{ "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": {} } ] }