The Kinetics of Potassium Ferrate Oxidation of Isopropyl Alcohol, Ethyl Alcohol, and Trimethylacetaldehyde

Authors

Ronald Lee Bartzatt, University of Nebraska-LincolnFollow

First Advisor

James Carr

Date of this Version

5-1980

Document Type

Thesis

Citation

A thesis presented to the faculty of the Graduate College in the University of Nebraska in partial fulfillment of requirements for the degree of Master of Science

Major: Chemistry

Under the supervision of Professor James Carr

Lincoln, Nebraska, May 1980

Comments

Copyright 1980, Ronald Lee Bartzatt. Used by permission

Abstract

Conclusion

The main emphasis of this thesis is to describe the reaction rates of potassium ferrate oxidation of various substrates, in relation to solution pH.

For trimethylacetaldehyde the order of dependence (slope ) from pH 7.85 to pH 9.45 is 0.951. This is calculated from - Log k_s versus pH graphing of the data shown in the esults. The correlation coefficient of this graphing is equal to 0.989 and y-intercept = −7.95.

For isopropanol the order of dependence from pH 8.23 to pH 8.44 is 2.45. The correlation coefficient = 0.973, and y-intercept = −18.77. From pH 8.44 to pH 9.02 the order of dependence is 0.834. The correlation coefficient = 0.998, and y-intercept is −5.19.

For ethyl alcohol the order of dependence from pH 8.12 to pH 8.46 is 2.28. The correlation coefficient = 0.953 and y-intercept = −17.43. From pH 8.46 to pH 9.01 the order of dependence is 0.783. The correlation coefficient = 0.890 and y-intercept is −4.77.

For the substrate ethanol a further observation can be made. Being an alcohol it is possible that a protonated and deprotonated species can exist in the reaction solutions. It would be useful to ascertain if the protonated or deprotonated species of ethanol contributes significantly to the rate of reaction. Therefore the total concentration of ethanol can be represented as in equation (1).

C_ETOH = [ETOH⁺₂] + [ETOH] + [ETO⁻] (1)

The deprotonation of ethanol along the route shown below has an equilibrium constant of K = 7.94 × 10⁻²⁰ (log K = 10^−19.1).

2ROH = ROH₂⁺ + RO⁻

Therefore only at a high pH (pH greater than 14) is the concentration of the deprotonated form significant and can be ignored for the pH range discussed here. Then equation (1) becomes,

C_ETOH = [ETOH₂⁺] + [ETOH] (2)

If the conditional rate constant k_s is multiplied by C_ETOH it becomes the sum of k_s(A)[ETOH₂⁺] and k_s(B)[ETOH] (see below).

k_sC_ETOH = k_s(A)[ETOH₂⁺] + k_s(B)[ETOH] (3)

Where k_s(A) is the conditional rate constant for the protonated acid (A for acid) form of ethanol and k_s(B) is the conditional rate constant for the base (B for base) form of protonated ethanol.

Protonated ethanol loses its extra hydrogen in a manner shown in equation (4), where the equilibrium constant K equals 250.

ROH₂⁺ + H₂O = ROH + H₃O⁺ (4)

We can express equation (4) in a different form.

K = ([ROH] [H₃O⁺]) / ([ROH₂⁺][H₂O]) = 250

We can obtain the expression for the acid dissociation of protonated ethanol K_a from K by multiplying K by [H₂0]. This procedure looks as follows:

K[H₂O] = ([H₃O⁺)[ETOH][H₂O]) / ([ETOH₂⁺][H₂O]) = K_a = 250(55.5) = 13875

Which will be called equation (5).

From equation (5) it is possible to obtain an expression for [ETOH₂⁺] which can be substituted into equation (3) becoming

k_s[ETOH] = (k_s(A)[H⁺][ETOH]) / K_a + k_s(B)[ETOH] (6)

Where [ETOH] = C_ETOH^•

Simplifying equation (6), it becomes,

k_s = (k_s(A) / K_a)[H⁺] + k_s(B)

Therefore by plotting k_s values for the substrate versus [H⁺], the k_s(B) value can be obtained from the y-intercept and (k_s(A) / K_a) from the slope. This has been done for ethanol and isopropanol (see Figures IX and X). For ethanol (k_s(A) / K_a) = 3.96 × 10⁶ and k_s(B) = 1.53 × 10⁻³. For isopropanol (k_s(A) / K_a)= 3.93 × 10⁶ and k_s(B) = 6,11 × 10⁻⁴.

For ethanol the ks(A) value can be found by the process shown below,

k_s(A) = (slope)K_a = slope (K) (55.5)

k_s(A) = 3.96 × 10⁶ (250) (55.5)

k_s(A) = 5.49 × 10¹⁰

Since the conditional rate constant k_s(A) is larger than the rate of diffusion of the species in the solution the protonated form of ethanol cannot contribute to the rate of reaction. The species CH₃CH₂OH is responsible. It is probable that this same argument holds for isopropanol. The value of K for isopropanol is not available.