# Chapter 17 - Resolving Kinetic data¶

## Example 1 - pg 446¶

In :
#calculate the Order of the reaction
#Initialization of variablesx1=5
import math
from math import log
x2=20.
x1=5.
n1=7.49
n2=5.14
#calculations
n=(log(n1)-log(n2))/(log(100-x1) - log(100-x2))
#results
print '%s %.2f' %("Order of the reaction  = ",n)

Order of the reaction  =  2.19


## Example 2 - pg 448¶

In :
#calculate the Order of the reaction
#Initialization of variablesx1=5
import math
from math import log
p2=169.
p1=363.
t1=410.
t2=880.
#calculations
ndash=(log(t2) - log(t1))/(log(p1) - log(p2))
n=ndash+1
#results
print '%s %.2f' %("Order of the reaction  = ",n)

Order of the reaction  =  2.00


## Example 3 - pg 454¶

In :
#calculate the Activation energy and Z
#Initialization of variables
import math
R=1.987 #cal/deg/mol
k1=4.45*10**-5
k2=2.52*10**-6
T1=283+273.2 #K
T2=356+273.2 #K
#calculations
Ea=2.303*R*1.7530 /(1/T1 - 1/T2)
logZ= math.log10(k1) +Ea/(2.303*R*T1)
Z=10**logZ
#results
print '%s %d %s' %("Activation energy =",Ea,"cal/mol")
print '%s %.1e %s' %("\n Z =",Z,"lt /mol sec")

Activation energy = 38456 cal/mol

Z = 5.7e+10 lt /mol sec


## Example 5 - pg 456¶

In :
#calculate the Equlibrium constant
#Initialization of variables
g1=0.661
g2=0.899
g3=0.405
g4=0.803
g5=0.946
g6=0.614
k=1.33
#calculations
k0=k*g3/(g1*g2)
k2=k0*g4*g5/g6
#results
print '%s %.2f %s' %("Equlibrium constant =",k2,"lt/mol min")

Equlibrium constant = 1.12 lt/mol min