{
"metadata": {
"name": "",
"signature": "sha256:b3b8e3a1df1ec221a5596f2fd615d0e7df4ba35a4012aeb734d5820e6983cd64"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13 - Thermodynamic changes accompanying chemical reaction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - pg 320"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the dHp value\n",
"#Initialization of variables\n",
"n1=10 #mol\n",
"n2=12 #mol\n",
"#calculations\n",
"dn=n1-n2\n",
"#results\n",
"print '%s %d %s' %(\"dHp = dEv-\",dn,\"*RT\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"dHp = dEv- -2 *RT\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - pg 322"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Enthalpy\n",
"#Initialization of variables\n",
"Ht1=-22063 #cal\n",
"T=298.15 #K\n",
"#calculations\n",
"H=Ht1 +0.5293*T + 0.3398*10**-3 *T**2 - 2.039*10**-7 *T**3\n",
"#results\n",
"print '%s %d %s' %(\"Enthalpy =\",H,\"cal\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Enthalpy = -21880 cal\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - pg 326"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Entropy\n",
"#Initialization of variables\n",
"Cp=0.797 #cal/deg/mol\n",
"#calculations\n",
"S=Cp/3.\n",
"#results\n",
"print '%s %.3f %s' %(\"Entropy =\",S,\" eu/mol\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Entropy = 0.266 eu/mol\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - pg 328"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Change in entropy\n",
"#Initialization of variables\n",
"T1=77.32 #K\n",
"P=1 #atm\n",
"T2=126 #K\n",
"Pc=33.5 #atm\n",
"#calculations\n",
"dS=27/32. *1.987*P/Pc *(T2/T1)**3\n",
"#results\n",
"print '%s %.2f %s' %(\"Change in entropy =\",dS,\"eu/mol\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in entropy = 0.22 eu/mol\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5 - pg 330"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Change in entropy, enthalpy and free energy\n",
"#Initialization of variables\n",
"S1=57.47\n",
"S2=50.34\n",
"S3=49\n",
"H1=8.09\n",
"H2=21.06\n",
"H3=0\n",
"F1=12.39\n",
"F2=20.72\n",
"F3=0\n",
"#calculations\n",
"dS=S1-S2-0.5*S3\n",
"dH=H1-H2-0.5*H3\n",
"dF=F1-F2-0.5*F3\n",
"#results\n",
"print '%s %.2f %s' %(\"Change in entropy =\",dS,\" eu\")\n",
"print '%s %.2f %s' %(\"\\n Change in enthalpy =\",dH,\" kcal\")\n",
"print '%s %.2f %s' %(\"\\n Change in free energy =\",dF,\"kcal\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Change in entropy = -17.37 eu\n",
"\n",
" Change in enthalpy = -12.97 kcal\n",
"\n",
" Change in free energy = -8.33 kcal\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6 - pg 334"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the change in free energy\n",
"#Initialization of variables\n",
"import math\n",
"P1=0.01\n",
"P2=0.1\n",
"P3=0.01\n",
"dF0=-54640 #cal\n",
"T=298.15 #K\n",
"R=1.987 #cal/deg\n",
"#calculations\n",
"Qp=P1/(P2*P3**0.5)\n",
"dF=dF0+R*T*math.log(Qp)\n",
"#results\n",
"print '%s %d %s' %(\"change in free energy =\",dF,\"cal\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"change in free energy = -54640 cal\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7 - pg 335"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Equilibrium constant\n",
"#Initialization of variables\n",
"print \"From table 13.4 \"\n",
"logKfwater=40.04724\n",
"logKfH2=0\n",
"logKfO2=0\n",
"#calculations\n",
"logK=logKfwater-logKfH2-0.5*logKfO2\n",
"K=10**logK\n",
"#results\n",
"print '%s %.4e' %(\"Equilibrium constant = \",K)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 13.4 \n",
"Equilibrium constant = 1.1149e+40\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8 - pg 339"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the value of Kc\n",
"#Initialization of variables\n",
"Kp=1.1*10**40 #atm**-0.5\n",
"dn=-0.5\n",
"R=0.08206 #lt atm/deg mol\n",
"T=298.15 #K\n",
"#calculations\n",
"Kc=Kp*(R*T)**(-dn)\n",
"#results\n",
"print '%s %.1e %s' %(\"Kc =\",Kc,\" (mol/lt)^-0.5\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Kc = 5.4e+40 (mol/lt)^-0.5\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9 - pg 339"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Density of equilibrium mixture\n",
"#Initialization of variables\n",
"import numpy\n",
"Kp=0.141 #atm\n",
"P=1 #atm\n",
"nu=2\n",
"R=0.08206 #lt atm/deg mol\n",
"T=298.15 #K\n",
"M=92.02 #g/mol\n",
"#calculations\n",
"p=([Kp+ 4*P,0, -Kp])\n",
"z=numpy.roots(p)\n",
"alpha=z[0]\n",
"wbyV=P*M/(R*T*(1+(nu-1)*alpha))\n",
"#results\n",
"print '%s %.2f %s' %(\"Density of the equilibrium mixture =\",wbyV,\" g/lt\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Density of the equilibrium mixture = 3.18 g/lt\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10 - pg 340"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Final pressure after equilibrium\n",
"#Initialization of variables\n",
"x=0.5\n",
"P=0.468 #atm\n",
"#calculations\n",
"P1=x*P\n",
"P2=x*P\n",
"Kp=P1*P2\n",
"#results\n",
"print '%s %.4f %s' %(\"Final pressure after equilibrium =\",Kp,\" atm^2\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Final pressure after equilibrium = 0.0548 atm^2\n"
]
}
],
"prompt_number": 12
}
],
"metadata": {}
}
]
}