The half-life time (t1/2) is the time necessary for the initial concentration to reduce to half of its value. To apply this method, a series of data oft1/2 versusCAoare required (Figure 2.18).
The procedure to determine the reaction order and the reaction rate coefficient is similar to that of the integral method. The only difference is that, when integrating the reaction rate equation, the limits of the integral are fromt= 0 andCAo, tot = t1/2andCA= CAo/2.
Thus, starting with the reaction rate equation for irreversible reactions of one component at constant density, Eq. (2.3):
−dCA dt =kCAn
CAo
(CAo)n
(CAo)n/2 (CAo)2/2 (CAo)1/2 (CAo)2
(CAo)1 ...
(t1/2)1 (t1/2)2...(t1/2)n t1/2 Figure 2.18 Data required to use the method of half-life time.
2.4.1 Reactions of Zero Order Forn= 0, Eq. (2.3) is:
−dCA
dt =k
Separating variables and integrating fromt= 0 andCA= CAotot = t1/2and CA= CAo/2:
−
CAo 2
CA o
dCA=k
t1 2
0
dt=kt1 2
−CACCAoAo2=kt1 2
−CAo
2 +CAo=kt1 2 CAo
2 =kt1 2 (2.61)
And the value ofkis evaluated with:
k= CAo
2t1 2 2 62
2.4.2 Reactions of the First Order
Forn= 1, Eq. (2.3) is:
−dCA
dt =kCA
Separating variables and integrating fromt= 0 andCA= CAotot = t1/2and CA= CAo/2:
−
CAo 2
CA o
dCA CA
=k
t1 2
0
dt=kt1 2
−lnCACAo2 CAo =kt1 2
−ln CAo
2 + lnCAo=kt1 2
ln 2 =kt1 2 (2.63)
And the value ofkis evaluated with:
k=ln 2
t1 2 2 64
2.4.3 Reaction of the Second Order Forn= 2, Eq. (2.3) is:
−dCA dt =kC2A
Separating variables and integrating fromt= 0 andCA= CAotot = t1/2and CA= CAo/2:
−
CAo 2
CA o
dCA
CA2 =k
t1 2
0
dt=kt1 2 1
CA CAo2 CAo
=kt1 2
2 CAo− 1
CAo=kt1 2 1
CAo
=kt1 2 (2.65)
And the value ofkis evaluated with:
k= 1
CAot1 2 2 66
2.4.4 Reaction of thenthOrder
Separating variables in Eq. (2.3) and integrating fromt= 0 andCA= CAoto t = t1/2andCA= CAo/2:
−
CAo 2
CA o
dCA
CAn =k
t1 2
0
dt=kt1 2 CA1−n
n−1 CAo 2 CAo
=kt1 2 1
n−1 CAo
2
1−n
−C1Ao−n =kt1 2
CAo1−n 2n−1−1 =k n−1t1 2 (2.67)
And the value ofkis evaluated with:
k= 2n−1−1 CAo1−n
n−1t1 2 Forn 1 2 68
Figures 2.19 shows Eqs. (2.61), (2.63), (2.65) and (2.67) in graphical form.
Example 2.6 The thermal decomposition of nitrous oxide (N2O) was studied at 1030 K, and the data reported in Table 2.12 oft1/2were obtained at different initial partial pressures ofN2O. Find the kinetic model.
Solution
The initial concentration of N2O for t1/2= 860 sec and (pN2O)o= 82.5 mmHg is evaluated with:
CAo=pAo
RT = 82 5mmHg
62 36mmHg lt
gmolK 1030 K
= 1 284 × 10−3gmol lt k
k
k n= 0
n= 2
n= 1
Order n
(n– 1) k CAo/2
1/CAo
ln 2
t1/2
t1/2 t1/2
t1/2
CAo 1–n(2n–1– 1)
Figure 2.19 Method of half-life time for irreversible reactions of one component.
The integrated equations for the first and second orders of reaction are:
Eq. (2.64) forn= 1 is:
k=ln 2 t1 2 =ln 2
860 = 0 806 × 10−3sec−1 Eq. (2.64) forn= 2 is:
k= 1
CAot1 2= 1
1 284 × 10−3 860 = 0 9056 lt gmolsec
The complete results are summarized in Table 2.12. It is observed that for the first order of reaction, the value ofkhas an increasing tendency, while the second order of reaction accurately fits the experimental data, with an average value ofk= 0.8954 lt/gmol sec.
It is important to mention that for reactions following the first order, the half-life time is the same to calculatek; in other words, the initial concen- tration does not have an effect onn= 1.
Finally, the kinetic model is:
−rA = 0 8954lt gmol sec CA2
2.4.5 Direct Method to Calculatekandnwith Data oft1/2
A direct approach to calculatenandkfrom data of half-life time is by linearization of Eq. (2.68) with logarithms:
t1 2= 2n−1−1 CAo1−n
n−1k 2 69
lnt1 2= ln 2n−1−1
n−1k + 1−n lnCAo 2 70
Table 2.12 Data and results of Example 2.6.
(pN2O)o(mmHg) t1/2(sec)
CAo× 103
(gmol/lt) k× 103(n= 1) k(n= 2)
82.5 860 1.284 0.806 0.9056
139 470 2.164 1.475 0.9832
296 255 4.608 2.718 0.8510
360 212 5.605 3.270 0.8416
Average 0.8954
When plotting Eq. (2.70), a straight line is obtained (x = lnCAovs.y = lnt1/2).
With the slope (1−n) the reaction order is calculated, and with the intercept the value ofk, as illustrated in Figure 2.20.
Example 2.7 Solve Example 2.6 by using the direct method to calculate kandnwith data oft1/2.
Solution
The data of Example 2.6 are used withx = ln(CAo)versusy = ln(t1/2). Using Eq. (2.70), the following results are obtained (Figure 2.21):
Slope = 1 –n ln t1/2
ln CAo ln (2n–1– 1)
(n– 1)k
Figure 2.20 Direct method to calculatenandkwith data of half-life time.
5.0 5.5 6.0 6.5 7.0
–7.0
ln(CAo)
ln(t1/2) Slope = 1 –n
–6.5 –6.0 –5.5 –5.0
Figure 2.21 Application of the direct method with data of half-life time for Example 2.7.
Slope = 1−n =−0 9278
Intercept = 0 5342
Correlation coefficient =r = 0 9967
Therefore,n= 1.9278. The entire value of the reaction order is 2, and the reaction rate coefficient is evaluated with the integrated equations used in Example 2.6.
2.4.6 Extension of the Method of Half-Life Time (t1/2) to Any Fractional Life Time (t1/m)
The method of half-life time can be extended to any fractional life time of the limiting reactant. For instance, t1/3 is the time necessary for the initial concentration to reduce to one-third of its value. In general,t1/m is the time necessary for the initial concentration to reduce to 1/m of its value.
The integrated equations for any fractional life time are similar to those for half-life time:
For ordern(n 1):
k= mn−1−1 CAo1−n n−1t1 m
2 71 Forn= 1:
k=ln m
t1m 2 72
2.4.7 Calculation of Activation Energy with Data of Half-Life Time With the data oft1/2or any other fractional life time (t1/m), it is possible to determine the activation energy of a reaction if the data of time are reported at different temperatures. The procedure is the following:
Substituting the Arrhenius equation (Eq. 1.91) into Eq. (2.68):
t1 2= 2n−1−1 C1Ao−n n−1Ae−EA RT which can also be written as:
t1 2= 2n−1−1 CAo1−n n−1 A eEA RT
linearization with logarithms gives:
lnt1 2= ln 2n−1−1 CAo1−n n−1 A + EA
R 1
T 2 73
With the slope of this linear equation, the activation energy can be calculated. The intercept is useless because it is a function of the reaction ordernand frequency factorA, as illustrated in Figure 2.22. The value of Acan only be calculated if the reaction order has been determined with another method.
Example 2.8 The decomposition ofN2O5was carried out at different temperatures in an isothermal reactor at constant density. The initial con- centration ofN2O5was the same for all the experiments. The experimen- tal results are reported in Table 2.13. Calculate the activation energy.
Solution
Using Eq. (2.73) withx =1/T(Tis temperature in K) versusy = ln(t1/2), Figure 2.23 can be prepared. It is seen that the experimental data fit a straight line, from which the following values are obtained:
Slope =EA R = 12458 86
Intercept =−31 88
Correlation coefficient =r = 0 9999 1/T ln t1/2
Slope =EA/R
Intercept = f(n, A, CA0) = ln
(n– 1)AAo (2n–1– 1)C1–n
Figure 2.22 Calculation ofEAwith data oft1/2.
Therefore, the value of the activation energy is:
EA= 12548 86 R = 12548 86 1 987 = 24755 8Cal gmol
2.4.8 Some Observations of the Method of Half-Life Time
2.4.8.1 Calculation ofnwith Two Data oft1/2Measured with DifferentCAo
If at the same temperature, the half-life time is measured with two experi- ments using two different initial concentrations of the limiting reactant, (CAo)1and (CAo)2, it is possible the calculate the reaction order.
Table 2.13 Data and results of Example 2.8.
Temperature ( C) t1/2(sec) Temperature(K) x= 1/T× 103 y = ln(t1/2)
50 780 323.15 3.095 6.659
100 4.6 373.15 2.680 1.526
150 8.8 × 10−2 423.15 2.363 −2.430
200 3.9 × 10−3 473.15 2.113 −5.547
300 3.9 × 10−5 573.15 1.745 −10.152
–12 –8 –4 0 4 8
0.002 0.003 0.004
1/T ln(t1/2)
Slope = EA/R 0.001
Figure 2.23 Calculation ofEAfor Example 2.8.
For instance, using Eq. (2.65) for the second order of reaction for the two data:
k= 1
CAo 1 t1 2 1
k= 1
CAo 2 t1 2 2
Equaling these two expressions:
t1 2 1
t1 2 2 = CAo 2
CAo 1
Using Eq. (2.66) for the third order of reaction for the two data:
k= 1
CAo 1 2 t1 2 1
k= 1
CAo 2
2 t1 2 2 Equaling these two expressions:
t1 2 1
t1 2 2 = CAo 2 CAo 1
2
These equations can be generalized for any reaction order:
t1 2 1
t1 2 2 = CAo 2 CAo 1
n−1
The linearization by logarithms gives:
ln t1 2 1
t1 2 2 = n−1 ln CAo 2
CAo 1 2 74
And, finally, the reaction order is calculated with:
n=
ln t1 2 1 t1 2 2 ln CAo 2 CAo 1
+ 1 2 75