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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