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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4d58417f5757dcfb290ed1ee0e2191d724d6d044069d2c21c72e96a05505c726"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17 - Resolving Kinetic data"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1 - pg 446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Order of the reaction\n",
+ "#Initialization of variablesx1=5\n",
+ "import math\n",
+ "from math import log\n",
+ "x2=20.\n",
+ "x1=5.\n",
+ "n1=7.49\n",
+ "n2=5.14\n",
+ "#calculations\n",
+ "n=(log(n1)-log(n2))/(log(100-x1) - log(100-x2))\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Order of the reaction = \",n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Order of the reaction = 2.19\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2 - pg 448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Order of the reaction\n",
+ "#Initialization of variablesx1=5\n",
+ "import math\n",
+ "from math import log\n",
+ "p2=169.\n",
+ "p1=363.\n",
+ "t1=410.\n",
+ "t2=880.\n",
+ "#calculations\n",
+ "ndash=(log(t2) - log(t1))/(log(p1) - log(p2))\n",
+ "n=ndash+1\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Order of the reaction = \",n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Order of the reaction = 2.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3 - pg 454"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Activation energy and Z\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "R=1.987 #cal/deg/mol\n",
+ "k1=4.45*10**-5\n",
+ "k2=2.52*10**-6\n",
+ "T1=283+273.2 #K\n",
+ "T2=356+273.2 #K\n",
+ "#calculations\n",
+ "Ea=2.303*R*1.7530 /(1/T1 - 1/T2)\n",
+ "logZ= math.log10(k1) +Ea/(2.303*R*T1)\n",
+ "Z=10**logZ\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Activation energy =\",Ea,\"cal/mol\")\n",
+ "print '%s %.1e %s' %(\"\\n Z =\",Z,\"lt /mol sec\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy = 38456 cal/mol\n",
+ "\n",
+ " Z = 5.7e+10 lt /mol sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5 - pg 456"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Equlibrium constant\n",
+ "#Initialization of variables\n",
+ "g1=0.661\n",
+ "g2=0.899\n",
+ "g3=0.405\n",
+ "g4=0.803\n",
+ "g5=0.946\n",
+ "g6=0.614\n",
+ "k=1.33\n",
+ "#calculations\n",
+ "k0=k*g3/(g1*g2)\n",
+ "k2=k0*g4*g5/g6\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Equlibrium constant =\",k2,\"lt/mol min\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equlibrium constant = 1.12 lt/mol min\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file