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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e681b00ee14cb4ba49bab5814ba58d8e293c97db9c72b39fdd9bbdd0d2223049"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Balances on Reactive Processes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9-1, page no. 443"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "ndot=2400.0 #mol/s\n",
+      "Hr1= -2878 #Kj/mol\n",
+      "HvWater=44.0 #Kj/mol\n",
+      "HvButane=19.2 #Kj/mol\n",
+      "\n",
+      "#Calculations and print '%s %.3f' %ing :\n",
+      "\n",
+      "print (\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print (\"Part 1\")\n",
+      "E1= ndot/4\n",
+      "deltaH1=E1*Hr1\n",
+      "print '%s %.3E' %(\"enthalpy change (KJ/s) = \",deltaH1)\n",
+      "print (\"part2\")\n",
+      "Hr2=2*Hr1\n",
+      "E2=ndot/8\n",
+      "deltaH2=E2*Hr2\n",
+      "print '%s %.3E' %(\"Enthalpy change (kJ/s) = \",deltaH2)\n",
+      "print (\"part 3\")\n",
+      "Hr3=Hr1+5*HvWater+HvButane\n",
+      "deltaH3=E1*Hr3\n",
+      "print '%s %.3E' %(\"Enthalpy change (kJ/s) = \",deltaH3)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        "Part 1\n",
+        "enthalpy change (KJ/s) =  -1.727E+06\n",
+        "part2\n",
+        "Enthalpy change (kJ/s) =  -1.727E+06\n",
+        "part 3\n",
+        "Enthalpy change (kJ/s) =  -1.583E+06\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1-2, page no. 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "deltaHr= -420.8 #kj/mol\n",
+      "R=8.314\n",
+      "T=298.0 #K\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"From reaction, only gaseous are counted\")\n",
+      "left=1+2\n",
+      "right=1+1\n",
+      "deltaUr=deltaHr-R*T*(right-left)/math.pow(10,3)\n",
+      "print '%s %.3f' %(\"deltaUr (KJ/mol) = \",deltaUr)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        "From reaction, only gaseous are counted\n",
+        "deltaUr (KJ/mol) =  -418.322\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 2,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3-1, page no. 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "HCO2= -393.5 #KJ/mol\n",
+      "HH2O= -285.84 #KJ/mol\n",
+      "HC5H12= -173. #KJ/mol\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "Hr=5*HCO2+6*HH2O-HC5H12\n",
+      "print '%s %.3f' %(\" \\n Heat of the rxn (KJ/mol) = \",Hr)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        " \n",
+        " Heat of the rxn (KJ/mol) =  -3509.540\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 3,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4-1, page no. 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "Hethane= -1559.9 #Kj/mol\n",
+      "Hethene= -1411 #Kj/mol\n",
+      "Hhydrogen= -285.84 #Kj/mol\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "Hr=Hethane-Hethene-Hhydrogen\n",
+      "print '%s %.3f' %(\" \\n Heat of the rxn (KJ/mol) = \",Hr)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        " \n",
+        " Heat of the rxn (KJ/mol) =  136.940\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 4,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5-1, page no. 453"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "nNH3=100.0 #mol/s\n",
+      "nO2in=200.0 #mol/s\n",
+      "H1=8.470 #Kj/mol\n",
+      "H3=9.570 #Kj/mol\n",
+      "T1=25.0\n",
+      "T2=300.0\n",
+      "Hr= -904.7 #Kj/mol\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "def fun(T):\n",
+      "    fun=29.50*math.pow(10,-3)+ T*0.8188*math.pow(10,-5) - math.pow(T,2) * 0.2925 *math.pow(10,-8) + math.pow(T,3) * 0.3652 * math.pow(10,-12)\n",
+      "    return fun\n",
+      "\n",
+      "H2, err=scipy.integrate.quad(fun,T1,T2)  #scipy.integrate.quad is an inbuilt function which can calculate definite integrals\n",
+      "E=nNH3/4.\n",
+      "nO2out=nO2in-nNH3*5./4.\n",
+      "nNO=nNH3\n",
+      "nH2O=nNH3*6./4.\n",
+      "deltaH=E*Hr+(nO2out*H1+nNO*H2+nH2O*H3)\n",
+      "Qdot=deltaH\n",
+      "print '%s %.3f' %(\" \\n Heat Transferred (kW) = \",Qdot)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        " \n",
+        " Heat Transferred (kW) =  -19701.467\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 5,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5-2, page no. 454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "import scipy\n",
+      "from scipy import integrate\n",
+      "\n",
+      "NinMethane=100.0 #mol\n",
+      "NinOxygen=100.0 #mol\n",
+      "NinNitrogen=376.0 #mol\n",
+      "NoutMethane=60.0 #mol\n",
+      "NoutOxygen=50.0 #mol\n",
+      "NoutNitrogen=376.0 #mol\n",
+      "NoutFormal=30.0 #mol\n",
+      "NoutCarbon=10.0 #mol\n",
+      "NoutWater=50.0 #mol\n",
+      "H1= -74.85 #Kj/mol\n",
+      "H2= 2.235 #Kj/mol\n",
+      "H3= 2.187 #Kj/mol\n",
+      "H5=3.758#Kj/mol\n",
+      "H6=3.655 #Kj/mol\n",
+      "H8= -393.5+4.75 #Kj/mol\n",
+      "H9= -241.83+4.27 #Kj/mol\n",
+      "T1=25.0 #C\n",
+      "T2=150.0 #C\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "def fun1(T):\n",
+      "    fun1=34.31*math.pow(10,-3)+ T*5.469*math.pow(10,-5) + math.pow(T,2) * 0.3661 *math.pow(10,-8) - math.pow(T,3) * 11 * math.pow(10,-12)\n",
+      "    return fun1\n",
+      "\n",
+      "HoutMethane, err=scipy.integrate.quad(fun1,T1,T2)\n",
+      "H4= -74.85 + HoutMethane\n",
+      "def fun2(T):\n",
+      "    fun2=34.28*math.pow(10,-3)+ T*4.268*math.pow(10,-5) - math.pow(T,3) * 8.694 * math.pow(10,-12)\n",
+      "    return fun2\n",
+      "\n",
+      "HoutFormal, err2=scipy.integrate.quad(fun2,T1,T2) #scipy.integrate.quadis an inbuilt function which can calculate definite integrals\n",
+      "H7= -115.90+ HoutFormal\n",
+      "deltaH=NoutWater*H9+NoutCarbon*H8+NoutFormal*H7+NoutNitrogen*H6+NoutOxygen*H5+NoutMethane*H4-NinNitrogen*H3-NinOxygen*H2-NinMethane*H1\n",
+      "Q=deltaH\n",
+      "print '%s %.3f' %(\" \\n Q (KJ) = \",Q)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        " \n",
+        " Q (KJ) =  -15296.233\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 6,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5-4, page no. 458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "basis=150.0 #mol/s\n",
+      "x=0.9\n",
+      "HinEthanol= -212.19\n",
+      "HinEthanone= -147.07\n",
+      "HoutEthanol= -216.81\n",
+      "HoutEthanone= -150.9\n",
+      "HoutHydrogen=6.595\n",
+      "NinEthanol=135.0\n",
+      "NinEthanone=15.0\n",
+      "Q=2440.0 #KW\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Carbon Balance\")\n",
+      "print(\"basis*x *2 + basis* (1-x) *2=2 n1+2 n2\")\n",
+      "print(\"Hydrogen Balance\")\n",
+      "print(\"basis * x *6 + basis * (1-x)*4 = 6 n1+4 n2+2 n3\")\n",
+      "print(\"Energy Balance\")\n",
+      "print(\"Q= HoutEthanol n1  HoutEthanone n2 + HoutHydrogen n3 -NinEthanol * HinEthanol -NinEthanone* HinEthanone\")\n",
+      "A=([[1, 1, 0],[3, 2, 1],[216.81, 150.9, -6.595]])\n",
+      "b=([[150],[435],[28412]])\n",
+      "C=numpy.dot(linalg.inv(A),b)\n",
+      "n1=C[0,0]\n",
+      "print '%s %.3f' %(\" \\n n1 (mol Ethanol/s) = \",n1)\n",
+      "n2=C[1,0]\n",
+      "print '%s %.3f' %(\" \\n n2 (mol Ethanone/s) = \",n2)\n",
+      "n3=C[2,0]\n",
+      "print '%s %.3f' %(\" \\n n3 (mol Hydrogen/s) = \",n3)\n",
+      "print(\"The solutions in the Text are Wrong\")\n",
+      "fraction=(NinEthanol-n1)/NinEthanol\n",
+      "print '%s %.3f' %(\"Fractional conversion of Ethanol = \",fraction)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        "Carbon Balance\n",
+        "basis*x *2 + basis* (1-x) *2=2 n1+2 n2\n",
+        "Hydrogen Balance\n",
+        "basis * x *6 + basis * (1-x)*4 = 6 n1+4 n2+2 n3\n",
+        "Energy Balance\n",
+        "Q= HoutEthanol n1  HoutEthanone n2 + HoutHydrogen n3 -NinEthanol * HinEthanol -NinEthanone* HinEthanone\n",
+        " \n",
+        " n1 (mol Ethanol/s) =  91.957\n",
+        " \n",
+        " n2 (mol Ethanone/s) =  58.043\n",
+        " \n",
+        " n3 (mol Hydrogen/s) =  43.043\n",
+        "The solutions in the Text are Wrong\n",
+        "Fractional conversion of Ethanol =  0.319\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 7,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5-5, page no. 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "print(\"from tables,\")\n",
+      "HfAcid= -1294. #Kj/mol\n",
+      "HfBase= -469.1 #Kj/mol\n",
+      "HfSalt= -1974. #Kj/mol \n",
+      "HfWater= -285.8 #Kj/mol\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"part 1\")\n",
+      "Hr=HfSalt+3*HfWater-HfAcid-3*HfBase\n",
+      "print '%s %.3f' %(\"Hr of the rxn (KJ/mol) = \",Hr)\n",
+      "print(\"part 2\")\n",
+      "deltaH=Hr*5./3.\n",
+      "print '%s %.3f' %(\"deltaH (KJ) = \",deltaH)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "from tables,\n",
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        "part 1\n",
+        "Hr of the rxn (KJ/mol) =  -130.100\n",
+        "part 2\n",
+        "deltaH (KJ) =  -216.833\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 8,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5-6, page no. 461"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "x=0.1\n",
+      "y=0.2\n",
+      "MAcid=98.1\n",
+      "MS=32.0\n",
+      "MSalt=142.0\n",
+      "MBase=40.0\n",
+      "MWater=18.0\n",
+      "MNa=46.0\n",
+      "basis=1000.0 #g\n",
+      "T2=35.0\n",
+      "T1=25.0\n",
+      "T3=40.0\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "print(\"Using S balance, \")\n",
+      "m2=basis*x*MS*MSalt/(MAcid*MS)\n",
+      "print '%s %.3f' %(\" \\n m2  (g Na2SO4) = \",m2)\n",
+      "print(\"Using Na balance, \")\n",
+      "m1=2*MNa*m2*MBase/(y*MNa*MSalt)\n",
+      "print '%s %.3f' %(\" \\n m1  (g NaOH) = \",m1)\n",
+      "print(\"Total mass balance, \")\n",
+      "m3=basis+m1-m2\n",
+      "print '%s %.3f' %(\" \\n m3  (g H2O) = \",m3)\n",
+      "print '%s %.3f' %(\" \\n Mass of product solution = \",m2+m3)\n",
+      "m=m2+m3\n",
+      "Water=m2*2/MSalt\n",
+      "print '%s %.3f' %(\" \\n Water Formed in the reaction  (mol H2O) = \",Water)\n",
+      "print(\"H2SO4(aq):\")\n",
+      "a1=basis*(1-x)/MWater\n",
+      "b1=basis*x/MAcid\n",
+      "rAcid=a1/b1\n",
+      "print '%s %.3f' %(\" \\n rAcid  (mol Water/mol Acid) = \",rAcid)\n",
+      "print(\"NaOH(aq):\")\n",
+      "a2=m1*(1-y)/MWater\n",
+      "b2=m1*y/MBase\n",
+      "rBase=a2/b2\n",
+      "print '%s %.3f' %(\" \\n rBase  (mol Water/mol Base) = \",rBase)\n",
+      "print(\"Na2SO4(aq):\")\n",
+      "a3=m3/MWater\n",
+      "b3=m2/MSalt\n",
+      "rSalt=a3/b3\n",
+      "print '%s %.3f' %(\" \\n rSalt  (mol Water/mol Salt) =\",rSalt)\n",
+      "E=b1\n",
+      "print '%s %.3f' %(\" \\n Extent of reaction  (mol) = \",E)\n",
+      "nHAcid=basis*3.85*(T3-T1)/1000\n",
+      "nHSalt=m*4.184*(T2-T1)/1000\n",
+      "nHBase=0\n",
+      "HfSalt= -1384\n",
+      "HfAcid= -884.6\n",
+      "HfBase= -468.1\n",
+      "HfWater= -285.84\n",
+      "deltaHr=HfSalt+ 2*HfWater - HfAcid - 2*HfBase\n",
+      "print '%s %.3f' %(\" \\n Entahlpy change in the rxn  (KJ/mol) = \",deltaHr)\n",
+      "Q=E*deltaHr + (nHSalt-nHAcid-nHBase)\n",
+      "print '%s %.3f' %(\" \\n Q of the rxn  (KJ) = \",Q)\n",
+      "print(\"The answer in the Text is wrong.\")\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        "Using S balance, \n",
+        " \n",
+        " m2  (g Na2SO4) =  144.750\n",
+        "Using Na balance, \n",
+        " \n",
+        " m1  (g NaOH) =  407.747\n",
+        "Total mass balance, \n",
+        " \n",
+        " m3  (g H2O) =  1262.997\n",
+        " \n",
+        " Mass of product solution =  1407.747\n",
+        " \n",
+        " Water Formed in the reaction  (mol H2O) =  2.039\n",
+        "H2SO4(aq):\n",
+        " \n",
+        " rAcid  (mol Water/mol Acid) =  49.050\n",
+        "NaOH(aq):\n",
+        " \n",
+        " rBase  (mol Water/mol Base) =  8.889\n",
+        "Na2SO4(aq):\n",
+        " \n",
+        " rSalt  (mol Water/mol Salt) = 68.833\n",
+        " \n",
+        " Extent of reaction  (mol) =  1.019\n",
+        " \n",
+        " Entahlpy change in the rxn  (KJ/mol) =  -134.880\n",
+        " \n",
+        " Q of the rxn  (KJ) =  -136.342\n",
+        "The answer in the Text is wrong.\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 9,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6-1, page no. 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "#Initialization of variables\n",
+      "import math\n",
+      "import numpy\n",
+      "from numpy import linalg\n",
+      "\n",
+      "x=0.85\n",
+      "H1= 802 #KJ/mol\n",
+      "H2= 1428 #kJ/mol\n",
+      "M1=16.0\n",
+      "M2=30.0\n",
+      "\n",
+      "#Calculations and printing :\n",
+      "\n",
+      "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n",
+      "y1=x*16\n",
+      "y2=(1-x)*30\n",
+      "xCH4=y1/(y1+y2)\n",
+      "HHVMethane=(H1+ 2*44.013)/M1\n",
+      "HHVEthane=(H2+ 3*44.013)/M2\n",
+      "HHV=xCH4*HHVMethane + (1-xCH4)*HHVEthane\n",
+      "print '%s %.3f' %(\" \\n HHV of Fuel (KJ/g) = \",HHV)\n",
+      "raw_input('press enter key to exit')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n",
+        " \n",
+        " HHV of Fuel (KJ/g) =  54.725\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "press enter key to exit\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 10,
+       "text": [
+        "''"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file