A reaction follows the stoichiometry: A + P rightarrow AP. The concentration mea

Question

A reaction follows the stoichiometry: A + P rightarrow AP. The concentration measured after mixing is to the right: (a) There is no dependence on P. What is the rate of the reaction? Calculate the rate coefficient, including units. (b) Propose a mechanism consistent with the kinetics and stoichiometry of the reaction. Write differential equations for your mechanism, and relate the measured s to the k's to the k's in your equation. (c) If the data are measured at 0 degree C, and if the rate doubles at 10 degree C, calculate the activation energy in kJ/mole. Consider ingestion of a drug dose, and assume that ingestion and absorption are rapid compared with utilization and elimination. Suppose that this is represented by a process: A rightarrow B rightarrow C where [A] is the concentration of the ingested drug, [B] is the concentration in the blood, and [C] represents elimination from the blood. We looked at "compartment" models like this earlier in the semester. (a) Assume that the drug delivery, with rate k_1, is zero order, and elimination, with rate k_2, is first order. Write the rate law in terms of d[B]/dt. (b) If the drug is introduced at time t = 0, draw a curve showing how [B] will change with time (if the drug is continuously administered. (c) Suppose the drug is stopped after some time - draw how [B] will change with time thereafter (i.e. make a second plot showing first the change under continuous administration and then what happens, in this second plot, after the administration is stopped. (d) If you had measurements of [B] as a function of time, how would you determine k_1 and k_2? (e) Derive a mathematical expression for [B] as a function of time from your equation in part (a). Plot this solution as well as the numerical solution to the equation of part (a) using Mathematica. (f) To what level does [B] rise after a long period of continuous administration?

Explanation / Answer

for zero order reaction, -dCA/dt=K

CA= CAO-Kt, the plot of CA vs t has to be straight line. The plot of CA vs t is drawn and shown below. The plot is not a straight line suggesting the reaction is not zero order.

for 1st order reaction, CA= CAO*e(-Kt),

so a plot of CA vs t will be exponential. This is found to be the trend and is shown below

from the plot K= rate constant = 0.002/sec.

Mechanism A K1------> A* slowest step

dCA/dt= K1[A]

A*+P<---K2/K-2->AP fast and equilibrium

dCA*/dt= K2[A*] [P]- K-2[AP] =0

[A*]= K-2[AP]/[P]

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

ln2= (Ea/8.314)(1/273-1/283)

Ea= 44523 Joules/mole= 44.523 KJ/mole

2. A--K1--->B--K2--->C

A to B reaction is zero order and B to C is 1st order.

dB/dt= K1- K2[B]

dB/(K1-K2[B]= dt

-ln(K1-K2[B])/K2= t+C

C is integration constant

at t=0, [B]=0

-lnK1/K2=C, -ln(K1-K2[B])/K2= t-ln(K1)/K2

lnK1- ln (K1-K2[B] = K2t

ln {K1/(K1-K2[B]} = K2t

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