Question

10 points av UESTION 3 The table below contains concentration time data for drug LIU-555. The study was carried out atroom temperature (250C) and pH 7.0. Based on this information, calculate the degradation rate constant for LIU-555 The units for the answer should be mg/(L-day) 4. Time (days) 0 1 A (mg/L) 1000 2 800 600 400

Answer 1

Given:-

Time (days)	0	1	2	3	4
A (mg / L)	1000	800	600	400	200

Above data shows that the reaction follows the Zero order reaction therefore rate constant of degradation is as follows:-

As we know that rate constant for zero order reaction is as follows:-

rate constant (k₀) =  [A₀] -  [A] / t

where

[A₀] = initial concentration of reactant

[A] = final concentration of reactant after t time

t = time

therefore

(i) time (t) = 1 day

initial concentration of [A₀] = 1000 mg / L

final concentration of [A] after 1 day = 800 mg / L

rate constant (k₀) =  [A₀] -  [A] / t

rate constant (k₀) = 1000 mg / L -  800 mg / L / 1day = 200 mg / L-day

(ii) time (t) = 2 day

initial concentration of [A₀] = 1000 mg / L

final concentration of [A] after 2 day = 600 mg / L

therefore

rate constant (k₀) =  [A₀] -  [A] / t

rate constant (k₀) = 1000 mg / L -  600 mg / L / 2 = 400 mg / L / 2 day = 200 mg / L-day

(iii) time (t) = 3 day

initial concentration of [A₀] = 1000 mg / L

final concentration of [A] after 3 day = 400 mg / L

therefore

rate constant (k₀) =  [A₀] -  [A] / t

rate constant (k₀) = 1000 mg / L -  400 mg / L / 3 day = 600 mg / L / 3 day = 200 mg / L-day

(iv) time (t) = 4 day

initial concentration of [A₀] = 1000 mg / L

final concentration of [A] after 4 day = 200 mg / L

therefore

rate constant (k₀) =  [A₀] -  [A] / t

rate constant (k₀) = 1000 mg / L -  200 mg / L / 4 day = 800 mg / L / 4 day = 200 mg / L-day

Above value of rate constant of is fixed which shows that reaction is zero order therefore the degradation rate constant (k₀) for LIU-555 is 200 mg / L-day (i.e the answer)