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diff --git a/Physical_Chemsitry/Chapter19.ipynb b/Physical_Chemsitry/Chapter19.ipynb new file mode 100755 index 00000000..5b2d735e --- /dev/null +++ b/Physical_Chemsitry/Chapter19.ipynb @@ -0,0 +1,356 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f00e66b23ba11c1502ba60271f6e9b549183dbf77dfd61ae21a7394bb7c1f4ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 19 - Photochemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1 - pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Overall transmittance\n",
+ "#Initialization of variables\n",
+ "r1=0.727\n",
+ "r2=0.407\n",
+ "#calculations\n",
+ "r3=r1*r2\n",
+ "#results\n",
+ "print '%s %.3f' %(\"Overall transmittance = \",r3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Overall transmittance = 0.296\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2 - pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Extinction coefficient\n",
+ "import math\n",
+ "#Initialization of variables\n",
+ "r=0.450\n",
+ "c=0.02 #M\n",
+ "l=4 #cm\n",
+ "#calculations\n",
+ "e=-math.log10(r) /(c*l)\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Extinction coefficient =\",e,\"litres mole^-1 cm^-1\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extinction coefficient = 4.33 litres mole^-1 cm^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3 - pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Transmittance of the solution\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "from math import log10\n",
+ "r1=0.850\n",
+ "r2=0.50\n",
+ "#calculations\n",
+ "Da=-log10(r1)\n",
+ "Db=-log10(r2)\n",
+ "D=Da+Db\n",
+ "r3=10**(-D)\n",
+ "#results\n",
+ "print '%s %.3f' %(\"Transmittance of solution =\",r3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transmittance of solution = 0.425\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4 - pg 491"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Extinction coefficient\n",
+ "#Initialization of variables\n",
+ "c=0.000025 #M\n",
+ "l=2 #cm\n",
+ "D=0.417\n",
+ "#calculations\n",
+ "e=D/(c*l)\n",
+ "#result\n",
+ "print '%s %d %s' %(\"Extinction coefficient =\",e,\" liters mole^-1 cm^-1\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extinction coefficient = 8340 liters mole^-1 cm^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5 - pg 491"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Kc for dissociation\n",
+ "#Initialization of variables\n",
+ "c=0.5 #M\n",
+ "c1=0.000025 #M\n",
+ "D2=0.280\n",
+ "D1=0.417\n",
+ "#calculations\n",
+ "c2=D2*c1/(D1)\n",
+ "dC=c1-c2\n",
+ "SCN=c- 6*c2 -4*dC\n",
+ "K=dC*SCN**2 /c2\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Kc for dissociation =\",K,\" M^2\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kc for dissociation = 0.12 M^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6 - pg 492"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the value of m\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "from math import log\n",
+ "D2=0.249\n",
+ "D1=0.172\n",
+ "a2=0.00752\n",
+ "a1=0.00527\n",
+ "#calculations\n",
+ "m=(log(D2) -log(D1))/(log(a2) - log(a1))\n",
+ "#results\n",
+ "print '%s %.2f' %(\"m = \",m)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "m = 1.04\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7 - pg 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Increase in optical density\n",
+ "#Initialization of variables\n",
+ "c=0.1 #M\n",
+ "V=100 #ml\n",
+ "v1=25 #ml\n",
+ "D=0.980\n",
+ "d1=0.090\n",
+ "d2=0.150\n",
+ "#calculations\n",
+ "a=v1*c/V\n",
+ "b=(V-v1)*c/V\n",
+ "Da=a*d1/c\n",
+ "Db=b*d2/c\n",
+ "Ddash=Da+Db\n",
+ "dD=D-Ddash\n",
+ "#results\n",
+ "print '%s %.3f' %(\"Increase in optical density =\",dD)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Increase in optical density = 0.845\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8 - pg 496"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the value of lambda for the reaction to occur\n",
+ "#Initialization of variables\n",
+ "E=50000. #cal/mol\n",
+ "#calculations\n",
+ "lam=2.8593/E\n",
+ "#results\n",
+ "print '%s %d %s' %(\"For the reaction to occur lambda <\",lam*10**8,\"A\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For the reaction to occur lambda < 5718 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9 - pg 497"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Amount of reactant disappeared\n",
+ "#Initialization of variables\n",
+ "lam=3000*10**-8 #cm\n",
+ "ield=0.420\n",
+ "Et=70000 #cal\n",
+ "#calculations\n",
+ "E=2.8593/lam\n",
+ "n=ield*Et/E\n",
+ "#results\n",
+ "print '%s %.3f %s' %(\"Amount of reactant disappeared =\",n,\" mol\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of reactant disappeared = 0.308 mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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