From c7fe425ef3c5e8804f2f5de3d8fffedf5e2f1131 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Physical_Chemsitry/Chapter_10.ipynb | 635 ++++++++++++++++++++++++++++++++++++ 1 file changed, 635 insertions(+) create mode 100755 Physical_Chemsitry/Chapter_10.ipynb (limited to 'Physical_Chemsitry/Chapter_10.ipynb') diff --git a/Physical_Chemsitry/Chapter_10.ipynb b/Physical_Chemsitry/Chapter_10.ipynb new file mode 100755 index 00000000..4ee0e2f1 --- /dev/null +++ b/Physical_Chemsitry/Chapter_10.ipynb @@ -0,0 +1,635 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a8663b753e53365cf46aa7f5948fd23b365bccc405a9c5b9305fa79e49b0f6dc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 - Chemical Kinetics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1 - pg 543" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Pressure \n", + "#initialisation of variables\n", + "t= 3 #sec\n", + "P0= 200 #mm\n", + "k= 17.3 #mm/sec\n", + "P1= 104 #mm\n", + "#CALCULATIONS\n", + "P= P0-k*t\n", + "P2= P+P1\n", + "#RESULTS\n", + "print '%s %d %s' % (' Pressure=',P2,' mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure= 252 mm of Hg\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2 - pg 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Half time\n", + "#initialisation of variables\n", + "k= 2.63*10**-3 #min^-1\n", + "#CALCULATIONS\n", + "t1= 0.693/k\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' Half time=',t1,'min')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Half time= 263.5 min\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3 - pg 546" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Partial Pressure of the reactant\n", + "#initialisation of variables\n", + "P= 200. #mm\n", + "t= 30. #min\n", + "k= 2.5*10**-4 #sec^-1\n", + "#CALCULATIONS\n", + "P0= P/(10**(k*t*60/2.303))\n", + "P1= P-P0\n", + "#RESULTS\n", + "print '%s %d %s' % (' Partial Pressure of reactant=',P1,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Partial Pressure of reactant= 72 mm of Hg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4 - pg 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the no of atoms\n", + "#initialisation of variables\n", + "t= 5600*365*24*60.\n", + "x= 5 #atoms\n", + "#CALCULATIONS\n", + "k= 0.693/t\n", + "N= x/k\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' No of atoms=',N, 'atoms')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " No of atoms= 2.12e+10 atoms\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5 - pg 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the time passed\n", + "#initialisation of variables\n", + "import math\n", + "t= 5600 #sec\n", + "r= 0.256\n", + "#CALCULATIONS\n", + "t1= (t/0.693)*2.303*math.log10(1/r)\n", + "#RESULTS\n", + "print '%s %d %s' % (' Time=',t1,'years ago')\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Time= 11012 years ago\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6 - pg 549" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the first order rate constant and half life\n", + "#initialisation of variables\n", + "import math\n", + "t= 25.1 #hr\n", + "C= 0.004366 \n", + "C1= 0.002192\n", + "C2= 0.006649\n", + "#CALCULATIONS\n", + "r= (C-C1)/(C2-C1)\n", + "k= 2.303*math.log10(1/r)/t\n", + "t1= 0.693/k\n", + "#RESULTS\n", + "print '%s %.1f %s' %(' Time=',t1,' hr')\n", + "print '%s %.2e %s' %(' Time=',k,' hr')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Time= 24.2 hr\n", + " Time= 2.86e-02 hr\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7 - pg 552" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Rate constant\n", + "#initialisation of variables\n", + "s= 18.6*10**4 #mm of hg\n", + "#CALCULATIONS\n", + "k= 1./s\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Rate constant=',k,' (mm Hg)^-1 sec^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Rate constant= 5.38e-06 (mm Hg)^-1 sec^-1\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8 - pg 552" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the requried Pressure\n", + "#initialisation of variables\n", + "k= 1.14*10**-4 #sec^-1\n", + "k1= 5.38*10**-6 #sec^-1\n", + "#CALCULATIONS\n", + "P= k/k1\n", + "P2=0.01*P\n", + "#RESULTS\n", + "print '%s %.3f %s' % (' Pressure=',P2,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure= 0.212 mm of Hg\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9 - pg 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the no of molecules\n", + "#initialisation of variables\n", + "T= 600 #K\n", + "P= 1 #atm\n", + "R= 0.082 #atm lit/mol K\n", + "#CALCULATIONS\n", + "C= P/(R*T)\n", + "r= C**2*4*10**-6 \n", + "r1= 6*10**23*r\n", + "#RESULTS\n", + "print '%s %.1e %s' % (' No of molecules=',r1,'molecules l^-1 sec^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " No of molecules= 9.9e+14 molecules l^-1 sec^-1\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10 - pg 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the time required\n", + "#initialisation of variables\n", + "k= 6.3*10**2 #ml mole^-1 sec^-1\n", + "P= 400. #mm\n", + "T= 600. #K\n", + "R= 82.06\n", + "#CALCULATIONS\n", + "C= (P/760.)/(R*T)\n", + "t= 1/(9.*C*k)\n", + "#RESULTS\n", + "print '%s %.1f %s' % (' time=',t,' sec')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " time= 16.5 sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11 - pg 556" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the pressure of No2 in both cases\n", + "#initialisation of variables\n", + "pf2= 2.00 #mm Hg\n", + "y= 0.96 #mm Hg\n", + "Pn= 5 #mm Hg\n", + "#CALCULATIONS\n", + "pF2= pf2-y\n", + "pNO2= Pn-2*y\n", + "pNO2F= 2*y\n", + "#RESULTS\n", + "print '%s %.2f %s' % (' pressure of NO2=',pNO2,'mm of Hg')\n", + "print '%s %.2f %s' % (' \\n pressure of NO2 after 30 sec=',pNO2F,'mm of Hg')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " pressure of NO2= 3.08 mm of Hg\n", + " \n", + " pressure of NO2 after 30 sec= 1.92 mm of Hg\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13 - pg 561" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Rate constant\n", + "#initialisation of variables\n", + "k= 4*10**-6 #mol^-1 sec^-1\n", + "Kc= 73\n", + "#CALCULATIONS\n", + "K1= k*Kc/2\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Rate constant=',K1,'l mol^-1 sec^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Rate constant= 1.46e-04 l mol^-1 sec^-1\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14 - pg 568" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the activation energy\n", + "#initialisation of variables\n", + "import math\n", + "R= 1.987 #atm lit/mol K\n", + "T= 573.2 #K\n", + "T1= 594.6 #K\n", + "k= 3.95*10**-6 #mol^-1 sec^-1\n", + "k1= 1.07*10**-6 #mol^-1 sec^-1\n", + "#CALCULATIONS\n", + "H= R*T*T1*2.303*math.log10((k/k1))/(T1-T)\n", + "#RESULTS\n", + "print '%s %d %s' %(' activation energy=',H,'calmol^-1')\n", + "print 'The answers in the texbook are a bit different due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " activation energy= 41338 calmol^-1\n", + "The answers in the texbook are a bit different due to rounding off error\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15 - pg 568" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the time required\n", + "#initialisation of variables\n", + "import math\n", + "H= 41300. #cal\n", + "T= 673. #K\n", + "T1= 595. #K\n", + "R= 1.987 #cal/mol K\n", + "K= 3.95*10**-6\n", + "P= 1 #atm\n", + "R1= 0.08205 #j/mol K\n", + "#CALCULATIONS\n", + "k2= math.e**(H*(T-T1)/(R*T*T1))*K\n", + "C= P/(R1*T)\n", + "t= 44.8/C\n", + "t2=R1*T*10**-2 /k2\n", + "#RESULTS\n", + "print '%s %d %s' %(' time =',t,'sec')\n", + "print '%s %d %s' %('Time required in case 2 = ',t2,'sec')\n", + "print 'The answers in the texbook are a bit different due to rounding off error'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " time = 2473 sec\n", + "Time required in case 2 = 2438 sec\n", + "The answers in the texbook are a bit different due to rounding off error\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16 - pg 569" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the collision diameter\n", + "#initialisation of variables\n", + "import math\n", + "H= 41300.\n", + "R= 1.987 #atm lit/mol K\n", + "T= 595. #K\n", + "M= 128. #gm\n", + "R1= 8.314*10**7 #atm lit/mol K\n", + "N= 6.02*10**23 #moleccules\n", + "k= 3.95*10**-6 #sec**-1\n", + "#CALCULATIONS\n", + "s= math.sqrt((k*10**3/(4*N))*(128/(math.pi*R1*T))**0.5*math.e**(H/(R*T)))\n", + "#RESULTS\n", + "print '%s %.3e %s' % (' collision diameter=',s,' cm')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " collision diameter= 8.356e-09 cm\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18 - pg 577" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the Concentration of A and B\n", + "#initialisation of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "p= 20.3 #percent\n", + "p1= 1.77 #percent\n", + "I= 100.\n", + "n= 2.\n", + "l= 300. #l mol^-1 cm^-1\n", + "l1= 30. #l mol^-1 cm^-1\n", + "l2= 10. #l mol^-1 cm^-1\n", + "l3= 200. #l mol^-1 cm^-1\n", + "#CALCULATIONS\n", + "A= ([[n*l, n*l1],[n*l2, n*l3]])\n", + "b= ([[math.log10(I/p1)],[math.log10(I/p)]])\n", + "c= numpy.dot(numpy.linalg.inv(A),b)\n", + "R1=c[0]\n", + "R2=c[1]\n", + "#RESULTS\n", + "print '%s %.2e %s' % (' Concentration of A =',R1,' mole l^-1')\n", + "print '%s %.2e %s' % (' \\n Concentration of B =',R2,' mole l^-1')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Concentration of A = 2.76e-03 mole l^-1\n", + " \n", + " Concentration of B = 1.59e-03 mole l^-1\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit