The rate constants for the first-order decomposition of acetone dicarboxylic aci

Question

The rate constants for the first-order decomposition of acetone dicarboxylic acid in aqueous solution were found to be 4.75 times 10^-4 s^-1 at 293 K and 5.48 times 10^-2 s^-1 at 333 K. Calculate (i) the Arrhenius parameters A and E_a, and (ii) the Eyring transition-state parameters Delta H and Delta S. (b) A small quantity of the alpha-particle emitting radioactive isotope^210 Po (t_1/2 = 138 days) was injected into the bloodstream of a patient, blood samples were then taken at intervals and counted with these results. Calculate the effective half-life and hence obtain the biological half-life (resulting only from the loss of^210 Po by excretion and absorption according to first-order kinetics) of^210 Po in the bloodstream. Determine the initial specific activity of^210 Po in the bloodstream at zero time and thereby also deduce the initial concentration of^210 expressed in atoms per ml.

Explanation / Answer

a)

claculate A and Ea

K = A*exp(-E/(RT))

for...

ln(K2/K1) = E/R*(1/T1-1/T2)

ln((5.48*10^-2)/(4.75*10^-4)) = E/8.314*(1/(333) - 1/(293))

E = ln((5.48*10^-2)/(4.75*10^-4)) *8.314 / (1/(333) - 1/(293))

E= 96290.74 J/mol = 96.29 kJ/mol

for

A:

choose any point

4.75*10^-4 = A*exp(-96290.74 /(8.314*293))

A = (4.75*10^-4) / exp(-96290.74 /(8.314*293)) = 6.975*10^13

For the parameters:

ln(K) = -H/(RT) + S/R

choose point 1

ln(4.75*10^-4) = -H/(8.31*293) + S/8.314

choose point 2

ln(5.48*10^-2) = -H/(8.31*333) + S/8.314

solve

ln((4.75*10^-4)/(5.48*10^-2)) =  -H/(8.31*293) + H/(8.31*333)

-4.7481 = (-0.000410)H + 0.0003613H

H(0.0003613+-0.000410) = -4.7481

H = -4.7481/(0.0003613+-0.000410) =97496.9199 J/mol = 97.4 kJ/mol

ForS:

ln(5.48*10^-2) = -H/(8.31*333) + S/8.314

ln(5.48*10^-2) = -97496.9199 /(8.31*333) + S/8.314

S = (ln(5.48*10^-2) + 97496.9199 /(8.31*333) )*8.314

S = 268.7800 J/molK

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