diff options
Diffstat (limited to 'Physical_Chemsitry/Chapter_10.ipynb')
| -rwxr-xr-x | Physical_Chemsitry/Chapter_10.ipynb | 635 |
1 files changed, 0 insertions, 635 deletions
diff --git a/Physical_Chemsitry/Chapter_10.ipynb b/Physical_Chemsitry/Chapter_10.ipynb deleted file mode 100755 index 4ee0e2f1..00000000 --- a/Physical_Chemsitry/Chapter_10.ipynb +++ /dev/null @@ -1,635 +0,0 @@ -{
- "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 |