of the rnoies of product and the sum of the moles of Answer(L. /^,Z7, f^.zf, t^.^f j>.liftinzu Nov 25-10:39 PM