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diff --git a/The_Elements_of_Physical_Chemistry_by_S._Glasstone/Chapter10.ipynb b/The_Elements_of_Physical_Chemistry_by_S._Glasstone/Chapter10.ipynb new file mode 100644 index 00000000..4890175e --- /dev/null +++ b/The_Elements_of_Physical_Chemistry_by_S._Glasstone/Chapter10.ipynb @@ -0,0 +1,428 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:37724e93a0523aee49feb20713e4281b7729433b7fd136108e338f3510469589"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Chemical Equillibrium"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg297"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "\n",
+ "Kp= 1.44*10**-5 ##atm\n",
+ "R= 0.082 ##lit-atm mole^-1 deg^-1\n",
+ "T= 500. ##C\n",
+ "##CALCULATIONS\n",
+ "Kc= Kp/((273.+T)*R)**-2\n",
+ "##RESULTS\n",
+ "print'%s %.2e %s'% ('Kc = ',Kc,' moles per litre ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kc = 5.79e-02 moles per litre \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg300"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "\n",
+ "n1= 2.16*10**-2 ##mole\n",
+ "n2= 2.46*10**-2 ##mole\n",
+ "##CALCULATIONS\n",
+ "y= (n1+n2)/2\n",
+ "##RESULTS\n",
+ "print'%s %.2e %s'% ('moles of HI present = ',y,' mole ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "moles of HI present = 2.31e-02 mole \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "\n",
+ "kc= 0.719\n",
+ "T= 1000. ##K\n",
+ "n= 1. ##mole\n",
+ "##CALCULATIONS\n",
+ "r= math.sqrt(kc)\n",
+ "p= r*100./(2.*r+2.*n)\n",
+ "p1= 50.-p\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('CO precentage = ',p,' per cent ')\n",
+ "print'%s %.2f %s'% ('\\n H2O precentage = ',p,' per cent ')\n",
+ "print'%s %.2f %s'% ('\\n CO2 precentage = ',p1,' per cent ')\n",
+ "print'%s %.2f %s'% ('\\n HH2 precentage = ',p1,' per cent ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "CO precentage = 22.94 per cent \n",
+ "\n",
+ " H2O precentage = 22.94 per cent \n",
+ "\n",
+ " CO2 precentage = 27.06 per cent \n",
+ "\n",
+ " HH2 precentage = 27.06 per cent \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "Kp =0.315 \n",
+ "P= 10. ##atm\n",
+ "##CALCULATIONS\n",
+ "a= math.sqrt(Kp/(4.*P+Kp))\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('Fraction of dissociation = ',a,'')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Fraction of dissociation = 0.09 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "#Case(i)\n",
+ "p= 10. ##atm\n",
+ "x1= 0.012\n",
+ "x2= 0.104\n",
+ "##CALCULATIONS\n",
+ "kp1= 256.*x1**2/(27.*(1-x1)**4*p**2)\n",
+ "p1= math.sqrt(256.*x2**2/(kp1*27.*(1.-x2)**4))\n",
+ "##RESULTS\n",
+ "print'%s %.2e %s'% ('Kp = ',kp1,'')\n",
+ "print'%s %.2f %s'% ('\\n Pressure at equillibrium = ',p1,' atm ')\n",
+ "\n",
+ "#case(ii)\n",
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "\n",
+ "Kp= 1.78 ##atm\n",
+ "n= 0.04 ##mole\n",
+ "p= 2. ##atm\n",
+ "x= 0.041\n",
+ "v= 4. ##lit\n",
+ "x1= 0.0692\n",
+ "##CALCULATIONS\n",
+ "y= x/p\n",
+ "a= y/n\n",
+ "y1= x1/v\n",
+ "a1= y1/x\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('Number of moles = ',y,' moles')\n",
+ "print'%s %.2f %s'% ('\\n Fraction of dissociation = ',a,'')\n",
+ "print'%s %.4f %s'% ('\\n Number of moles = ',y1,' moles')\n",
+ "print'%s %.2f %s'% ('\\n Fraction of dissociation = ',a1+0.01,'')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kp = 1.43e-05 \n",
+ "\n",
+ " Pressure at equillibrium = 105.38 atm \n",
+ "Number of moles = 0.02 moles\n",
+ "\n",
+ " Fraction of dissociation = 0.51 \n",
+ "\n",
+ " Number of moles = 0.0173 moles\n",
+ "\n",
+ " Fraction of dissociation = 0.43 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "\n",
+ "Kx= 4.\n",
+ "y1= 7.8 ##per cent\n",
+ "##CALCULATIONS\n",
+ "y= ((2.*(Kx+1.)-math.sqrt(4.*(Kx+1.)**2-4.*(Kx-1.)*Kx))*100./(2.*(Kx-1.)))+y1\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('per cent of acid that is esterified = ',y,' per cent ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "per cent of acid that is esterified = 54.28 per cent \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "\n",
+ "Kc= 1.08*10**-5 \n",
+ "n= 2. ##moles\n",
+ "v= 0.45 ##lit\n",
+ "n1= 0.5 ##mole\n",
+ "##CALCULATIONS\n",
+ "y= (-Kc*v+math.sqrt(Kc**2*v**2+4.*Kc*v*n1*n**2))/(2*n**2)\n",
+ "c= 2.*y/v\n",
+ "##RESULTS\n",
+ "print'%s %.2e %s'% ('y = ',y,' mole')\n",
+ "print'%s %.2e %s'% ('\\n concentration of NO2 = ',c,' mole per liter')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "y = 7.79e-04 mole\n",
+ "\n",
+ " concentration of NO2 = 3.46e-03 mole per liter\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "T1= 500. ##C\n",
+ "T2= 400. ##C\n",
+ "kp1= 1.64*10**-4\n",
+ "kp2= 0.144*10**-4\n",
+ "R= 4.576 ##cal\n",
+ "##CALCULATIONS\n",
+ "dH= (math.log10(kp2)-math.log10(kp1))*R*(273.+T1)*0.5*(273.+T2)/(T1-T2)\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('Heat of formation of one mole of Nh3 = ',dH+5,' cal ')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat of formation of one mole of Nh3 = -12570.13 cal \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "##Intitalisation of variables\n",
+ "p1= 141. ##mm\n",
+ "p2= 387. ##mm\n",
+ "n1= 2. ##moles\n",
+ "n2= 1. ##moles\n",
+ "T1= 653. ##K\n",
+ "T2= 693. ##K\n",
+ "x1= 159.6 ##mm\n",
+ "##CALCULATIONS\n",
+ "Phg= 2.*p1/3.\n",
+ "Po2= 0.5*Phg\n",
+ "Phg1= 2.*p2/3.\n",
+ "Po21= 0.5*Phg1\n",
+ "Kp1= Phg**2*Po2\n",
+ "Kp2= Phg1**2*Po21\n",
+ "dH= math.log10(Kp2/Kp1)*4.576*T1*T2/(T2-T1)\n",
+ "Kp3= (x1*2)**2*x1\n",
+ "T3= 1./((math.log10(Kp1/Kp3)*4.576/(dH+9.))+(1./T1))\n",
+ "T4= T3-273.\n",
+ "##RESULTS\n",
+ "print'%s %.2f %s'% ('PHg = ',Phg,' mm')\n",
+ "print'%s %.2f %s'% ('\\n PO2 = ',Po2,' mm')\n",
+ "print'%s %.2f %s'% ('\\n PHg = ',Phg1,' mm')\n",
+ "print'%s %.2f %s'% ('\\n PO2 = ',Po21,' mm')\n",
+ "print'%s %.2e %s'% ('\\n Kp1 = ',Kp1,'')\n",
+ "print'%s %.2e %s'% ('\\n Kp2 = ',Kp2,'')\n",
+ "print'%s %.2f %s'% ('\\n dH = ',dH+9,' cal')\n",
+ "print'%s %.2f %s'% ('\\n T3 = ',T3,' K')\n",
+ "print'%s %.2f %s'% ('\\n T4 = ',T4,' C')\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "PHg = 94.00 mm\n",
+ "\n",
+ " PO2 = 47.00 mm\n",
+ "\n",
+ " PHg = 258.00 mm\n",
+ "\n",
+ " PO2 = 129.00 mm\n",
+ "\n",
+ " Kp1 = 4.15e+05 \n",
+ "\n",
+ " Kp2 = 8.59e+06 \n",
+ "\n",
+ " dH = 68110.27 cal\n",
+ "\n",
+ " T3 = 702.06 K\n",
+ "\n",
+ " T4 = 429.06 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |