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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter12 - Chemical reaction equillibria"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.1 Page: 471"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.1 - Page: 471\n",
+ "\n",
+ "\n",
+ " This problem involves proving a relation in which no mathematics and no calculations are involved.\n",
+ "\n",
+ "\n",
+ " For prove refer to this example 12.1 on page 471 of the book.\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.1 - Page: 471\\n\\n\"\n",
+ "\n",
+ "# This problem involves proving a relation in which no mathematics and no calculations are involved.\n",
+ "# For prove refer to this example 12.1 on page number 471 of the book.\n",
+ "\n",
+ "print \" This problem involves proving a relation in which no mathematics and no calculations are involved.\\n\\n\"\n",
+ "print \" For prove refer to this example 12.1 on page 471 of the book.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.2 Page: 473"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.2 - Page: 473\n",
+ "\n",
+ "\n",
+ " This problem involves proving a relation in which no mathematics and no calculations are involved.\n",
+ "\n",
+ "\n",
+ " For prove refer to this example 12.2 on page 473 of the book.\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.2 - Page: 473\\n\\n\"\n",
+ "\n",
+ "# This problem involves proving a relation in which no mathematics and no calculations are involved.\n",
+ "# For prove refer to this example 12.2 on page number 473 of the book.\n",
+ "\n",
+ "print \" This problem involves proving a relation in which no mathematics and no calculations are involved.\\n\\n\"\n",
+ "print \" For prove refer to this example 12.2 on page 473 of the book.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.3 Page: 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.3 - Page: 479\n",
+ "\n",
+ "\n",
+ "Reaction is feasible\n",
+ "\n",
+ "Equilibium Constant is 7.964e+32\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import exp\n",
+ "from __future__ import division\n",
+ "print \"Example: 12.3 - Page: 479\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: C2H5OH(g) + (1/2)O2(g) = CH3CHO(g) + H2O(g)\n",
+ "Temp = 298## [K]\n",
+ "G_CH3CHO = -133.978## [kJ]\n",
+ "G_H2O = -228.60## [kJ]\n",
+ "G_C2H5OH = -174.883## [kJ]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#***************#\n",
+ "\n",
+ "G_O2 = 0## [kJ]\n",
+ "G_rkn = G_CH3CHO + G_H2O -(G_C2H5OH + G_O2)## [kJ]\n",
+ "G_rkn = G_rkn*1000## [J]\n",
+ "if G_rkn < 0 :\n",
+ " print \"Reaction is feasible\\n\"\n",
+ " K = exp(-(G_rkn/(R*Temp)))#\n",
+ " print \"Equilibium Constant is %.3e\"%(K)#\n",
+ "else:\n",
+ " print \"Reaction is not feasible\\n\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.4 Page: 479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.4 - Page: 479\n",
+ "\n",
+ "\n",
+ "Vaporisation of liquid water is not spontaneous\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.4 - Page: 479\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# H2O(l) = H2O(g)\n",
+ "deltaH = 9710## [cal]\n",
+ "deltaS = 26## [e.u.]\n",
+ "Temp = 27 + 273## [K]\n",
+ "P = 1## [atm]\n",
+ "#**************#\n",
+ "\n",
+ "deltaG = deltaH - Temp*deltaS## [cal]\n",
+ "if deltaG > 0:\n",
+ " print \"Vaporisation of liquid water is not spontaneous\\n\"\n",
+ "else:\n",
+ " print \"Vaporisation of liquid water is spontaneous\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.5 Page: 481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.5 - Page: 481\n",
+ "\n",
+ "\n",
+ "Standard Gibbs free energy change is -33.5 kJ/mol\n",
+ "\n",
+ "Equilibrium constant is 7.45e+05\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.5 - Page: 481\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: N2(g) + 3H2(g) = 2NH3(g)\n",
+ "Temp = 298## [K]\n",
+ "G_NH3 = -16.750## [kJ/mol]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#***************#\n",
+ "\n",
+ "G_N2 = 0## [kJ/mol]\n",
+ "G_H2 = 0## [kJ/mol]\n",
+ "G_rkn = 2*G_NH3 - G_N2 - 3*G_H2## [kJ/mol]\n",
+ "print \"Standard Gibbs free energy change is %.1f kJ/mol\\n\"%(G_rkn)#\n",
+ "G_rkn = G_rkn*1000## [J/mol]#\n",
+ "K = exp(-(G_rkn/(R*Temp)))#\n",
+ "print \"Equilibrium constant is %.2e\"%(K)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.6 Page: 481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.6 - Page: 481\n",
+ "\n",
+ "\n",
+ "Equilibrium Constant is 1.005e+05\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.6 - Page: 481\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: CO(g) + H2O(g) = CO2(g) + H2(g)\n",
+ "G_CO = -32.8## [kcal]\n",
+ "G_H2O = -54.64## [kcal]\n",
+ "G_CO2 = -94.26## [kcal]\n",
+ "Temp = 273 + 25## [K]\n",
+ "R = 1.987## [cal/mol.K]\n",
+ "#***************#\n",
+ "\n",
+ "G_H2 = 0## [kcal]\n",
+ "G_rkn = G_CO2 + G_H2 - (G_CO + G_H2O)## [kcal]\n",
+ "G_rkn = G_rkn*1000## [cal]\n",
+ "K = exp(-(G_rkn/(R*Temp)))#\n",
+ "print \"Equilibrium Constant is %.3e\"%(K)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.7 Page: 485"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.7 - Page: 485\n",
+ "\n",
+ "\n",
+ "Equilibrium constant at 1000 K is 0.9190\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.7 - Page: 485\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: CO(g) + H2O(g) = CO2(g) + H2(g)\n",
+ "T1 = 298## [K]\n",
+ "T2 = 1000## [K]\n",
+ "P = 1## [bar]\n",
+ "K1 = 1.1582*10**5#\n",
+ "H_CO = -110.532## [kJ]\n",
+ "H_H2O = -241.997## [kJ]\n",
+ "H_CO2 = -393.978## [kJ]\n",
+ "R = 8.314## [J/mol.K]\n",
+ "#***************#\n",
+ "\n",
+ "H_H2 = 0## [kJ]\n",
+ "H_rkn = H_CO2 + H_H2 - (H_CO + H_H2O)## [kJ]\n",
+ "H_rkn = H_rkn*1000## [J]\n",
+ "# From Van't Hoff Equation,\n",
+ "K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)))#\n",
+ "print \"Equilibrium constant at 1000 K is %.4f\"%(K2)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.8 Page: 485"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.8 - Page: 485\n",
+ "\n",
+ "\n",
+ "Equilibrium constant at 700 K is 3.8878e-04\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.8 - Page: 485\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: N2(g) + 3H2(g) = 2NH3(g)\n",
+ "T1 = 298## [K]\n",
+ "T2 = 700## [K]\n",
+ "H_NH3 = -46.1## [kJ]\n",
+ "G_NH3 = -16.747## [kJ]\n",
+ "R = 8.314## [J/mol.K]\n",
+ "#**************#\n",
+ "\n",
+ "H_N2 = 0## [kJ]\n",
+ "H_H2 = 0## [kJ]\n",
+ "G_N2 = 0## [kJ]\n",
+ "G_H2 = 0## [kJ]\n",
+ "H_rkn = 2*H_NH3 - (H_N2 + 3*H_H2)## [kJ]\n",
+ "G_rkn = 2*G_NH3 - (G_N2 + 3*G_H2)## [kJ]\n",
+ "H_rkn = H_rkn*1000## [J]\n",
+ "G_rkn = G_rkn*1000## [J]\n",
+ "K1 = exp(-(G_rkn/(R*T1)))#\n",
+ "K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)))#\n",
+ "print \"Equilibrium constant at 700 K is %.4e\"%(K2)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.9 Page: 486"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.9 - Page: 486\n",
+ "\n",
+ "\n",
+ "Standard Gibbs free energy at 1000 K -6.055 kJ\n",
+ "\n",
+ "Standard Equilibrium Constant is 2.1\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log\n",
+ "print \"Example: 12.9 - Page: 486\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: CO(g) + H2O(g) ----> CO2(g) + H2(g)\n",
+ "T1 = 298## [K]\n",
+ "T2 = 1000## [K]\n",
+ "deltaH_298 = -41450## [J/mol]\n",
+ "deltaGf_298 = -28888## [J/mol]\n",
+ "alpha_CO2 = 45.369## [kJ/kmol K]\n",
+ "alpha_H2 = 27.012## [kJ/kmol K]\n",
+ "alpha_CO = 28.068## [kJ/kmol K]\n",
+ "alpha_H2O = 28.850## [kJ/kmol K]\n",
+ "beeta_CO2 = 8.688*10**(-3)## [kJ/kmol square K]\n",
+ "beeta_H2 = 3.509*10**(-3)## [kJ/kmol square K]\n",
+ "beeta_CO = 4.631*10**(-3)## [kJ/kmol square K]\n",
+ "beeta_H2O = 12.055*10**(-3)## [kJ/kmol square K]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#*************#\n",
+ "\n",
+ "delta_alpha = alpha_CO2 + alpha_H2 - (alpha_CO + alpha_H2O)#\n",
+ "delta_beeta = beeta_CO2 + beeta_H2 - (beeta_CO + beeta_H2O)#\n",
+ "# To obtain the standard heat of reaction:\n",
+ "deltaH_0 = deltaH_298 - (delta_alpha*T1 + (delta_beeta*T1**2)/2)## [kJ/mol]\n",
+ "# From Eqn. 12.52:\n",
+ "IR = (deltaH_0 - delta_alpha*T1*log(T1) - (delta_beeta*T1**2)/2 - deltaGf_298)/T1## [kJ/mol K]\n",
+ "# Substituting T = T2 and IR in Eqn. 12.51:\n",
+ "deltaG_1000 = deltaH_0 - delta_alpha*T2*log(T2) - (delta_beeta*T2**2)/2 - IR*T2## [kJ/mol]\n",
+ "print \"Standard Gibbs free energy at 1000 K %.3f kJ\\n\"%(deltaG_1000/1000)#\n",
+ "\n",
+ "# Standard Equilibrium Constant at 1000 K\n",
+ "K_1000 = exp(-(deltaG_1000)/(R*T2))#\n",
+ "print \"Standard Equilibrium Constant is %.1f\"%(K_1000)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.10 Page: 489"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.10 - Page: 489\n",
+ "\n",
+ "\n",
+ "Composition of the gas leaving the mixture\n",
+ "\n",
+ "yCO = yH20 = 0.2247\n",
+ "\n",
+ "yCO2 = yH2 = 0.2753\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.10 - Page: 489\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: CO(g) + H2O(g) ------> CO2(g) + H2(g)\n",
+ "P = 10## [bar]\n",
+ "T = 1000## [K]\n",
+ "K_1000 = 1.5#\n",
+ "#***********#\n",
+ "\n",
+ "# Moles in feed:\n",
+ "nCO_feed = 1#\n",
+ "nH20_feed = 1#\n",
+ "# Let e be the degree of completion at equilibrium.\n",
+ "# Moles at Equilibrium:\n",
+ "# nCO_eqb = 1 - e#\n",
+ "# nH20_eqb = 1 - e#\n",
+ "# nCO2_eqb = e#\n",
+ "# nH2_eqb = e#\n",
+ "# Total moles at equilibrium = 1 - e + 1 - e + e + e = 2\n",
+ "# Mole Fractions at Equilibrium:\n",
+ "# yCO_eqb = (1 - e)/2#\n",
+ "# yH20_eqb = (1 - e)/2#\n",
+ "# yCO2_eqb = e/2#\n",
+ "# yH2_eqb = e/2#\n",
+ "# Sum of stoichometric coeffecient:\n",
+ "v = 1 + 1 - 1 - 1#\n",
+ "K = K_1000*P**v#\n",
+ "#deff('[y] = f(e)','y = K - (e/2)*(e/2)/(((1 - e)/2)*(1 - e)/2)')#\n",
+ "def f(e):\n",
+ " y = K - (e/2)*(e/2)/(((1 - e)/2)*(1 - e)/2)\n",
+ " return y\n",
+ "from scipy.optimize import fsolve\n",
+ "e = fsolve(f, 0.5)#\n",
+ "print \"Composition of the gas leaving the mixture\\n\"\n",
+ "print \"yCO = yH20 = %.4f\\n\"%((1 - e)/2)#\n",
+ "print \"yCO2 = yH2 = %.4f\\n\"%(e/2)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.11 Page: 489"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.11 - Page: 489\n",
+ "\n",
+ "\n",
+ "Molar Composition of the gases\n",
+ "\n",
+ "yN2 = 0.1103\n",
+ "\n",
+ "yH2 = 0.7207\n",
+ "\n",
+ "yNH3 = 0.1690\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.11 - Page: 489\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: N2(g) + 3H2(g) --------> 2NH3\n",
+ "nN2_feed = 1#\n",
+ "nH2_feed = 5#\n",
+ "T = 800## [K]\n",
+ "P = 250## [bar]\n",
+ "K = 1.106*10**(-5)#\n",
+ "#**************#\n",
+ "\n",
+ "# Let e be the degree of completion at equilibrium.\n",
+ "# Moles at Equilibrium:\n",
+ "# nN2_eqb = 1 - e#\n",
+ "# nH2_eqb = 5 - 3*e#\n",
+ "# nNH3_eqb = 2*e#\n",
+ "# Total moles at equilibrium = 1 - e + 5 - 3*e + 2*e = 2*(3 - e)\n",
+ "# Mole Fractions at Equilibrium:\n",
+ "# yN2_eqb = (1 - e)/(2*(3 - e))#\n",
+ "# yH2_eqb = (5 - 3*e)/(2*(3 - e))#\n",
+ "# yNH3_eqb = 2*e/(2*(3 - e))#\n",
+ "# Sum of stoichometric coeffecient:\n",
+ "v = 2 - 3 - 1#\n",
+ "Ky = K*P**(-v)#\n",
+ "def f(e):\n",
+ " y = Ky - ((2*e/(2*(3 - e)))**2)/(((1-e)/((2*(3 - e))))*((5 - 3*e)/(2*(3 - e)))**3)\n",
+ " return y\n",
+ "e = fsolve(f, 0.5)#\n",
+ "print \"Molar Composition of the gases\\n\"\n",
+ "print \"yN2 = %.4f\\n\"%((1 - e)/(2*(3 - e)))#\n",
+ "print \"yH2 = %.4f\\n\"%((5 - 3*e)/(2*(3 - e)))#\n",
+ "print \"yNH3 = %.4f\\n\"%((2*e)/(2*(3 - e)))#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.12 Page: 490"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.12 - Page: 490\n",
+ "\n",
+ "\n",
+ "Molar Composition of the gases\n",
+ "\n",
+ "nSO2 = 0.52\n",
+ "\n",
+ "nO2 = 3.26\n",
+ "\n",
+ "nSO3 = 9.48\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.12 - Page: 490\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: SO2(g) + (1/2)O2 ------> SO3(g)\n",
+ "P = 1## [bar]\n",
+ "T = 750## [K]\n",
+ "K = 74#\n",
+ "#************#\n",
+ "\n",
+ "# Moles in Feed:\n",
+ "nSO2 = 1#\n",
+ "nO2 = 0.5#\n",
+ "# Let e be the degree of completion at equilibrium.\n",
+ "# Moles at Equilibrium:\n",
+ "# nSO2_eqb = 10 - e#\n",
+ "# nO2_eqb = 8 - 0.5*e#\n",
+ "# nSO3_eqb = e#\n",
+ "# Total no. of moles = 10 - e + 8 - 0.5*e + e = 18 -0.5*e#\n",
+ "# Mole fraction at Equilibrium:\n",
+ "# ySO2_eqb = (10 - e)/(18 - 0.5*e)#\n",
+ "# yO2_eqb = (8 - 0.5*e)/(18 - 0.5*e)#\n",
+ "# ySO3_eqb = e/(18 - 0.8*e)#\n",
+ "# Sum of stoichometric coeffecient:\n",
+ "v = 1 - 1 -0.5#\n",
+ "Ky = K*P**(-v)#\n",
+ "def f(e):\n",
+ " y = Ky - (e*(18 - (0.5*e)))/((10 - e)*(8 - 0.5*e))\n",
+ " return y\n",
+ "e = fsolve(f, 7)#\n",
+ "print \"Molar Composition of the gases\\n\"\n",
+ "print \"nSO2 = %.2f\\n\"%((10 - e))#\n",
+ "print \"nO2 = %.2f\\n\"%((8 - 0.5*e))#\n",
+ "print \"nSO3 = %.2f\\n\"%(e)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.13 Page: 492"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.13 - Page: 492\n",
+ "\n",
+ "\n",
+ "Total Pressure Required for 50 % conversion of PCl5 is 5.4 atm\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.13 - Page: 492\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: PCl5 = PCl3 + Cl2\n",
+ "T = 250## [OC]\n",
+ "Kp = 1.8#\n",
+ "e = 0.5#\n",
+ "#**************#\n",
+ "\n",
+ "# Basis: 1 mol of PCl5\n",
+ "# At Equilibrium:\n",
+ "n_PCl5 = 1 - e#\n",
+ "n_PCl3 = e#\n",
+ "n_Cl2 = e#\n",
+ "n_total = n_PCl5 + n_PCl3 + n_Cl2#\n",
+ "# Patrial Pressures:\n",
+ "# P_PCl5 = (n_PCl5/n_total)*P\n",
+ "# P_PCl3 = (n_PCl3/n_total)*P\n",
+ "# P_Cl2 = (n_Cl2/n_total)*P\n",
+ "def f(P):\n",
+ " y = Kp - ((n_PCl3/n_total)*P)*((n_Cl2/n_total)*P)/((n_PCl5/n_total)*P)\n",
+ " return y\n",
+ "P = fsolve(f, 7)## [atm]\n",
+ "print \"Total Pressure Required for 50 %% conversion of PCl5 is %.1f atm\"%(P)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.14 Page: 494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.14 - Page: 494\n",
+ "\n",
+ "\n",
+ "Degree of Conversion is 0.802\n",
+ "\n",
+ "Equilibrium Composition of the reaction Mixture\n",
+ "\n",
+ "ySO2 = 0.0639\n",
+ "\n",
+ "yO2 = 0.0319\n",
+ "\n",
+ "ySO3 = 0.2588\n",
+ "\n",
+ "yAr = 0.6454\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.14 - Page: 494\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: SO2(g) + (1/2)O2(g) -------> SO3(g)\n",
+ "# Moles in Feed:\n",
+ "nSO2_feed = 1#\n",
+ "nO2_feed = 0.5#\n",
+ "nAr_feed = 2#\n",
+ "P = 30## [bar]\n",
+ "T = 900## [K]\n",
+ "K = 6#\n",
+ "#*************#\n",
+ "\n",
+ "# Let e be the degree of completion at equilibrium.\n",
+ "# nSO2_eqb = 1 - e#\n",
+ "# nO2_eqb = 0.5*(1 - e)#\n",
+ "# nSO3_eqb = e#\n",
+ "# nAr_eqb = 2#\n",
+ "# Total moles at equilibrium = 1 - e + 0.5*(1 - e) + e + 2 = (7 - e)/2\n",
+ "# Mole fractions:\n",
+ "# ySO2_eqb = 2*(1 - e)/(7 - e)\n",
+ "# yO2_eqb = (1 - e)/(7 - e)\n",
+ "# ySO3_eqb = 2*e/(7 - e)\n",
+ "# yAr_eqb = 4/(7 - e)\n",
+ "# Sum of Stoichiometric Coeffecient:\n",
+ "v = 1 - 1 - 1/2#\n",
+ "Ky = K*P**(-v)#\n",
+ "e = 0.8#\n",
+ "err = 1#\n",
+ "while err > 0.2:\n",
+ " Ky_new = (2*e/(7 - e))/(((2*(1 - e))/(7 - e))*(((1 - e)/(7 - e))**0.5))#\n",
+ " err = abs(Ky - Ky_new)#\n",
+ " Ky = Ky_new#\n",
+ " e = e + 0.001#\n",
+ "print \"Degree of Conversion is %.3f\\n\"%(e)#\n",
+ "ySO2_eqb = 2*(1 - e)/(7 - e)#\n",
+ "yO2_eqb = (1 - e)/(7 - e)#\n",
+ "ySO3_eqb = 2*e/(7 - e)#\n",
+ "yAr_eqb = 4/(7 - e)#\n",
+ "print \"Equilibrium Composition of the reaction Mixture\\n\"\n",
+ "print \"ySO2 = %.4f\\n\"%(ySO2_eqb)#\n",
+ "print \"yO2 = %.4f\\n\"%(yO2_eqb)#\n",
+ "print \"ySO3 = %.4f\\n\"%(ySO3_eqb)#\n",
+ "print \"yAr = %.4f\\n\"%(yAr_eqb)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.15 Page: 498"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.15 - Page: 498\n",
+ "\n",
+ "\n",
+ "Mole fraction of ethyl acetate is 0.344\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.15 - Page: 498\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: CH3COOH(l) + C2H5OH(l) --------> CH3COOC2H5(l) + H2O(l)\n",
+ "T = 373.15## [K]\n",
+ "nCH3COOH_feed = 1#\n",
+ "nC2H5OH_feed = 1#\n",
+ "deltaHf_CH3COOH = -484.5## [kJ]\n",
+ "deltaHf_C2H5OH = -277.69## [kJ]\n",
+ "deltaHf_CH3COOC2H5 = -480## [kJ]\n",
+ "deltaHf_H2O = -285.83## [kJ]\n",
+ "deltaGf_CH3COOH = -389.9## [kJ]\n",
+ "deltaGf_C2H5OH = -174.78## [kJ]\n",
+ "deltaGf_CH3COOC2H5 = -332.2## [kJ]\n",
+ "deltaGf_H2O = -237.13## [kJ]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#******************#\n",
+ "\n",
+ "deltaH_298 = deltaHf_CH3COOC2H5 + deltaHf_H2O - deltaHf_CH3COOH - deltaHf_C2H5OH## [kJ]\n",
+ "deltaG_298 = deltaGf_CH3COOC2H5 + deltaGf_H2O - deltaGf_CH3COOH - deltaGf_C2H5OH## [kJ]\n",
+ "T0 = 298## [K]\n",
+ "K_298 = exp(-(deltaG_298*1000/(R*T0)))#\n",
+ "K_373 = K_298*exp((deltaH_298*1000/R)*((1/T0) - (1/T)))#\n",
+ "# Let e be the degree of completion at equilibrium.\n",
+ "# nCH3COOH_eqb = 1 - e#\n",
+ "# nC2H5OH_eqb = 1 - e#\n",
+ "# nCH3COOC2H5_eqb = e#\n",
+ "# nH2O_eqb = e#\n",
+ "# Total moles at equilibrium = 1 - e + 1 - e + e + e = 2\n",
+ "# Mole fractions:\n",
+ "# ySO2_eqb = (1 - e)/2\n",
+ "# yO2_eqb = (1 - e)/2\n",
+ "# ySO3_eqb = e/2\n",
+ "# yAr_eqb = e/2\n",
+ "# Sum of Stoichiometric Coeffecient:\n",
+ "v = 1 + 1 - 1 - 1#\n",
+ "def f(e):\n",
+ " y = K_373 - ((e/2)*(e/2))/(((1 - e)/2)*((1 - e)/2))\n",
+ " return y\n",
+ "e = fsolve(f, 0.5)#\n",
+ "print \"Mole fraction of ethyl acetate is %.3f\"%(e/2)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.16 Page: 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.16 - Page: 501\n",
+ "\n",
+ "\n",
+ "The decomposition pressure of limestone is 0.0493 bar at 1000 K\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.16 - Page: 501\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: CaCO3(s) = CaO(s) + CO2(g)\n",
+ "T = 1000## [K]\n",
+ "deltaH_1000 = 1.7533*10**5## [J]\n",
+ "deltaS_1000 = 150.3## [J/mol K]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#****************#\n",
+ "\n",
+ "deltaG_1000 = deltaH_1000 - T*deltaS_1000## [J]\n",
+ "K_1000 = exp(-(deltaG_1000/(R*T)))## [bar]\n",
+ "P = K_1000#\n",
+ "print \"The decomposition pressure of limestone is %.4f bar at 1000 K\"%(P)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.17 Page: 501"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.17 - Page: 501\n",
+ "\n",
+ "\n",
+ "The decomposition pressure is 0.0093 bar at 400 K\n",
+ "\n",
+ "The decomposition pressure is 343.20 bar at 700 K\n",
+ "\n",
+ "The decomposition temperature is 493.450 K\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.17 - Page: 501\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "# Reaction: A(s) ---------> B(s) + C(g)\n",
+ "def f1(T):\n",
+ " deltaG = 85000 - 213.73*T + 6.71*T*log(T) - 0.00028*T**2\n",
+ " return deltaG\n",
+ "T1 = 400## [K]\n",
+ "T2 = 700## [K]\n",
+ "Pc = 1## [bar]\n",
+ "R = 8.314## [J/mol K]\n",
+ "#**************#\n",
+ "\n",
+ "deltaG_400 = f1(400)## [J]\n",
+ "deltaG_700 = f1(700)## [J]\n",
+ "K_400 = exp(-(deltaG_400/(R*T1)))## [bar]\n",
+ "K_700 = exp(-(deltaG_700/(R*T2)))## [bar]\n",
+ "print \"The decomposition pressure is %.4f bar at 400 K\\n\"%(K_400)#\n",
+ "print \"The decomposition pressure is %.2f bar at 700 K\\n\"%(K_700)#\n",
+ "\n",
+ "# Equilibrium constant for solid - gas reaction is:\n",
+ "# K = aB*aC/aA = aC = fC = Pc\n",
+ "def f2(T):\n",
+ " y = Pc - exp(-f1(T)/(R*T))\n",
+ " return y\n",
+ "T = fsolve(f2,900)## [K]\n",
+ "print \"The decomposition temperature is %.3f K\"%(T)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.18 Page: 502"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.18 - Page: 502\n",
+ "\n",
+ "\n",
+ "Equilibrium Composition\n",
+ "\n",
+ "yH2 = 0.34\n",
+ "\n",
+ "y_H2O = 0.66\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.18 - Page: 502\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*****Data******#\n",
+ "T = 875## [K]\n",
+ "K = 0.514#\n",
+ "P = 1## [bar]\n",
+ "#*************#\n",
+ "\n",
+ "# Reaction: Fe(s) + H2O(g) --------->FeO(s) + H2(g)\n",
+ "# K = a_FeO*a_H2/(a_Fe*a_H2O)\n",
+ "# Since the activities of the solid components Fe & FeO at equilibrium may be taken as unity.\n",
+ "# a_Fe = a_FeO = 1\n",
+ "# Ka = a_H2/a_H2O#\n",
+ "# Feed:\n",
+ "nH2O_feed = 1#\n",
+ "nH2feed = 0#\n",
+ "# Let e be the degree of completion at equilibrium.\n",
+ "# Moles at Equilibrium:\n",
+ "# nH2O_eqb = 1 - e#\n",
+ "# nH2_eqb = e#\n",
+ "# Total moles at equilibrium = 1 - e + e = 1\n",
+ "# Mole Fractions at Equilibrium:\n",
+ "# yH20_eqb = 1 - e#\n",
+ "# yH2_eqb = e#\n",
+ "# Sum of stoichometric coeffecient:\n",
+ "v = 1 - 1#\n",
+ "Ky = K*P**(-v)#\n",
+ "# Ky = e/(1 - e)\n",
+ "e = Ky/(Ky + 1)#\n",
+ "yH2_eqb = e#\n",
+ "yH2O_eqb = 1 - e#\n",
+ "print \"Equilibrium Composition\\n\"\n",
+ "print \"yH2 = %.2f\\n\"%(yH2_eqb)#\n",
+ "print \"y_H2O = %.2f\\n\"%(yH2O_eqb)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.19 Page: 503"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.19 - Page: 503\n",
+ "\n",
+ "\n",
+ " Mathematics is involved in proving but just that no numerical computations are involved.\n",
+ "\n",
+ "\n",
+ " For prove refer to this example 12.19 on page 503 of the book.\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.19 - Page: 503\\n\\n\"\n",
+ "\n",
+ "# Mathematics is involved in proving but just that no numerical computations are involved.\n",
+ "# For prove refer to this example 12.19 on page number 503 of the book.\n",
+ "\n",
+ "print \" Mathematics is involved in proving but just that no numerical computations are involved.\\n\\n\"\n",
+ "print \" For prove refer to this example 12.19 on page 503 of the book.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.20 Page: 505"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.20 - Page: 505\n",
+ "\n",
+ "\n",
+ "No. of independent reaction that occur is 2\n",
+ "\n",
+ "No. of Degree of freedom is 4\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.20 - Page: 505\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#Reactions:\n",
+ "# CO + (1/2)O2 ------------> CO2 ......................................(1)\n",
+ "# C + O2 ------------------> CO2 ......................................(2)\n",
+ "# H2 + (1/2)O2 ------------> H2O ......................................(3)\n",
+ "# C + 2H2 -----------------> CH4 ......................................(4)\n",
+ "\n",
+ "# Elimination of C:\n",
+ "# Combining Eqn. (2) with (1):\n",
+ "# CO + (1/2)O2 ------------> CO2 ......................................(5)\n",
+ "# Combining Eqn. (2) with (4):\n",
+ "# CH4 + O2 ----------------> 2H2 + CO2 ................................(6)\n",
+ "\n",
+ "# Elimination of O2:\n",
+ "# Combining Eqn. (3) with (4):\n",
+ "# CO2 + H2 ----------------> CO + H2O .................................(7)\n",
+ "# Combining Eqn. (3) with (6):\n",
+ "# CH4 + 2H2O -------------> CO2 + 4H2 .................................(8)\n",
+ "\n",
+ "# Equations 7 & 8 are independent sets. Hence\n",
+ "r = 2## [No. of independent rkn.]\n",
+ "C = 5## [No. of component]\n",
+ "P = 1## [No. of phases]\n",
+ "s = 0## [No special constraint]\n",
+ "# Applying Eqn. 12.81\n",
+ "F = C - P + 2 - r - s## [Degree of freedom]\n",
+ "print \"No. of independent reaction that occur is %d\\n\"%(r)#\n",
+ "print \"No. of Degree of freedom is %d\"%(F)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.21 Page: 506"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.21 - Page: 506\n",
+ "\n",
+ "\n",
+ "No. of Degree of freedom is 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.21 - Page: 506\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "# Reaction: CaCO3 -----------> CaO + CO2\n",
+ "r = 1## [No. of independent rkn.]\n",
+ "C = 3## [No. of component]\n",
+ "P = 3## [No. of phases, solid CaO, solid CaCO3, gaseous CO2]\n",
+ "s = 0## [No special constraint]\n",
+ "# Applying Eqn. 12.81\n",
+ "F = C - P + 2 - r - s## [Degree of freedom]\n",
+ "print \"No. of Degree of freedom is %d\"%(F)#"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example: 12.22 Page: 508"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example: 12.22 - Page: 508\n",
+ "\n",
+ "\n",
+ "At Equilibrium\n",
+ "\n",
+ "yC4H10 = 0.0009\n",
+ "\n",
+ "yC2H4 = 0.0535\n",
+ "\n",
+ "yC2H6 = 0.0535\n",
+ "\n",
+ "yC3H6 = 0.4461\n",
+ "\n",
+ "yCH4 = 0.4461\n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Example: 12.22 - Page: 508\\n\\n\"\n",
+ "\n",
+ "# Solution\n",
+ "\n",
+ "#*********Data*********#\n",
+ "# Reaction: \n",
+ "# C4H10 -----------> C2H4 + C2H6 ....................................(A)\n",
+ "# C4H10 -----------> C3H6 + CH4 .................................... (B)\n",
+ "T = 750## [K]\n",
+ "P = 1.2## [bar]\n",
+ "Ka = 3.856#\n",
+ "Kb = 268.4#\n",
+ "#************************#\n",
+ "\n",
+ "# Let\n",
+ "# ea = Degree of conversion of C4H10 in reaction (A)\n",
+ "# eb = Degree of conversion of C4H10 in reaction (B)\n",
+ "\n",
+ "# Moles in Feed:\n",
+ "nC4H10_feed = 1#\n",
+ "nC2H4_feed = 0#\n",
+ "nC2H6_feed = 0#\n",
+ "nC3H6_feed = 0#\n",
+ "nCH4_feed = 0#\n",
+ "\n",
+ "# Moles at Equilibrium:\n",
+ "# nC4H10_eqb = 1 - ea - eb\n",
+ "# nC2H4_eqb = ea\n",
+ "# nC2H6_eqb = ea\n",
+ "# nC3H6_eqb = eb\n",
+ "# nCH4_eqb = eb\n",
+ "\n",
+ "# Total moles at equilibrium = 1 - ea - eb + ea + eb + eb = 1 + ea + eb\n",
+ "\n",
+ "# Mole Fraction:\n",
+ "# yC4H10_eqb = (1 - ea - eb)/(1 + ea + eb)\n",
+ "# yC2H4_eqb = ea/(1 + ea + eb)\n",
+ "# yC2H6_eqb = ea/(1 + ea + eb)\n",
+ "# yC3H6_eqb = eb/(1 + ea + eb)\n",
+ "# yCH4_eqb = eb/(1 + ea + eb)\n",
+ "\n",
+ "# Sum of the stoichometric coeffecient:\n",
+ "va = 1 + 1 - 1#\n",
+ "vb = 1 + 1 - 1#\n",
+ "\n",
+ "# e = [ea eb]\n",
+ "# Solution of simultaneous equation\n",
+ "def F(e):\n",
+ " f1 = (((e[0]/(1 + e[0] + e[1]))*(e[0]/(1 + e[0] + e[1])))/((1 - e[0] - e[1])/(1 + e[0] + e[1])))*P**va - Ka#\n",
+ " f2 = (((e[1]/(1 + e[0] + e[1]))*(e[1]/(1 + e[0] + e[1])))/((1 - e[0] - e[1])/(1 + e[0] + e[1])))*P**vb - Kb#\n",
+ " return [f1, f2]\n",
+ "\n",
+ "# Initial guess:\n",
+ "e = [0.1, 0.8]#\n",
+ "y = fsolve(F, e)#\n",
+ "ea = y[0]#\n",
+ "eb = y[1]#\n",
+ "yC4H10_eqb = (1 - ea - eb)/(1 + ea + eb)#\n",
+ "yC2H4_eqb = ea/(1 + ea + eb)#\n",
+ "yC2H6_eqb = ea/(1 + ea + eb)#\n",
+ "yC3H6_eqb = eb/(1 + ea + eb)#\n",
+ "yCH4_eqb = eb/(1 + ea + eb)#\n",
+ "\n",
+ "print \"At Equilibrium\\n\"\n",
+ "print \"yC4H10 = %.4f\\n\"%(yC4H10_eqb)#\n",
+ "print \"yC2H4 = %.4f\\n\"%(yC2H4_eqb)#\n",
+ "print \"yC2H6 = %.4f\\n\"%(yC2H6_eqb)#\n",
+ "print \"yC3H6 = %.4f\\n\"%(yC3H6_eqb)#\n",
+ "print \"yCH4 = %.4f\\n\"%(yCH4_eqb)#"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
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generated by cgit (git 2.34.1) at 2024-12-20 10:54:24 +0000