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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Chemical Equilibrium"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.1   Page: 311\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find The chemical equilibrium composition of the gaseous mixture contains\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 298.15      #[K] temperature\n",
+      "P = 1.          #[atm] pressure \n",
+      "R = 8.314*10**(-3)      #[kJ/(mol*K)]\n",
+      "\n",
+      "# Calculations\n",
+      "# For an ideal binary solution the Gi# Calculationss free energy is given by\n",
+      "# g_mix = summation(x_i*g_i_0) + R*T*summation(x_i*math.log(x_i))\n",
+      "# Differentiating the above equation with respect to x_a , remembering that for a binary mixture dx_b = dx_a, finding\n",
+      "\n",
+      "# dg_mix/dx_a = g_a_0-g_b_0+R*T*[math.log(x_a)+1-(math.log(x_b)+1)]\n",
+      "# and x_a+x_b = 1\n",
+      "# so\n",
+      "# dg_mix/dx_a = g_a_0-g_b_0+R*T*[math.log(x_a/(1-x_a))]\n",
+      "\n",
+      "# setting up this equal to zero ( to find the minimum on the g-x curve ) and solving gives\n",
+      "# x_a/(1-x_a) = exp((g_b_0-g_a_0)/(R*T))\n",
+      "\n",
+      "# From the table A.8 (page 427) reported in the book, pure component Gibbs free energies for isobumath.tane,a,and n-bumath.tane,b, we find\n",
+      "g_a_0 = -20.9       #[kJ/mol]\n",
+      "g_b_0 = -17.2       #[kJ/mol]\n",
+      "\n",
+      "# Now solving the above equation for x_a, we have\n",
+      "x_a = math.exp((g_b_0-g_a_0)/(R*T))/(1+math.exp((g_b_0-g_a_0)/(R*T)))\n",
+      "x_b = 1-x_a\n",
+      "\n",
+      "# Results\n",
+      "print \" The chemical equilibrium composition of the gaseous mixture contains %f mol fraction isobutane \\t\\t\\t\\t\\t\\t\\t\\tan %f mol fraction n-bumath.tane\"%(x_a,x_b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The chemical equilibrium composition of the gaseous mixture contains 0.816475 mol fraction isobutane \t\t\t\t\t\t\t\tan 0.183525 mol fraction n-bumath.tane\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.2   Page: 319\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "The equilibrium consmath.tant for the reaction\n",
+      "The concentration of NO at equilibrium\n",
+      "The equilibrium constant\n",
+      "The concentration of NO \n",
+      "'''\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 298.15              #[K] temperature\n",
+      "P = 0.987               #[atm] pressure\n",
+      "g_0_NO = 86.6           #[kJ/mol] Free energy of formation the NO from elements\n",
+      "R = 8.314               #[J/(mol*K)]\n",
+      "\n",
+      "# And the corresponding values for the elements N2 and O2 are\n",
+      "g_0_O2 = 0.00\n",
+      "g_0_N2 = 0.00\n",
+      "\n",
+      "# The reaction of the nitrogen and oxygen to form nitric oxide at 298.15 K is\n",
+      "# N2 + O2 = NO\n",
+      "\n",
+      "# Calculations\n",
+      "#  Here \n",
+      "delta_g_0 = 2*g_0_NO - g_0_O2 - g_0_N2          #[kJ/mol]\n",
+      "# Changing in the J/mol \n",
+      "delta_g_01 = delta_g_0*1000                     #[J/mol]\n",
+      "\n",
+      "# hence \n",
+      "K_298 = math.exp((-delta_g_01)/(R*T))\n",
+      "\n",
+      "# The activities are all \n",
+      "# a_i = f_i/f_i_0\n",
+      "# f_i_0 correspond to the smath.radians(numpy.arcmath.tan(ard state, which for gas at idael gas state are \n",
+      "f_0_N2 = 1.             #[bar]\n",
+      "f_0_O2 = 1.             #[bar]\n",
+      "f_0_NO = 1.             #[bar]\n",
+      "\n",
+      "# If we make the most general statement of the activities (for gases ) we would have\n",
+      "# a_i = y_i*v_i*Y_i*P/f_i_0 = y_i*phi*P/f_i_0\n",
+      "\n",
+      "# At this low pressure we may safely asssume that the NO,O2 and N2 behave as ideal gases for which v_i*Y_i = phi = 1.00 and substituting these we find\n",
+      "# K_298 = [a_NO]**(2)/([a_N2]*[a_O2]) = [y_NO]**(2)/([y_N2]*[y_O2])\n",
+      "\n",
+      "# Now umath.sing this equilibrium consmath.tant we can calculare he equilibrium concentratin of NO in the air sample in which \n",
+      "#oxygen = 21%, nitrogen = 78% and argon = 1% ,so\n",
+      "y_N2 = 0.78\n",
+      "y_O2 = 0.21\n",
+      "\n",
+      "# Hence From above expression, we have\n",
+      "y_NO_298 = math.sqrt(K_298*y_N2*y_O2)\n",
+      "\n",
+      "# Making the similar calculations for the temperature 2000 K, we  have\n",
+      "T_1 = 2000                  #[K]\n",
+      "K_2000 = 4.0*10**-4\n",
+      "\n",
+      "# So,\n",
+      "y_NO_2000 = math.sqrt(K_2000*y_N2*y_O2)*10**(6)         #[ppm]\n",
+      "\n",
+      "# Results\n",
+      "print \" The equilibrium constant for the reaction at 298.15 K is \\t\\t\\t %.1e\"%(K_298)\n",
+      "print \" The concentration of NO at equilibrium at temperature 298.15 K is \\t\\t %.1e\"%(y_NO_298)\n",
+      "print \" The equilibrium consmath.tant for the reaction at 2000 K is \\t\\t\\t %.1e\"%(K_2000)\n",
+      "print \" The concentration of NO at equilibrium at temperature 2000 K is \\t\\t %.0f ppm\"%(round(y_NO_2000,-2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The equilibrium constant for the reaction at 298.15 K is \t\t\t 4.5e-31\n",
+        " The concentration of NO at equilibrium at temperature 298.15 K is \t\t 2.7e-16\n",
+        " The equilibrium consmath.tant for the reaction at 2000 K is \t\t\t 4.0e-04\n",
+        " The concentration of NO at equilibrium at temperature 2000 K is \t\t 8100 ppm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.3   Page: 321\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find The calculated NO cocentration\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Temp = 2000.                #[K]\n",
+      "n_air = 1.                  #[mol] no of moles of the air\n",
+      "\n",
+      "# Let the moless of the NO formed be 2*x\n",
+      "# Then at equilibrium the unreacted moles of the N2 and O2 will be (0.78-x) and (0.21-x) respectively\n",
+      "\n",
+      "# Calculations\n",
+      "# from the previous example, we have \n",
+      "# [y_NO]**(2) = K_298*[y_N2]*[y_O2]\n",
+      "# here \n",
+      "K_2000 = 4*10**(-4)\n",
+      "# Substituting all the values, we have \n",
+      "# (2*x)**(2) = K_2000*(0.78-x)*(0.21-x)\n",
+      "\n",
+      "#Now \n",
+      "def f(x): \n",
+      "\t return  (2*x)**(2) - K_2000*(0.78-x)*(0.21-x)\n",
+      "#def f(x): \n",
+      "\t #return  (K_2000-2)*x**(2)-K_2000*(0.78+0.21)*x+K_2000*0.78*0.21\n",
+      "x = fsolve(f,0)\n",
+      "# Here negative root is meaningless,so\n",
+      "# concentration of NO\n",
+      "c_NO = 2*x*10**(6)              #[ppm]\n",
+      "# now\n",
+      "p = c_NO/8100.*100\n",
+      "\n",
+      "# Results\n",
+      "print \" The calculated NO cocentration is %f ppm, which %f%% of the value computed in example 12.1\"%(c_NO,p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The calculated NO cocentration is 7996.442873 ppm, which 98.721517% of the value computed in example 12.1\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.5   Page: 324\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "Concentration of the ethylene \n",
+      "Concentration of the water \n",
+      "Concentration of the ethanol\n",
+      "'''\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Temp = 298.             #[K]\n",
+      "K = 29.6                # equilibrium consmath.tant at 298 K \n",
+      "P = 1.                  #[bar]\n",
+      "n_water_0 = 0.833       #[mol]\n",
+      "n_ethylene_0 = 1.       #[mol]\n",
+      "n_ethanol_0 = 0.        #[mol]\n",
+      "\n",
+      "# Calculations\n",
+      "n_T_0 = (n_water_0+n_ethylene_0+n_ethanol_0)        #[mol]\n",
+      "\n",
+      "# From the previous example, we have \n",
+      "# [(0+e)/(1.833-e)]/([(1-e)/(1.833-e)]*[(0.833-e)/(1.833-e)]) = K*P/(1 bar)\n",
+      "# let y = [(0+e)/(1.833-e)]/([(1-e)/(1.833-e)]*[(0.833-e)/(1.833-e)])- K*P/(1 bar)\n",
+      "def f(e): \n",
+      "\t return  ((0+e)/(1.833-e))/(((1-e)/(1.833-e))*((0.833-e)/(1.833-e)))-K*P/(1)\n",
+      "e_1 = fsolve(f,0)\n",
+      "e_2 = fsolve(f,0.5)\n",
+      "\n",
+      "# Here the root 'e_2' is meaningless, Then \n",
+      "y_ethanol = ((0+e_2)/(1.833-e_2))\n",
+      "y_ethylene = ((1-e_2)/(1.833-e_2))\n",
+      "y_water = ((0.833-e_2)/(1.833-e_2))\n",
+      "\n",
+      "# Results\n",
+      "print \"Concentration of the ethylene at the equilibrium is %f\"%(y_ethylene)\n",
+      "print \" Concentration of the water at the equilibrium is    %f\"%(y_water)\n",
+      "print \" Concentration of the ethanol at the equilibrium is  %f\"%(y_ethanol)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Concentration of the ethylene at the equilibrium is 0.243702\n",
+        " Concentration of the water at the equilibrium is    0.092079\n",
+        " Concentration of the ethanol at the equilibrium is  0.664219\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.6   Page: 324\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "The equilibrium mol fraction of the hydrogen \n",
+      "And the equilibrium mol fraction of the oxygen\n",
+      "'''\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Temp = 273.15+25            #[C]\n",
+      "P = 1.                      #[bar]\n",
+      "R = 8.314                   #[J/(mol*K)]\n",
+      "\n",
+      "# We have the reaction as \n",
+      "#  H2 + 0.5O2 = H2O\n",
+      "# Using values of the Gibbs free energies of formation in the Table A.8(page 427) we have\n",
+      "g_H2O_0 = -237.1            #[kJ/mol]\n",
+      "g_O2_0 = 0                  #[kJ/mol]\n",
+      "g_H2_0 = 0                  #[kJ/mol]\n",
+      "\n",
+      "# Calculations\n",
+      "# now\n",
+      "delta_g_0 = g_H2O_0 - 0.5*g_O2_0-g_H2_0         #[kJ/mol]\n",
+      "# expresmath.sing delta_g_0 in [J/mol] \n",
+      "delta_g_1 = delta_g_0*1000                      #[J/mol]\n",
+      "\n",
+      "# and \n",
+      "K = math.exp((-delta_g_1)/(R*Temp))\n",
+      "# And we have \n",
+      "# K = [a_H2O]/([a_H2]*[a_O2]**(0.5))\n",
+      "# Here we will again assume as in the previous example that we have an ideal solution of the ideal gases for which in equation 12.18 (page 320),we have \n",
+      "# v_i*Y_i = phi = 1.00 , and that for each reacmath.tant or product f_i_0 = 1 bar, putting all the values and simplifying \n",
+      "\n",
+      "# K = [y_H2O]/([y_H2]*[y_O2]**(0.5))*((1 bar)/P)**(0.5)\n",
+      "# Choomath.sing oxygen as the selected reacmath.tant, and assuming that we begin with 0.5 mol of oxygen and 1 mol of hydrogen,\n",
+      "# we have the stoichiometric coefficients of -1, -0.5 and +1 \n",
+      "# and \n",
+      "n_T_0 = 1.5#[mol]\n",
+      "# Also summation(v_i) = -0.5\n",
+      "\n",
+      "# Thus \n",
+      "# K = [e/(1.5-0.5*e)]/([(1-e)/(1.5-0.5*e)]*[(0.5-0.5*e)/(1.5-0.5*e)]**(0.5))\n",
+      "\n",
+      "# Now \n",
+      "def f(e): \n",
+      "\t return (e/(1.5-0.5*e))/(((1-e)/(1.5-0.5*e))*((0.5-0.5*e)/(1.5-0.5*e))**(0.5))\n",
+      "#   e = fsolve(f,.99999)\n",
+      "# e = (1-2.4e-28)\n",
+      "\n",
+      "# So the equilibrium concentration of the hydrogen and oxygen are as\n",
+      "# y_H2 = [(1-e)/(1.5-0.5*e)]\n",
+      "# y_O2 = [(0.5-0.5*e)/(1.5-0.5*e)]\n",
+      "# These values are so less that scilab consol is print laying them zero, however we get\n",
+      "y_H2 = 2.4e-28\n",
+      "y_O2 = 0.5*y_H2\n",
+      "\n",
+      "# Results\n",
+      "print \" The equilibrium mol fraction of the hydrogen is   %0.3e\"%(y_H2)\n",
+      "print \" And the equilibrium mol fraction of the oxygen is %e \"%(y_O2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The equilibrium mol fraction of the hydrogen is   2.400e-28\n",
+        " And the equilibrium mol fraction of the oxygen is 1.200000e-28 \n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.7   Page: 327\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "Total change in the free energy of water going under given conditions \n",
+      "and the free energy change\n",
+      "'''\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "Temp = 298.15           #[K]\n",
+      "Press = 1*10**(5)       #[Pa]\n",
+      "R = 8.314               #[J/(mol*K)]\n",
+      "\n",
+      "# Calculations\n",
+      "# We will calculate the free energy change from liquid to hypothetical gas in three steps\n",
+      "# 1) The liquid is reduced in pressure from the smath.radians(numpy.arcmath.tan(ard pressure of 1 bar to its vapour pressure at 298.15K and for this cange in the state we have \n",
+      "v_liquid = 1.805*10**(-5)           #[m**(3)/mol] this liquid specific volume and we will treat is as a consmath.tant\n",
+      "\n",
+      "# The vapour preesure of the water 25C is given as\n",
+      "P_vapour_25 = 0.0317*10**(5)            #[Pa]\n",
+      "\n",
+      "# thus change in the Gibbs free energy is \n",
+      "\n",
+      "def f0(P): \n",
+      "\t return v_liquid*P**(0)\n",
+      "\n",
+      "delta_g_0_1 =  quad(f0,Press,P_vapour_25)[0]\n",
+      "\n",
+      "\n",
+      "# 2) In the second step the liquid is vaporized at that pressure, for which\n",
+      "delta_g_0_2 = 0             #[J/mol]\n",
+      "# because this is an equilibrium vaporization.\n",
+      "\n",
+      "# 3) And in this last step the vapour is replaced by an ideal gas, which  will not condence, and compressed from the vapour pressure at 298.15K to 1 bar\n",
+      "# In this case the specific volume v_ideal of the ideal gas is replaced by the ideal gas law viz. (R*T)/P\n",
+      "\n",
+      "def f1(P): \n",
+      "\t return 1./P\n",
+      "\n",
+      "delta_g_0_3 = (R*Temp)* quad(f1,P_vapour_25,Press)[0]\n",
+      "\n",
+      "\n",
+      "# Thus total change in free energy is \n",
+      "delta_g_0 = delta_g_0_1+delta_g_0_2+delta_g_0_3         #[J/mol]\n",
+      "#expresmath.sing the result in kJ/mol\n",
+      "delta_g_1 = delta_g_0/1000                              #[kJ/mol]\n",
+      "\n",
+      "# Results\n",
+      "print \" Total change in the free energy of water going under given conditions is %0.2f kJ/mol\"%(delta_g_1)\n",
+      "\n",
+      "# From Table A.8 we find \n",
+      "delta_g_0_ideal_gas = -228.6                            #[kJ/mol]\n",
+      "delta_g_0_liquid = -237.1                               #[kJ/mol]\n",
+      "# So \n",
+      "delta_g_o = delta_g_0_ideal_gas-delta_g_0_liquid        #[kJ/mol]\n",
+      "\n",
+      "print \" From the values of Table A.8 given in the book the free energy change is %0.2f kJ/mol\"%(delta_g_o)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Total change in the free energy of water going under given conditions is 8.55 kJ/mol\n",
+        " From the values of Table A.8 given in the book the free energy change is 8.50 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.8   Page: 330\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find Equilibrium constants for the formation of ammonia from hydrogen and nitrogen are\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "T1 = 273.15+25              #[K]\n",
+      "T2 = 273.15+400             #[K]\n",
+      "R = 8.314                   #[J/(mol*K)]\n",
+      "\n",
+      "# Umath.sing the table A.8, we have\n",
+      "# Gibb's free energy of the various species at 298.15 K are\n",
+      "g0_NH3 = -16.5              #[kJ/mol]\n",
+      "g0_N2 = 0                   #[kJ/mol]\n",
+      "g0_H2 = 0                   #[kJ/mol]\n",
+      "\n",
+      "# We have the reaction as \n",
+      "# 0.5N2 + 1.5H2 = NH3\n",
+      "\n",
+      "# Calculations\n",
+      "# So, Gibb's free energy change in the reaction is given as \n",
+      "delta_g_0 = g0_NH3 - 0.5*g0_N2 - 1.5*g0_H2          #[kJ/mol]\n",
+      "\n",
+      "# and\n",
+      "K_1 = math.exp(-delta_g_0*1000/(R*T1))          # Equilibrium consmath.tant of the reaction at temperature 298.15 K\n",
+      "\n",
+      "# Similarly enthalpy of the various species are \n",
+      "h0_NH3 = -46.1          #[kJ/mol]\n",
+      "h0_N2 = 0               #[kJ/mol]\n",
+      "h0_H2 = 0               #[kJ/mol]\n",
+      "\n",
+      "# So, enthalpy change of the reaction is given as \n",
+      "del_h_1 = h0_NH3 - 0.5*h0_N2 - 1.5*h0_H2        #[kJ/mol]\n",
+      "\n",
+      "# Now, from Table 12.3( page 332 )\n",
+      "a_NH3 = 3.578\n",
+      "a_H2 = 3.249\n",
+      "a_N2 = 3.280\n",
+      "b_NH3 = 3.020*10**(-3)      #[1/K]\n",
+      "b_H2 = 0.422*10**(-3)\n",
+      "b_N2 = 0.593*10**(-3)\n",
+      "c_NH3 = 0                   #[1/K**(2)]\n",
+      "c_H2 = 0                    #[1/K**(2)]\n",
+      "c_N2 = 0                    #[1/K**(2)]\n",
+      "d_NH3 = -0.186*10**(5)      #[K**(2)]\n",
+      "d_H2 = 0.083*10**(5)        #[K**(2)]\n",
+      "d_N2 = 0.040*10**(5)        #[K**(2)]\n",
+      "\n",
+      "# So,\n",
+      "del_a = a_NH3 - 0.5*a_N2 - 1.5*a_H2\n",
+      "del_b = b_NH3 - 0.5*b_N2 - 1.5*b_H2\n",
+      "del_c = c_NH3 - 0.5*c_N2 - 1.5*c_H2\n",
+      "del_d = d_NH3 - 0.5*d_N2 - 1.5*d_H2\n",
+      "\n",
+      "# Now, enthalpy change of the reaction at any other temparature is given by\n",
+      "# Integrating and putting the limits, we have\n",
+      "#  del_h = del_h_1 + R*( del_a*T + del_b*T**(2)/2 + del_c*T**(3)/3 - del_d/T) - R*( del_a*T_1 + del_b*T_1**(2)/2 + del_c*T_1**(3)/3 - del_d/T_1)\n",
+      "# let\n",
+      "I = R*( del_a*T1 + del_b*T1**(2)/2 + del_c*T1**(3)/3 - del_d/T1)        #[J/mol]\n",
+      "\n",
+      "# From equation 12.28 and above relations we have\n",
+      "# Let math.log(K_2/K_1) = X, So,\n",
+      "\n",
+      "def f5(T): \n",
+      "\t return (del_h_1*1000 - I + R*(del_a*T + del_b*T**(2)/2 + del_c*T**(3)/3 - del_d/T))/T**(2)\n",
+      "\n",
+      "X = (1/R)* quad(f5,T1,T2)[0]\n",
+      "\n",
+      "\n",
+      "# So, \n",
+      "K_2 = K_1*math.exp(X)\n",
+      "\n",
+      "# Results\n",
+      "print \" Equilibrium consmath.tants for the formation of ammonia from hydrogen and nitrogen are \"\n",
+      "print \" K = %0.0f at temperature 25 deg C\"%(K_1)\n",
+      "print \" K = %f at temperature 400 deg C\"%(K_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Equilibrium consmath.tants for the formation of ammonia from hydrogen and nitrogen are \n",
+        " K = 778 at temperature 25 deg C\n",
+        " K = 0.013588 at temperature 400 deg C\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.9   Page: 335\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "The mole fraction of NH3 in the equilibrium\n",
+      "The mole fraction of NH3 in the equilibrium\n",
+      "'''\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "# Initial moles of the gases are \n",
+      "n_H2_0 = 1.5            #[mol]\n",
+      "n_N2_0 = 0.5            #[mol]\n",
+      "n_NH3_0 = 0             #[mol]\n",
+      "T_1 = 298.15            #[K]\n",
+      "T_2 = 673.15            #[K]\n",
+      "P = 1.                  #[bar]\n",
+      "\n",
+      "# We start with the equation as \n",
+      "# [f_NH3/f_0_NH3]/([f_N2/f_0_N2]**(0.5)*[f_H2/f_0_H2]**(1.5)) = K\n",
+      "\n",
+      "# For a pressure of 1 bar with the assumption of ideal solution of ideal gases and smath.radians(numpy.arcmath.tan(ard state fugacities of 1 bar,\n",
+      "# a_i = [f_i/f_0_i] = [P*y_i/(1 bar)] = y_i \n",
+      "# The equilibrium relation is given by \n",
+      "# K = [y_NH3]/([y_N2]**(0.5)*[y_H2]**(1.5))\n",
+      "\n",
+      "# We have the stoichiometric coefficient of N2, H2 and NH3 as -0.5, -1.5 and +1 respectively, so summation(v_i) = -1\n",
+      "# Now umath.sing the equilibrium relations which are Equations 12.W, 12.X and 12.Y ( page 322 ), we have \n",
+      "\n",
+      "# K = ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))\n",
+      "# Form the example 12.8 of this book we know that \n",
+      "K_298 = 778.            # at temperature 298.15K\n",
+      "K_673 = 0.013           # at temperature 673.15K\n",
+      "\n",
+      "# Calculations\n",
+      "# Solving for temperature 298.15\n",
+      "def g(e_1): \n",
+      "\t return  ((0+e_1)/(2-e_1))/(((0.5-0.5*e_1)/(2-e_1))**(0.5)*((1.5-1.5*e_1)/(2-e_1))**(1.5))-K_298\n",
+      "     \n",
+      "e_1 = fsolve(g,0.97)\n",
+      "y_NH3_298 = e_1/(2-e_1)\n",
+      "\n",
+      "# Similarily solving for temperature 673.15K\n",
+      "def h(e_2): \n",
+      "\t return  ((0+e_2)/(2-e_2))/(((0.5-0.5*e_2)/(2-e_2))**(0.5)*((1.5-1.5*e_2)/(2-e_2))**(1.5))-K_673\n",
+      "\n",
+      "e_2 = fsolve(h,0)\n",
+      "y_NH3_673 = e_2/(2-e_2)\n",
+      "\n",
+      "# Results\n",
+      "print \" The mole fraction of NH3 in the equilibrium at the temperature 298.15K and 1 bar is %f\"%(y_NH3_298)\n",
+      "print \" The mole fraction of NH3 in the equilibrium at the temperature 673.15K and 1 bar is %f\"%(y_NH3_673)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mole fraction of NH3 in the equilibrium at the temperature 298.15K and 1 bar is 0.939036\n",
+        " The mole fraction of NH3 in the equilibrium at the temperature 673.15K and 1 bar is 0.004187\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.10  Page: 337\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find The mole fraction of the ammonia in the equilibrium\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Temp = 273.15+400           #[K]\n",
+      "P = 150*1.01325             #[bar]\n",
+      "\n",
+      "# Comparing this with the example 12.9, we see that we can use the same equation , but K_673 is replaced by K_673*(P/(1bar))**(1.5+0.5-1)\n",
+      "K_673 = 0.013\n",
+      "\n",
+      "# So \n",
+      "K = K_673*(P/1)**(1.5+0.5-1)\n",
+      "\n",
+      "# Calculations\n",
+      "# We have \n",
+      "# K = ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))\n",
+      "def f(e): \n",
+      "\t return  ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))-K\n",
+      "e=fsolve(f,0.5)\n",
+      "\n",
+      "# Thus mole fraction of the ammonia in the gas is given by \n",
+      "y_NH3 = (0+e)/(2-e)\n",
+      "\n",
+      "# Results\n",
+      "print \"The mole fraction of the ammonia in the equilibrium is %0.2f\"%(y_NH3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mole fraction of the ammonia in the equilibrium is 0.31\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.11   Page: 338\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find The mole fraction of the ammonia in the equilibrium\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "# The data used in this example will e same as in the example 12.10\n",
+      "T = 273.15+400              #[K] given temperature\n",
+      "P = 150*1.01325             #[bar] given pressure\n",
+      "\n",
+      "# Here again the equation will be same as in the example 12.9 like we used in the example 12.10 only K_673 is replaced by (K/K_v)*[P/(1 bar)]**(1.5+0.5-1)\n",
+      "K_673 = 0.013\n",
+      "# The value of 'K_v' is calculated by the equation 12.BN, which is \n",
+      "# log10(1/K_v) = (0.1191849/T + 91.87212/T**(2) + 25122730/T**(4))*P\n",
+      "\n",
+      "# Calculations\n",
+      "# So \n",
+      "K_v = (10**((0.1191849/T + 91.87212/T**(2) + 25122730/T**(4))*P))**(-1)\n",
+      "\n",
+      "# Thus \n",
+      "K = (K_673/K_v)*(P/1)**(1.5+0.5-1)\n",
+      "\n",
+      "# Now from the previous example we have\n",
+      "# K = ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))\n",
+      "\n",
+      "def f(e): \n",
+      "\t return  ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))-K\n",
+      "e = fsolve(f,0.2)\n",
+      "\n",
+      "# Mol fraction of the ammonia in the gas phase in the equilibrium is given by\n",
+      "y_NH3 = (0+e)/(2-e)\n",
+      "\n",
+      "# Results\n",
+      "print \" The mole fraction of the ammonia in the equilibrium is %0.2f\"%(y_NH3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mole fraction of the ammonia in the equilibrium is 0.34\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.12   Page: 340\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "find\n",
+      "On increasing the pressure from 1 atm to 150 atm the reacted amount of the equimolar\n",
+      "reactants at equilibrium becomes times of initial\n",
+      "'''\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "p_i = 1.                #[atm] initial pressure \n",
+      "P = 150.                #[atm] final pressure\n",
+      "T = 273+25.             #[K] Given temperature\n",
+      "R = 8.314               #[J/(mol*K)]\n",
+      "\n",
+      "# Now ignoring the difference between 25C and 20C, we use the values given in the table A.8 (page 427) to get \n",
+      "delta_g_0 = 10.54*1000          #[J/mol]\n",
+      "# And thus \n",
+      "\n",
+      "# Calculations\n",
+      "K = math.exp((-delta_g_0)/(R*T))\n",
+      "\n",
+      "# Now the chemical reaction is given by\n",
+      "# C2H5OH + CH3COOH = C2H5OOC2H5 + H2O\n",
+      "\n",
+      "# Let we start with 1 mol each of ethanol and acetic acid, and at equilibrium 'e' moles each of the reacmath.tants reacted, then\n",
+      "# remaining amount of each of the two reacmath.tants will be (1-e) and that products formation will be 'e' mol each\n",
+      "\n",
+      "# We have \n",
+      "# K = (a_C2H5OOC2H5*a_H2O)/(a_C2H5OH*a_CH3COOH) = (x_C2H5OOC2H5*x_H2O)/(x_C2H5OH*x_CH3COOH) = (e*e)/((1-e)*(1-e))\n",
+      "# Now solving for 'e'\n",
+      "def f(e): \n",
+      "\t return  (e*e)/((1-e)*(1-e))-K\n",
+      "e = fsolve(f,0)\n",
+      "\n",
+      "# To see the effect of changing the pressure we first compute the volume increase of the reaction \n",
+      "# delta_v = v_C2H5OOC2H5 + v_H2O - v_C2H5OH - v_CH3COOH, where v_i is the molar volume of the ith component\n",
+      "# From the Table 12.4(page 340), we have\n",
+      "v_C2H5OOC2H5 = 97.67            #[ml/mol]\n",
+      "v_H2O = 18.03                   #[ml/mol]\n",
+      "v_C2H5OH = 58.30                #[ml/mol]\n",
+      "v_CH3COOH = 57.20               #[ml/mol]\n",
+      "\n",
+      "# Thus volume increase of the reaction is\n",
+      "delta_v = v_C2H5OOC2H5 + v_H2O - v_C2H5OH - v_CH3COOH       #[ml/mol]\n",
+      "\n",
+      "# So, from Le Chatelier's principal, on increamath.sing the pressure , the reaction is forced in the direction of the reacmath.tant or away from the product \n",
+      "# To calculate the extent of shifting we will take the help of the activity of each of the four component \n",
+      "# a_i = (f_i/f_i_0) = (x_i*Y_i*p_i)/p_i*exp(v/(R*T)*(P-p_i))\n",
+      "# we will assume that this is an ideal solution so that Y_i = 1.00, for every component\n",
+      "\n",
+      "# Now substituting the activity of each component in the expression of the equilibrium consmath.tant given above, we have\n",
+      "# K = (x_C2H5OOC2H5*x_H2O)/(x_C2H5OH*x_CH3COOH)*exp[(delta_v)/(R*T)*(P-p_i)]\n",
+      "# or\n",
+      "# K = (e_1*e_1)/((1-e_1)*(1-e_1))*exp[(delta_v)/(R*T)*(P-p_i)]\n",
+      "\n",
+      "# Solving for 'e_1'\n",
+      "def g(e_1): \n",
+      "\t return  (e_1*e_1)/((1-e_1)*(1-e_1))*math.exp((delta_v)/(R*T)*(P-p_i))-K\n",
+      "e_1 = fsolve(g,0.2)\n",
+      "\n",
+      "# Now if we carry out the calculation to enough significant figures then\n",
+      "a = e_1/e\n",
+      "\n",
+      "# It indicates that e_1 is 'a' times of that of the e\n",
+      "\n",
+      "# Results\n",
+      "print \"On increamath.sing the pressure from 1 atm to 150 atm the reacted amount of the equimolar \\\n",
+      "reacmath.tants at equilibrium becomes %f times of initial\"%(a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "On increamath.sing the pressure from 1 atm to 150 atm the reacted amount of the equimolar reacmath.tants at equilibrium becomes 0.994639 times of initial\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 12.13  Page: 342\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find Value of the K_p at the given condition\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 150.            #[atm] given pressure\n",
+      "T = 400.            #[C] temperature\n",
+      "# Umath.sing the values from the example 12.11, we know that\n",
+      "K = 0.013\n",
+      "K_v = 0.84\n",
+      "delta_v = 1.5+0.5-1\n",
+      "\n",
+      "# Calculations\n",
+      "# so \n",
+      "# K_p = (K/K_v)*[1/bar]**(-summation(v_i)) = (K/K_v)*[1/bar]**(delta_v)\n",
+      "\n",
+      "K_p = (K/K_v)*(1/1)**(delta_v)              #[1/bar]\n",
+      "\n",
+      "# Results\n",
+      "print \" Value of the K_p at the given condition is %f 1/bar)\"%(K_p)\n",
+      "print  \" The basic K is dimensionless%( but K_p has the dimensions of pressure to the power.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Value of the K_p at the given condition is 0.015476 1/bar)\n",
+        " The basic K is dimensionless%( but K_p has the dimensions of pressure to the power.\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file