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Experiment 21 Rates of Chemical Reactions, II. A Clock Reaction We need to determine the relative rate constant k' and the orders m, n, and p in such a way as to be consistent with the data in the table. The solution to this problem is quite simple, once you make a few observations on the reaction mixtures. Each mixture (2 to 4) differs from Reaction Mixture 1 in the concentration of only one species (see table). This means that for any pair of mixtures that includes Reaction Mixture 1, there is only one concentration that changes. From the ratio of the relative rates for such a pair of mixtures we can find the order for the reactant whose concentration was changed. Proceed as follows. Write Equation 5 below for Reaction Mixtures 1 and 2, substituting the relative rates and the concentrations of I-, BrO3-, and H+ ions from the table you have just completed. Relative Rate 1 = 6.94 = k'(0.0020)^m(0.0016)^n(0.020)^p Relative Rate 2 = 12.821 = k'(0.0040)^m(0.0016)^n(0.020)^p Divide the first equation by the second, noting that nearly all the terms cancel out. The result is simply Relative Rate 1 [0.0020]^m[0.0016]^n[0.020]^p 6.94 ---------------- = ---------------------------------------------------- = --------- = 0.541 = .5^m Relative Rate 2 [0.0040]^m[0.0016]^n[0.020]^p 12.821 If you have done this properly, you will have an equation involving only m as an unknown. Solve this equation for m, the order of the reaction with respect to I- ion. m = 0.87 (nearest integer) m log 0.5 = log 0.541 -0.26680 ----------------- = 0.87 -0.30103 Applying the same approach to Reaction Mixtures 1 and 3, find the value of n, the order of the reaction with respect to BrO3- ion. Relative Rate 1 = 6.94 = k'(0.0020)^m(0.0016)^n(0.020)^p Relative Rate 3 = 9.615 = k'(0.0020)^m(0.0016)^n(0.020)^p Dividing one equation by the other: 6.94 -------- = n = 9.615 Now that you have the idea, apply the method once again, this time to Reaction Mixtures 1 and 4, and find p, the order with respect to H+ ion. Relative Rate 4 = k'( )^m( )^n( )^p Dividing the equation for Relative Rate 1 by that for Relative Rate 4, we get p = Having found m, n, and p (nearest integers), the relative rate constant, k', can be calculated by substitution of m, n, p, and the known rates and reactant concentrations into Equation 5. Evaluate k' for Reaction Mixtures 1 to 4. Reaction 1 2 3 4 k' k' ave Standard deviation in k' (See Appendix VIII) Why should k' have nearly the same value for each of the above reactions? Using k' ave in Equation 5, predict the relative rate and time, tpred, for Reaction Mixture 5. Use the concentrations in the table. Relative rate pred tpred tobs

          Experiment 21 Rates of Chemical Reactions, II. A Clock Reaction We need to determine the relative rate constant k' and the orders m, n, and p in such a way as to be consistent with the data in the table. The solution to this problem is quite simple, once you make a few observations on the reaction mixtures. Each mixture (2 to 4) differs from Reaction Mixture 1 in the concentration of only one species (see table). This means that for any pair of mixtures that includes Reaction Mixture 1, there is only one concentration that changes. From the ratio of the relative rates for such a pair of mixtures we can find the order for the reactant whose concentration was changed. Proceed as follows. Write Equation 5 below for Reaction Mixtures 1 and 2, substituting the relative rates and the concentrations of I-, BrO3-, and H+ ions from the table you have just completed. Relative Rate 1 = 6.94 = k'(0.0020)^m(0.0016)^n(0.020)^p Relative Rate 2 = 12.821 = k'(0.0040)^m(0.0016)^n(0.020)^p Divide the first equation by the second, noting that nearly all the terms cancel out. The result is simply Relative Rate 1 [0.0020]^m[0.0016]^n[0.020]^p 6.94 ---------------- = ---------------------------------------------------- = --------- = 0.541 = .5^m Relative Rate 2 [0.0040]^m[0.0016]^n[0.020]^p 12.821 If you have done this properly, you will have an equation involving only m as an unknown. Solve this equation for m, the order of the reaction with respect to I- ion. m = 0.87 (nearest integer) m log 0.5 = log 0.541 -0.26680 ----------------- = 0.87 -0.30103 Applying the same approach to Reaction Mixtures 1 and 3, find the value of n, the order of the reaction with respect to BrO3- ion. Relative Rate 1 = 6.94 = k'(0.0020)^m(0.0016)^n(0.020)^p Relative Rate 3 = 9.615 = k'(0.0020)^m(0.0016)^n(0.020)^p Dividing one equation by the other: 6.94 -------- = n = 9.615 Now that you have the idea, apply the method once again, this time to Reaction Mixtures 1 and 4, and find p, the order with respect to H+ ion. Relative Rate 4 = k'( )^m( )^n( )^p Dividing the equation for Relative Rate 1 by that for Relative Rate 4, we get p = Having found m, n, and p (nearest integers), the relative rate constant, k', can be calculated by substitution of m, n, p, and the known rates and reactant concentrations into Equation 5. Evaluate k' for Reaction Mixtures 1 to 4. Reaction 1 2 3 4 k' k' ave Standard deviation in k' (See Appendix VIII) Why should k' have nearly the same value for each of the above reactions? Using k' ave in Equation 5, predict the relative rate and time, tpred, for Reaction Mixture 5. Use the concentrations in the table. Relative rate pred tpred tobs

Experiment 21 Rates of Chemical Reactions, II. A Clock Reaction We need to determine the relative rate constant k' and the orders m, n, and p in such a way as to be consistent with the data in the table. The solution to this problem is quite simple, once you make a few observations on the reaction mixtures. Each mixture (2 to 4) differs from Reaction Mixture 1 in the concentration of only one species (see table). This means that for any pair of mixtures that includes Reaction Mixture 1, there is only one concentration that changes. From the ratio of the relative rates for such a pair of mixtures we can find the order for the reactant whose concentration was changed. Proceed as follows. Write Equation 5 below for Reaction Mixtures 1 and 2, substituting the relative rates and the concentrations of I-, BrO3-, and H+ ions from the table you have just completed. Relative Rate 1 = 6.94 = k'(0.0020)^m(0.0016)^n(0.020)^p Relative Rate 2 = 12.821 = k'(0.0040)^m(0.0016)^n(0.020)^p Divide the first equation by the second, noting that nearly all the terms cancel out. The result is simply Relative Rate 1 [0.0020]^m[0.0016]^n[0.020]^p 6.94 —————- = —————————————————- = ——— = 0.541 = .5^m Relative Rate 2 [0.0040]^m[0.0016]^n[0.020]^p 12.821 If you have done this properly, you will have an equation involving only m as an unknown. Solve this equation for m, the order of the reaction with respect to I- ion. m = 0.87 (nearest integer) m log 0.5 = log 0.541 -0.26680 —————– = 0.87 -0.30103 Applying the same approach to Reaction Mixtures 1 and 3, find the value of n, the order of the reaction with respect to BrO3- ion. Relative Rate 1 = 6.94 = k'(0.0020)^m(0.0016)^n(0.020)^p Relative Rate 3 = 9.615 = k'(0.0020)^m(0.0016)^n(0.020)^p Dividing one equation by the other: 6.94 ——– = n = 9.615 Now that you have the idea, apply the method once again, this time to Reaction Mixtures 1 and 4, and find p, the order with respect to H+ ion. Relative Rate 4 = k'( )^m( )^n( )^p Dividing the equation for Relative Rate 1 by that for Relative Rate 4, we get p = Having found m, n, and p (nearest integers), the relative rate constant, k', can be calculated by substitution of m, n, p, and the known rates and reactant concentrations into Equation 5. Evaluate k' for Reaction Mixtures 1 to 4. Reaction 1 2 3 4 k' k' ave Standard deviation in k' (See Appendix VIII) Why should k' have nearly the same value for each of the above reactions? Using k' ave in Equation 5, predict the relative rate and time, tpred, for Reaction Mixture 5. Use the concentrations in the table. Relative rate pred tpred tobs

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