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Kinetics TUTORIAL 2 1. At 25 °C, the half-life period for the decomposition of N 205 is 5.7 hours and is independent of the initial pressure of N 205, calculate: a- The time for the reaction. b- The specific rate constant for the reaction. Answer:
a. The t1/2 is independent on the initial conc. So, it is st
1 order reaction, hence the reaction never goes to an end and the life time is infinity. b. t1/2 = k=
0.693 k
0.693 t half
0.693 =
5.7
=
0.1215 hr
-1
2. A drug product is known to be ineffective after it has decomposed 30%.The original concentration of one sample was 50 mg/ml. When assayed 20 months later the concentration was found to be 40 mg/ml, assuming that the decomposition is first order, what should be the expiration time on the label? What is the half-life of this product? Answer: -log Ct = log Co – (kt / 2.303) log 40 = log 50 – (k.20 / 2.303) 20k/2.303 = log 50 -log 40 = log50/40 k= log50/40 . 2.303/20 = 0.01116 m
62.1 month K 0.01116 3. An ophthalmic solution dispensed at 5 mg/ml conc. exhibits a first order reaction, with -1 a rate constant of 0.0005 day . How much drug is remaining after 120 days? & how long will it take for the drug to degrade to 90% of its initial concentration? Answer: =
-T90% =0.105/K = 0.105/0.0005 = 210 days 4. The initial concentration of both ethyl acetate and sodium hydroxide in a reaction mixture was 0.01 M each. The change in the concentration "x" of alkali during 20 minutes was 0.00566 mole/liter. Compute a: the second order rate constant, b: the half life of the reaction. Answer:
When a = b x kt a (a x) =
−
K
x =
0.00566
at (a x) −
t1/2 =
1 ak
=
=
0.01 * 20(0.01 0.00566)
=
6.52
-1
-1
L.mole min
−
1 0.01 * 6.52
=
15.34 min
5. Tetracycline undergo epimerization. At the equilibrium point the distribution consisted of 32% of tetracycline and 68% of epi-tetracycline. Measuring the concentration of tetracycline at different time interval and plotting the data according to the following equation lead to a straight line the slope of which equals 0.01 min-1, calculate both Kf and kr
5. Answer:
slope = (Kf + Kr)/2.303 Also: at equilibrium Kf *Aeq = Kr *Beq So, now you have 2 equations and 2 unknowns, the 2 equations can be solved simultaneously. K f K r
6. The rate of decomposition of 0.056 molar glucose at 140 °c in an aqueous solution containing 0.35 HCI was found to be: Time (hr) Glucose Remaining 2 (mole/10 )
0.5
2
3
4
6
8
12
5.52
5.42
5.32
5.02
4.80
4.52
4.10
What is the order of glucose decomposition and the specific reaction rate constant? Determine the half-life period. Does the available data allow an accurate calculation of the kinetic parameters? Answer:
We cannot determine the order or t 1/2 because the readings are incomplete. To determine the order of a reaction, you must be given data until Ct = 1/2Co.( i.e the initial concentration reaches half.) Using the graphical method: No.6 time (h) 0.5 2 3 4 6 8 12
k 0.028783 0.016338 0.017101 0.027339 0.025696 0.026787 0.025986
average k=
0.024004
t1/2 = 0.693/k= 28.86979 h
7.
The following are the concentrations of Salphadiazien remaining at various times o during the thermal decomposition at 40 C.
Time (months)
0
4
8
12
16
20
24
Conc. (mg/5ml)
100
81.4
68.8
56.4
44.6
36.2
31.2
Using the above data determine:
a. The order of the decomposition reaction using two different methods. b. The rate constant of the reaction. c. The shelf life of the product. Answer: No7 graphical method: t
n = [log (15/15)/log (100/80)]+1 =1 B) slope= -k/2.303= -0.0215 therefore k= 0.0495145month-1
C) t 90% : 0.105/k = 0.105/0.0495145 = 2.12 m
8. A fast reaction in solution between compounds A and B was followed over a period of 60 s at 3I0 K by assaying the concentrations of A and B that remained at various time intervals. The following results were obtained: 3
3
Time (s)
Concentration of A (mol /dm )
Concentration of B (mol /dm )
0
0.2
0.1
10
0.166
0.066
20
0.146
0.046
30
0.134
0.034
60
0.114
0.014
a. Determine the order of the reaction and b. Calculate the value of its rate constant. c. Calculate the half life of the product. Answer:
No.8 A) Graphical method: time (s)
remaining conc of A 0 10 20 30 60
0.2 0.166 0.146 0.134 0.114
Log remaining conc of A -0.698970004 -0.779891912 -0.835647144 -0.872895202 -0.943095149
Try second order plot when the initial concentrations are not equal. remaining remaining time (s) conc of A conc of B b a 0 0.2 0.1 0.1 10 0.166 0.066 0.1 20 0.146 0.046 0.1 30 0.134 0.034 0.1 60 0.114 0.014 0.1 x=aab-x( remaining x(remaining) remaining) b(a-x) 0 0.2 0.1 0.034 0.166 0.066 0.054 0.146 0.046 0.066 0.134 0.034 0.086 0.114 0.014
B) k= slope * 2.303/(a-b) k=0.0101*2.303/(0.2-0.1) k= 0.232603 dm3/mol.s C) Half life: The half-life or shelf life of a second–order reaction can Only be calculated when the initial concentration of both Reactants are equal (a = b). T1/2= 1/ak
