From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter7.ipynb | 346 +++++++++++++++++++++ 1 file changed, 346 insertions(+) create mode 100755 Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb (limited to 'Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb') diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb new file mode 100755 index 00000000..db391867 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter7.ipynb @@ -0,0 +1,346 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 - Principles of chemical equilibrium" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E1 - Pg 147" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the reaction gibbs energy\n", + "#Initialization of variables\n", + "import math\n", + "G=-31. #kJ/mol\n", + "T=37+273. #K\n", + "Cadp=1/1000. #mmol/L\n", + "Cp=8/1000. #mmol/L\n", + "Catp=8/1000. #mmol/L\n", + "R=8.314 #J/K mol\n", + "#calculations\n", + "Q=Cadp*Cp/Catp\n", + "deltaG=G+R*T*math.log(Q) /1000.\n", + "#results\n", + "print '%s %.1f %s' %(\"Reaction Gibbs energy =\",deltaG,\" kJ/mol\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reaction Gibbs energy = -48.8 kJ/mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E2 - Pg 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the gibbs energy\n", + "#Initialization of variables\n", + "Hr=-285.83 #kJ/mol\n", + "Sr=-163.34 #J/ K mol\n", + "T=298.15 #K\n", + "#calculations\n", + "Gr=Hr-T*Sr/1000.\n", + "#results\n", + "print '%s %.2f %s' %('Gibbs energy =',Gr,'kJ/mol')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Gibbs energy = -237.13 kJ/mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E3 - Pg 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the change in gibbs energy\n", + "#Initialization of variables\n", + "import math\n", + "aADP=1 #mol/L\n", + "aP=1 #mol/L\n", + "aATP=1 #mol/L\n", + "aH2O=1 #mol/L\n", + "aH=math.pow(10,-7) #mol/L\n", + "G=10 #kJ/mol\n", + "T=298. #K\n", + "R=8.314 #J/K mol\n", + "#calculations\n", + "Q=aADP*aP*aH/(aATP*aH2O)\n", + "Gr=G+R*T*math.log(Q)/1000.\n", + "#results\n", + "print '%s %.1f %s' %('Change in nGibbs energy =',Gr,'kJ/mol')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in nGibbs energy = -29.9 kJ/mol\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E4 - Pg 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the equivalent fration\n", + "#Initialization of variables\n", + "Gr=1.7*1000 #J/mol\n", + "T=298. #K\n", + "R=8.314 #J/K mol\n", + "K=0.5\n", + "#calculations\n", + "GbyRT=Gr/(R*T)\n", + "feq=K/(K+1)\n", + "#results\n", + "print '%s %.2f' %(\"Equivalent fraction = \",feq)\n", + "print '%s' %(\"For the second part, Gr=1.7 + 2.48 ln(f/1-f)\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent fraction = 0.33\n", + "For the second part, Gr=1.7 + 2.48 ln(f/1-f)\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E5 - Pg 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the pressure of N2,H2,NH3 gases\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "species=(['N2', 'H2', 'NH3'])\n", + "change=(['-x', '-3x', '2x'])\n", + "E=(['1-x', '3-3x', '2x'])\n", + "print '%s' %(\"Concentration table\")\n", + "print '%s' %(\"species\")\n", + "print '%s' %(\"change\")\n", + "print(E)\n", + "K=977.\n", + "#Calculations\n", + "g=math.sqrt(27*K/4.)\n", + "vector=([g, -(2*g +1), g])\n", + "sol=numpy.roots(vector)[1]\n", + "\n", + "PN2=1-sol\n", + "PH2=3-3*sol\n", + "PNH3=2*sol\n", + "K=math.pow(PNH3,2)/(math.pow(PH2,3) *PN2)\n", + "#results\n", + "print '%s %.2f %s' %(\"Pressure of N2 gas = \",PN2,\"bar\")\n", + "print '%s %.2f %s' %(\"\\n Pressure of H2 gas =\",PH2,\"bar\")\n", + "print '%s %.2f %s' %(\"\\n Pressure of NH3 gas =\",PNH3,\"bar\")\n", + "print '%s %.2e %s' %(\"\\n K final =\",K,\"> it is close to original value.\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Concentration table\n", + "species\n", + "change\n", + "['1-x', '3-3x', '2x']\n", + "Pressure of N2 gas = 0.10 bar\n", + "\n", + " Pressure of H2 gas = 0.31 bar\n", + "\n", + " Pressure of NH3 gas = 1.79 bar\n", + "\n", + " K final = 9.77e+02 > it is close to original value.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example I2 - Pg 148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the equilibrium constant\n", + "#Initialization of variables\n", + "import math\n", + "Gr=-3.40 #kJ/mol\n", + "R=8.314 #J/k mol\n", + "T=298. #K\n", + "#calculations\n", + "lnK=Gr*1000./(R*T)\n", + "K=math.exp(lnK)\n", + "#results\n", + "print '%s %.2f' %('Equilibrium constant K= ',K)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equilibrium constant K= 0.25\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example I3 - Pg 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the decomposition temperature\n", + "#Initialization of variables\n", + "Hr=178. #kJ/mol\n", + "Sr=161. #J/K mol\n", + "#calculations\n", + "T=Hr*1000 /Sr\n", + "#results\n", + "print '%s %.2e %s' %(\"Decompostion temperature =\",T,\"K\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Decompostion temperature = 1.11e+03 K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example I4 - Pg 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the standard reaction gibbs energy\n", + "#Initialization of variables\n", + "GCO2=-394. #kJ/mol\n", + "GCO=-137. #kJ/mol\n", + "GO2=0\n", + "#calculations\n", + "deltaG=2*GCO2-2*GCO+GO2\n", + "#results\n", + "print '%s %d %s' %('Standard reaction gibbs energy =',deltaG,' kJ/mol')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Standard reaction gibbs energy = -514 kJ/mol\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit