(i) Half-Life Period Method
It is the time needed for the reactant concentration to drop to 50% of its initial
value.
A number of experimental runs are carried out at different initial concentrations,
c A 0.
m=1
1
-ln c A /c A 0 = kt 0.9
0.8
At t =t 1/ 2, c A =c A 0 /2 0.7
cA0 0.6
1
-ln 2 = -ln 2 = kt 12
Concentration, cA(mol/L)
0.5
cA 0.4
0.3
t1/2 = ln k/2 0.2
0.1
0
0 0.5 1 1.5 2 2.5 3 3.5 4
m=2 t1/2 Time, t (min)
1 1
c A c A 0 = kt
-
2 1
At t = t 12 t
c A 0 c A 0 = k 1/ 2
-
1 1
t 1/ 2 = .
c A0 k
m=m
c A1−m -c A 01−m = (m-1).kt
¿ - c A 01−m = (m-1).kt 1/ 2
c A 01−m[ 2m−1 – 1] = (m-1).kt 1/ 2
m −1
t 1/ 2 = 2 – 1 . c A 01−m
k (m−1)
ln t 1/ 2 = (1-m) ln c A 0 + ln k '
' 2m −1 – 1
k =
k (m−1)
1
ln t 1/ 2 = (1-m) ln c A 0 + ln k '
m<1 m>1 m =1
ln t1/2 is plotted vs lncA0. 5
4.5
Slope = 1-m 4
m is obtained from the slope 3.5
3
k is obtained from the intercept 2.5
lnt1/2
2
1.5
1
0.5
0
0 1 2 3 4 5 6 7 8 9 10
ln cA0
Summary of Integral Methods:
(A) General Procedure:
cA
dcA
F(cA) = −∫
cA 0 f (c A )
m f(cA) F(cA)
1 cA -ln cA / cA0
2 cA2 1/ cA – 1 / cA0
2 (Bimolecular) cA cB [1/(m-1)cA]*ln[M-XA/M(1-XA)]
m cA m [1/(m-1)][cA 1-m - cA0 1-m ]
(B) Half-life Period:
m t1/2
1 ln 2/ k
m (2m-1 -1)/[k(m-1)].cA0 1-m
2
Constant Volume Gas-phase Reactions:
As the volume is kept constant, the total pressure π may change.
Partial pressure p A is used in place ofc A .
−dc A −1 dp A k
– r A= = =k c A m= p m
m A
dt RT dt ( RT )
−dp A k
= p m=k ' p A m
m−1 A
dt ( RT )
m=m, k ' =k /( RT )m −1
m=0 , k ' =kRT
m=1, k ' =k
m=2, k ' =k / RT
Reaction: aA +bB+…=rR +sS +…
At t=0, nA 0
nB 0
nR 0
nS 0
n Inert
At t=0, n0 =n A + nB + …+n R +nS + …+n Inert
0 0 0 0
At t=t , n A −ax ; n B −bx ; n R +rx ; n S +sx ;
0 0 0 0
n Inert
At t=t , n=n0 + x [ ( r+ s+ … )− ( a+b+ … ) ] =n 0+ x ∆ N
x=(n−n0) /∆ N
∆ N =[ ( r +s +… ) −( a+b+ … ) ]
n A n A −ax a (n−n0 )
c A= = 0
=c A −
V V ∆N V 0
a '
p A =c A RT= p A − ( π −π 0 )= p A −δ A ( π −π 0)
0
∆N 0
r
p R= p R + ( π−π 0 ) = pr + δ 'R (π −π 0 )
0
∆N 0
a r
δ 'A= , δ 'R =
∆N ∆N
3
PROBLEM #: Determine the rate equation for the gas-phase decomposition of di-
tert-butyl peroxide
(CH3)3COOC(CH3)3 C2H6 + 2 CH3-C-CH3
O
The reaction is carried out in a batch reactor in which the total pressure is recorded
with time. Pure reactant is present in the reactor initially.
Time (min) 0 2.5 5 10 15 20
Total Pressure,
7.5 10.5 12.5 15.8 17.9 19.4
π (mm Hg)
pA (mm Hg) 7.5 6.0 5.0 3.35 2.30 1.55
-ln PA/PAo 0 0.223 0.405 0.806 1.182 1.577
8.0 1.8
Soln: 7.0 pA 1.6
−d p A k 6.0 1.4
= p m =k ' . pmA
m−1 A 1.2
dt ( RT ) 5.0
pA (mm Hg)
-ln(pA/pA0)
1
4.0
0.8
3.0
A R + 2S 0.6
2.0 0.4
a = 1, r = 1, s=2
1.0 0.2
∆N = r + s – a = 1 + 2 – 1 = 2 0.0 0
0 2 4 6 8 10 12 14 16 18 20
a 1
A = ∆ N = 2 t (min)
pA0 = π0 = 7.5 mm Hg (Pure reactant present initially)
a 1
pA = pA0 – ∆ N (π - π0) = π0 - 2 (π - π0)
3 π 0 −π
pA =
2
pA is calculated at different time, t
−d p A
(i) If m = 0, = k ’. pA is plotted vs t. It is not a straight line. So , m 0
dt
−d p A
(ii) If m = 1, = k’ pA = k pA
dt
pA
-ln p = kt
A0
pA
-ln p = kt is calculated and plotted against t, It is a straight line.
A0
So, m = 1. The slope, k = k’ = 0.079 min-1
4
−d p A
= 0.079 pA
dt
(B) Differential Method of Analysis:
1.0
-rA = -dcA/dt = kcAm;
0.9
ln(-dcA/dt) = m.lncA +lnk 0.8
0.7
Get experimental data of cA vs t 0.6
0.5
cA
Calculate –dcA/dt at different cA 0.4
0.3
Plot ln(-dcA/dt) vs lncA 0.2
0.1
Slope = m; Intercept = lnk 0.0
0 1 2 3 4 5 6 7
t
If the rate equation is more complicated, Mathematical manipulation is required.
-dCA/dt may be obtained from graphical evaluation of slope or numerical technique.
More complicated rate equation: 0.0
-3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0
-0.5
-rA = -dcA/dt = (k1cA)/(1 + k2cA) [Adsorption eqn.]
f(x) = x − 0.69
-1.0
1/(-rA) = (1 + k2cA)/k1cA = (1/k1cA) + k2/k1 R² = 1
-1.5
1/(-rA) is plotted vs 1/CA
ln(-dcA/dt)
-2.0
Slope = 1/k1
-2.5
Intercept = k2/k1
-3.0
Advantage of Differential Method:
-3.5
No trial & error is needed Direct method -4.0
ln(cA)
Possible to determine non-integer order of reaction
One set of cA vs t data is sufficient
Disadvantage:
Estimation of the slope of cA vs t curve (-dcA/dt) is difficult & often erroneous
5
Problem #2: At room temperature, sucrose is hydrolyzed by catalytic action of the
enzyme sucrase as follows:
Starting with a sucrose concentration of 1.0 mmol/L and an enzyme concentration
of 0.01 mmol/L, the following kinetic data is obtained in a batch reactor.
t(h) cA (mmol/L) -rA 1/cA 1/-rA
0 1 0.16 1 6.25
1 0.84 0.16 1.19 6.25
2 0.68 0.155 1.471 6.45
3 0.53 0.15 1.887 6.67
4 0.38 0.13 2.631 7.69
5 0.37 0.11 3.704 9.09
6 0.16 0.09 6.25 11.11
7 0.09 0.06 11.11 16.67
8 0.04 0.045 25.0 22.22
9 0.018 0.017 55.56 58.82
10 0.006 0.007 166.67 142.86
Soln: cAO = 1mmol/L, cEO = 0.01mmol/L
is estimated numerically.
If independent variable data are equispaced (equal time intervals in this case) with N
time instants,
(at in interior time domain points)
(at the initial time instant)
(at the final time instant)
Mechaelis-Menten equation (for enzyme catalysed fermentation reactions):
-1/rA is plotted against 1/cA . Slope = M/kcEO = M/(k * 0.01) = 0.827
Intercept = 1/kcEO = 1/(k * 0.01) = 6.0
k = 16.67 h-1; M = 0.138
-r A = 16.67. cEO cA /(0.138 + cA)
