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{
"metadata": {
"name": "",
"signature": "sha256:aba24805b98ee0919f2ef39d5927b5007d2ce98264adcd2e65263a9fbffd1f67"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15:CHEMICAL EQUILIBRIUM"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.1, Page No:676"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"# (b).Number of moles of each constituents\n",
"nCH4=2; # Number of moles of CH4\n",
"\n",
"#Calculation\n",
"E=3-nCH4; # Amount of reaction from (a) and refer example 15.1 (a)\n",
"nH2O=1-E;# Number of moles of H2O\n",
"nCO=1+E;# Number of moles of CO\n",
"nH2=4+3*E;# Number of moles of H2\n",
"\n",
"#Results\n",
"print \"(b).Number of moles of each constituents\",\"\\nNumber of moles of H2O = \",nH2O\n",
"print \"Number of moles of CO = \",nCO,\"\\nNumber of moles of H2 = \",nH2"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(b).Number of moles of each constituents \n",
"Number of moles of H2O = 0\n",
"Number of moles of CO = 2 \n",
"Number of moles of H2 = 7\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.2, Page No:680"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"T0=298; # Given temperature in kelvin\n",
"R_1=8.314; # Universal gas constant in kJ/kg mol K\n",
"# (a).CO+1/2 O2 = CO2\n",
"# From table of properties of combustion\n",
"del_hfco2=-393509;# Enthalpy of heat \n",
"del_hfco=-110525;# Enthalpy of heat \n",
"s_co2=213.795;# Entropy of heat \n",
"s_co=197.652;# Entropy of heat \n",
"s_o2=205.142;# Entropy of heat \n",
"\n",
"#Calculation for (a)\n",
"del_Ga=(del_hfco2-del_hfco-T0*(s_co2-s_co-(1/2*s_o2)));\n",
"Ka=math.exp (abs (del_Ga)/(R_1*1000*T0));\n",
"\n",
"#Result for (a)\n",
"print \"(a).CO+1/2 O2 = CO2\"\n",
"print (\" The equilibrium constant at 298 K = %0.3f (Error in textbook) \")%Ka\n",
"\n",
"#Calculation for (b)\n",
"# (b).2CO + O2 = 2CO2\n",
"Kb=math.exp (2*abs (del_Ga)/(R_1*1000*T0));\n",
"\n",
"#Result for (b)\n",
"print \"\\n(b).2CO + O2 = 2CO2\"\n",
"print (\"The equilibrium constant at 298 K = %0.3f (Error in textbook)\")%Kb\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a).CO+1/2 O2 = CO2\n",
" The equilibrium constant at 298 K = 1.123 (Error in textbook) \n",
"\n",
"(b).2CO + O2 = 2CO2\n",
"The equilibrium constant at 298 K = 1.262 (Error in textbook)\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3, Page No:686"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"T0=298; # Temperature of surroundings in kelvin\n",
"R_1=8.314; # Universal gas constant in kJ/kg mol K\n",
"T=2800; # Given Temperature in kelvin\n",
"\n",
"#Calculation\n",
"# From table of properties of combustion\n",
"del_hfco2=-393509; # Enthalpy of heat \n",
"del_hfco=-110525; # Enthalpy of heat \n",
"del_H=del_hfco2-del_hfco; # Standard enthalpy of reaction\n",
"Ka=1.202*10**45; # The equilibrium constant From the example 15.2\n",
"K1=math.log (Ka);\n",
"K=math.exp(-(del_H/R_1)*((1/T)-(1/T0))+K1);\n",
"\n",
"#Result\n",
"print\"K =\",round(K,3),\" (roundoff error)\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"K = 5.687 (roundoff error)\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.5, Page No:689"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"T=2800; # Temperature of combustion in kelvin\n",
"p=1; # Pressure of combustion in atm\n",
"# For this reverse reaction at 2800K and 1atm, from Table 15.1\n",
"K=44.168; # K=e^3.788;\n",
"\n",
"#Calculation\n",
"K=math.sqrt (K); # For stoichiometric equation CO+1/2 O2 = CO2 which is halved\n",
"# From equation 15.24a and by the iteration process we get the following\n",
"a=0.198;\n",
"b=(1+a)/2;\n",
"c=1-a;\n",
"\n",
"#Results\n",
"print \"The balance for the actual reaction equation CO + O2 \u2192 aCO + bO2 + cCO2 is given by \"\n",
"print \"a =\",a,\"\\nb =\",b,\"\\nc =\",c\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The balance for the actual reaction equation CO + O2 \u2192 aCO + bO2 + cCO2 is given by \n",
"a = 0.198 \n",
"b = 0.599 \n",
"c = 0.802\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.6, Page No:691"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"# By driving the equation for equilibrium constant as shown in example 15.6 we get 6.646(6)^(1/6)=((1-a)/a)((3+a)/(1+a))^1/2\n",
"# by simple iteration process we get\n",
"a=0.095;\n",
"\n",
"#Calculations\n",
"b=(1+a)/2;\n",
"c=1-a;\n",
"\n",
"#Results\n",
"print \"The equilibrium composition of CO = \",a,\"mol (roundoff error)\"\n",
"print \"The equilibrium composition of O2 = \",b,\"mol (roundoff error)\"\n",
"print \"The equilibrium composition of CO2 = \",c,\"mol (roundoff error)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The equilibrium composition of CO = 0.095 mol (roundoff error)\n",
"The equilibrium composition of O2 = 0.5475 mol (roundoff error)\n",
"The equilibrium composition of CO2 = 0.905 mol (roundoff error)\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.7, Page NO:691"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"T=2800; # Temperature of combustion in kelvin\n",
"p=1; # Pressure of combustion in atm\n",
"# For this reverse reaction at 2800K and 1atm, from Table 15.1\n",
"K=44.168; # K=e^3.788;\n",
"\n",
"#Calculations\n",
"K=math.sqrt (K); # For stoichiometric equation CO+1/2 O2 = CO2 which is halved\n",
"# From equation 15.24a and by the iteration process we get the following\n",
"a=0.302;\n",
"b=(1+a)/2;\n",
"c=1-a;\n",
"\n",
"#Results\n",
"print \"The balance for the actual reaction equation CO + 1/2O2 + 1.88N2 \u2194 aCO + bO2 + cCO2 +3.76N2 is given by \"\n",
"print \"a =\",a,\"\\nb =\",b,\"\\nc =\",c"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The balance for the actual reaction equation CO + 1/2O2 + 1.88N2 \u2194 aCO + bO2 + cCO2 +3.76N2 is given by \n",
"a = 0.302 \n",
"b = 0.651 \n",
"c = 0.698\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.8, Page No:693"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"T=3000; # Temperature of combustion in kelvin\n",
"p=1; # Pressure of combustion in atm\n",
"T0=298; # Temperature of surroundings in kelvin\n",
"R_1=8.314; # Universal gas constant in kJ/kg mol K\n",
"# Gibbs functions at 298K from Table 14.1\n",
"del_gNO=86550; # In kJ/kmol\n",
"del_gNO2=51310; # In kJ/kmol\n",
"# From table of properties of combustion\n",
"del_hfNO=90250; # Enthalpy of heat \n",
"del_hfNO2=33180; # Enthalpy of heat \n",
"\n",
"#Calculations\n",
"K1=math.exp (-(del_hfNO/R_1)*((1/T)-(1/T0))-((del_gNO)/(R_1*T0)));\n",
"K2=math.exp (-(del_hfNO2/R_1)*((1/T)-(1/T0))-((del_gNO2)/(R_1*T0)));\n",
"# By solving equilibrium equations by iteration method\n",
"E1=0.228; E2=0.0007;\n",
"yNO=E1/4.76; # Mole fraction of NO in exhaust gas\n",
"yNO2=E2/4.76; # Mole fraction of NO2 in exhaust gas\n",
"\n",
"#Results\n",
"print \"Percentage of NOx present in the exhaust gas \",\"\\nMole fraction of NO in exhaust gas = \",round(yNO*100,2),\"%\"\n",
"print \"Mole fraction of NO2 in exhaust gas = \",round(yNO2*100,4),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage of NOx present in the exhaust gas \n",
"Mole fraction of NO in exhaust gas = 4.79 %\n",
"Mole fraction of NO2 in exhaust gas = 0.0147 %\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|
