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-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:b7db16137d946fe9cbf59acb0ec0858aab53c92eb7e66cd0a1bfe1d955619ce0"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter 6 - Chemical Equilibrium"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 1 - pg 353"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the value of Kc\n",

-      "#initialisation of variables\n",

-      "d= 3.880 #g l^-1\n",

-      "M= 208.3 #gm\n",

-      "P= 1 #atm\n",

-      "R= 0.08205 #cal/mol K\n",

-      "T= 473.1 #K\n",

-      "#CALCULATIONS\n",

-      "d1= M*P/(R*T)\n",

-      "d2= (d1-d)/d\n",

-      "Kp= d2**2/(1-d2**2)\n",

-      "Kc= Kp/(R*T)\n",

-      "#RESULTS\n",

-      "print '%s %.3e %s' %(' Kc =',Kc,'moles l^-1')\n",

-      "print '%s %.4f %s' %(' Kp =',Kp,'atm')\n",

-      "print '%s %.4f' %('Fraction dissociated = ',d2)\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Kc = 4.429e-03 moles l^-1\n",

-        " Kp = 0.1719 atm\n",

-        "Fraction dissociated =  0.3830\n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 2 - pg 353"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the percentage of Pcl5 dissociated\n",

-      "#initialisation of variables\n",

-      "import math\n",

-      "P= 10 #atm\n",

-      "Kp= 0.1719\n",

-      "#CALCULATIONS\n",

-      "a= math.sqrt(Kp/(10+Kp))*100\n",

-      "#RESULTS\n",

-      "print '%s %.3f %s' % (' percentage =',a,'percent')"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " percentage = 13.000 percent\n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 3 - pg 354"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the value of Kp\n",

-      "#initialisation of variables\n",

-      "P= 0.3429 #atm\n",

-      "p0= 0.3153 #atm\n",

-      "#CALCULATIONS\n",

-      "Kp= (2*(P-p0))**2/(2*p0-P)\n",

-      "#RESULTS\n",

-      "print '%s %.2e %s' % (' Kp =',Kp,'atm')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Kp = 1.06e-02 atm\n"

-       ]

-      }

-     ],

-     "prompt_number": 3

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 4 - pg 355"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the pressure required\n",

-      "#initialisation of variables\n",

-      "Kp= 1.06*10**-2 #atm\n",

-      "a= 0.990\n",

-      "#CALCULATIONS\n",

-      "P= Kp*(1-a**2)/(4*a**2)\n",

-      "#RESULTS\n",

-      "print '%s %.2e %s' % (' pressure =',P,' atm')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " pressure = 5.38e-05  atm\n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 5 - pg 357"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the pressure required\n",

-      "#initialisation of variables\n",

-      "G= 2054.7 #cal\n",

-      "R= 1.9872 #cal/mol K\n",

-      "T= 298.16 #K\n",

-      "#CALCULATIONS\n",

-      "P= 10**(-G/(2.303*T*R))\n",

-      "#RESULTS\n",

-      "print '%s %.5f %s' % (' pressure =',P,'atm')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " pressure = 0.03120 atm\n"

-       ]

-      }

-     ],

-     "prompt_number": 5

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 7 - pg 359"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the solubility product constant\n",

-      "#initialisation of variables\n",

-      "T= 25 #C\n",

-      "H= 25.31 #cal\n",

-      "H1= -40.02 #cal\n",

-      "H2= -30.36 #cal\n",

-      "S1= 17.67 #cal deg^-1\n",

-      "S2= 13.17 #cal deg^-1\n",

-      "S3= -22.97 #cal deg^-1\n",

-      "R= 1.987 #cal/mol K\n",

-      "#CALCULATIONS\n",

-      "H3= (H+H1-H2)*1000\n",

-      "S4= S1+S2+S3\n",

-      "G= H3-(273.2+T)*S4\n",

-      "Ka= 10**(-G/(2.303*R*(273.2+T)))\n",

-      "#RESULTS\n",

-      "print '%s %.1e' %(' solubility product constant = ',Ka)"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " solubility product constant =  1.8e-10\n"

-       ]

-      }

-     ],

-     "prompt_number": 6

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 8 - pg 359"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the increase in free energy\n",

-      "#initialisation of variables\n",

-      "import math\n",

-      "T= 25 #C\n",

-      "H= -36430 #cal\n",

-      "S= -4.19 #cal deg^-1\n",

-      "a= 0.1\n",

-      "f= 0.2\n",

-      "R= 1.987 #cal/mol K\n",

-      "#CALCULATIONS\n",

-      "G= H-(273.2+T)*S\n",

-      "Q= a*f/a**2\n",

-      "G1= G+R*(273.2+T)*math.log(Q)\n",

-      "#RESULTS\n",

-      "print '%s %.1f %s' % (' increase in free energy =',G1, 'cal')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " increase in free energy = -34769.8 cal\n"

-       ]

-      }

-     ],

-     "prompt_number": 7

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 9 - pg 361"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the free energy of formation\n",

-      "#initialisation of variables\n",

-      "H= 21600. #cal\n",

-      "S= 50.339 #cal\n",

-      "S1= 49.003 #cal\n",

-      "S2= 45.767 #cal\n",

-      "T= 298.2 #K\n",

-      "#CALCULATIONS\n",

-      "H1= 2*H\n",

-      "S1= 2*S-S1-S2\n",

-      "G= H1-T*S1\n",

-      "Gj= G/(2*1000)\n",

-      "#RESULTS\n",

-      "print '%s %.3f %s' % (' free energy of formation =',Gj,'kcal')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " free energy of formation = 20.719 kcal\n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 10 - pg 361"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the range of humidity\n",

-      "#initialisation of variables\n",

-      "R= 1.987 #cal/mol K\n",

-      "T= 25. #C\n",

-      "G1= -193.8 #cal\n",

-      "G2= -54.6 #cal\n",

-      "G3= -253.1 #cal\n",

-      "G4= -253.1 #cal\n",

-      "G5= -54.6 #cal\n",

-      "G6= -309.7 #cal\n",

-      "#CALCULATIONS\n",

-      "G= G1+G2-G3\n",

-      "Ph= 10**(-G*10**3/(2.303*R*(273.2+T)))\n",

-      "G0= G4+G5-G6\n",

-      "Ph1= 10**(-G0*10**3/(2.303*R*(273.2+T)))\n",

-      "p= Ph*100./Ph1\n",

-      "#RESULTS\n",

-      "print '%s %.2f %s' %(' range of humidity =',p,'percent')\n",

-      "print 'The answers are a bit different due to rounding off error in textbook'"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " range of humidity = 1.05 percent\n",

-        "The answers are a bit different due to rounding off error in textbook\n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 11 - pg 362"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the Ksp\n",

-      "#initialisation of variables\n",

-      "m= 10**-2\n",

-      "m1= 10**-22\n",

-      "G= -22.15 #kcal\n",

-      "G1= -5.81 #kcal\n",

-      "G2= 20.6 #kcal\n",

-      "T= 25 #C\n",

-      "R= 1.987 #cal/mol K\n",

-      "#CALCULATIONS\n",

-      "G3= G-(G1+G2)\n",

-      "Ksp= 10**(G3*10**3/(2.303*R*(273+T)))\n",

-      "#RESULTS\n",

-      "print '%s %.0e' %(' Ksp = ',Ksp)\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Ksp =  8e-28\n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 12 - pg 366"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the heat of dissociation and standard free energy of iodine\n",

-      "import numpy\n",

-      "import math\n",

-      "import matplotlib\n",

-      "from matplotlib import pyplot\n",

-      "import warnings\n",

-      "#initialization of variables\n",

-      "R=1.987\n",

-      "T=1000 #K\n",

-      "T=numpy.array([973.,1073.,1173.,1274.])\n",

-      "kp=numpy.array([.175e-2,1.108e-2,4.87e-2,17.05e-2])\n",

-      "#calculations\n",

-      "Tx=1000./T\n",

-      "logkp=numpy.log10(kp)\n",

-      "slope, intercept = numpy.polyfit(Tx,logkp,1)\n",

-      "dH=-slope*2.303*R*1000.\n",

-      "dH0=dH-1.5*R*T\n",

-      "dG1=-R*T*logkp[1]*2.303\n",

-      "dGt=28720 #cal\n",

-      "dGI=(dGt/1000. + 4.63)/2\n",

-      "#results\n",

-      "print '%s %d %s' %('Heat of dissociation = ',dH,'cal')\n",

-      "print '%s %.2f %s' %('standard free energy of iodine atom = ',dGI,'kcal/mol')\n",

-      "pyplot.plot(Tx,logkp)\n",

-      "pyplot.xlabel('1000/T')\n",

-      "pyplot.ylabel('log Kp')\n",

-      "pyplot.title('Logarithm of Kp for dissociation of Iodine as a function of reciprocal temperature')\n",

-      "pyplot.show()\n",

-      "warnings.filterwarnings(\"ignore\")\n",

-      "print 'The answers are a bit different due to rounding off error in textbook'"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Heat of dissociation =  37460 cal\n",

-        "standard free energy of iodine atom =  16.68 kcal/mol\n"

-       ]

-      },

-      {

-       "metadata": {},

-       "output_type": "display_data",

-       "png": 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LJbr3Xjj+eNhlF7jwQlh22awjEpmfTr+npyoqdTPbi1Ct/w04EpgITDWzvcxs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\nxbe7AS8WGex5YG0z6wF8BOw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-       "text": [

-        "<matplotlib.figure.Figure at 0x70f3750>"

-       ]

-      },

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "The answers are a bit different due to rounding off error in textbook\n"

-       ]

-      }

-     ],

-     "prompt_number": 20

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 13 - pg 361"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the Partial pressure\n",

-      "#initialisation of variables\n",

-      "import math\n",

-      "T= 2000 #K\n",

-      "P= 1 #atm\n",

-      "G= 41438 #cal\n",

-      "R= 1.987 #cal/mol K\n",

-      "T2= 298.2 #K\n",

-      "T1= 2000 #K\n",

-      "H= 43200 #cal\n",

-      "#CALCULATIONS\n",

-      "Kp= 10**(-G/(2.303*R*T2))\n",

-      "Kp1= Kp*10**(H*(T-T2)/(2.303*R*T1*T2))\n",

-      "p= math.sqrt(Kp1*0.8*0.2)\n",

-      "#RESULTS\n",

-      "print '%s %.1e %s' % (' Partial pressure of NO =',p,'atm ')"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Partial pressure of NO = 7.7e-03 atm \n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 15 - pg 368"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the Temperature required\n",

-      "#initialisation of variables\n",

-      "G0 = 0 #cal\n",

-      "G= 13200. #cal\n",

-      "T1= 298.2\n",

-      "H1= 23100. #cal\n",

-      "#CALCULATIONS\n",

-      "T= H1/((H1/T1)-(G/T1)+(G0/T1))\n",

-      "#RESULTS\n",

-      "print '%s %.1f %s' % (' Temperature =',T,' K ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Temperature = 695.8  K \n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 16 - pg 373"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the Equilibrium constant\n",

-      "#initialisation of variables\n",

-      "T= 2000 #K\n",

-      "R= 1.987 #cal /mol K\n",

-      "G= 31160 #cal\n",

-      "#CALULATIONS\n",

-      "Kp= 10**(-G/(2.303*R*T))\n",

-      "#RESULTS\n",

-      "print '%s %.2e' % ('Equilibrium constant =',Kp  )\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "Equilibrium constant = 3.94e-04\n"

-       ]

-      }

-     ],

-     "prompt_number": 14

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 17 - pg 373"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the fraction of methane decomposed\n",

-      "#initialisation of variables\n",

-      "p= 0.08 #atm\n",

-      "#CALCULATIONS\n",

-      "a= (1-p)/(p+1)\n",

-      "#RESULTS\n",

-      "print '%s %.2f' % ('fraction =  ',a)\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "fraction =   0.85\n"

-       ]

-      }

-     ],

-     "prompt_number": 15

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 18 - pg 374"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the enthalpy of the reaction\n",

-      "#initialisation of variables\n",

-      "H= -57240. #cal\n",

-      "T= 2257. #C\n",

-      "Hh= -54.60 #cal\n",

-      "Ho= -38.56 #cal\n",

-      "HO= -57.08 #cal\n",

-      "#CALCULATIONS\n",

-      "H1= H-T*(2*Hh-2*Ho-HO)\n",

-      "#RESULTS\n",

-      "print '%s %.1f %s' % (' Enthalpy =',H1,'cal')\n",

-      "print 'The answers in the textbook are a different due to a rounding off error '"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Enthalpy = -113665.0 cal\n",

-        "The answers in the textbook are a different due to a rounding off error \n"

-       ]

-      }

-     ],

-     "prompt_number": 16

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example 19 - pg 375"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "#calculate the Enthalpy\n",

-      "#initialisation of variables\n",

-      "H= -57797 #cal\n",

-      "T= 25 #C\n",

-      "Hh= 7.934 #cal\n",

-      "Ho= -6.788 #cal\n",

-      "HO= 6.912 #cal\n",

-      "#CALCULATIONS\n",

-      "H1= 2*H-(T+273.16)*(2*Hh+2*Ho-HO)\n",

-      "#RESULTS\n",

-      "print '%s %.1f %s' % (' Enthalpy =',H1,'cal ')\n",

-      "print 'The answers in the textbook are a different due to a rounding off error '"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Enthalpy = -114216.5 cal \n",

-        "The answers in the textbook are a different due to a rounding off error \n"

-       ]

-      }

-     ],

-     "prompt_number": 17

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
\ No newline at end of file