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| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d (patch) | |
| tree | 22789c9dbe468dae6697dcd12d8e97de4bcf94a2 /Chemistry_by_Raymond_Chang/Chapter_14.ipynb | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
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diff --git a/Chemistry_by_Raymond_Chang/Chapter_14.ipynb b/Chemistry_by_Raymond_Chang/Chapter_14.ipynb new file mode 100755 index 00000000..29fa23a9 --- /dev/null +++ b/Chemistry_by_Raymond_Chang/Chapter_14.ipynb @@ -0,0 +1,359 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14:Chemical Equilibrium"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.2,Page no:622"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "NO=0.0542 #equilibrium conc of NO, M\n",
+ "O2=0.127 #equilibrium conc of O2, M\n",
+ "NO2=15.5 #equilibrium conc of NO2, M\n",
+ "\n",
+ "#Calculation\n",
+ "Kc=NO2**2/(O2*NO**2) #equilibrium constant for given reaction\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of the equilibrium constant of the reaction is %.2e\"%Kc\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of the equilibrium constant of the reaction is 6.44e+05\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.3,Page no:623"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "PCl3=0.463 #equilibrium pressure of PCl3, atm\n",
+ "PCl5=0.875 #equilibrium pressure of PCl5, atm\n",
+ "Kp=1.05 #equilibrium constant of the reaction\n",
+ "\n",
+ "#Calculation\n",
+ "Cl2=Kp*PCl5/PCl3 #equilibrium pressure of Cl2 in atm, formula from the definition of equilibrium constant\n",
+ "\n",
+ "#Result\n",
+ "print\"The value of the equilibrium pressure of the Cl2 gas is :\",round(Cl2,2),\"atm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of the equilibrium pressure of the Cl2 gas is : 1.98 atm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.4,Page no:623"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Kc=10.5 \n",
+ "delta_n=1-3 \n",
+ "T=273+220 \n",
+ "\n",
+ "#Calculation\n",
+ "Kp=Kc*(0.0821*T)**delta_n \n",
+ "\n",
+ "#Result\n",
+ "print\"The value of the equilibrium constant of the reaction is :%.2e\"%Kp"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of the equilibrium constant of the reaction is :6.41e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.6,Page no:626"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CO2=0.236 #pressure of CO2 gas, atm\n",
+ "T=273+800 \n",
+ "\n",
+ "#Calculation\n",
+ "#(a)\n",
+ "Kp=CO2 \n",
+ "#(b)\n",
+ "delta_n=1 \n",
+ "Kc=Kp*(0.0821*T)**-delta_n \n",
+ "\n",
+ "#Result\n",
+ "print\"(a) the value of Kp of the reaction is :\",Kp\n",
+ "print\"(b) the value of Kc of the reaction is %.2e\"%Kc"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) the value of Kp of the reaction is : 0.236\n",
+ "(b) the value of Kc of the reaction is 2.68e-03\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.8,Page no:633"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Kc=1.2 #equilibrium constant for the reaction\n",
+ "N2=.249/3.5 #conc of N2, M\n",
+ "H2=(3.21*10**-2)/3.5 #conc of H2, M\n",
+ "NH3=(6.42*10**-4)/3.5 #conc of NH3, M\n",
+ "\n",
+ "#Calculation\n",
+ "Qc=NH3**2/(N2*H2**3) #reaction quotient initial\n",
+ "print\"Qc=\",round(Qc,3),\"(approx)\"\n",
+ "\n",
+ "#Result\n",
+ "if(Qc==Kc):\n",
+ " d=\"the system is in equilibrium\" \n",
+ "elif(Qc<Kc):\n",
+ " d=\"the system is not in equilibrium and the reaction will move from left to right\" \n",
+ "else:\n",
+ " d=\"the system is not in equilibrium and the reaction will move from right to left\" \n",
+ "print d"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Qc= 0.613 (approx)\n",
+ "the system is not in equilibrium and the reaction will move from left to right\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.9,Page no:635"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Kc=54.3 \n",
+ "H2i=0.5 #initial moles of H2\n",
+ "I2i=0.5 #initial moles of I2\n",
+ "\n",
+ "#Calculation\n",
+ "#Let us assume that x moles have reacted, so, HI=2x, H2=0.5-x, I2=0.5-x, when we substitute in Kc=(HI)**2/(H2)*(I2) we get 54.3=(2x)**2/((0.5-x)*(0.5-x)) taking root we get 7.37=2*x/0.5-x\n",
+ "x=0.393 #from the above equation\n",
+ "H2=0.5-x \n",
+ "I2=0.5-x \n",
+ "HI=2*x \n",
+ "\n",
+ "#Result\n",
+ "print\"The equilibrium concentration of H2 is :\",H2,\"M\" \n",
+ "print\"The equilibrium concentration of I2 is :\",I2,\"M\"\n",
+ "print\"The equilibrium concentration of HI is :\",HI,\"M\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equilibrium concentration of H2 is : 0.107 M\n",
+ "The equilibrium concentration of I2 is : 0.107 M\n",
+ "The equilibrium concentration of HI is : 0.786 M\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.10,Page no:636"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Variable declaration\n",
+ "Kc=54.3 \n",
+ "HIo=0.0224 \n",
+ "H2o=0.00623 \n",
+ "I2o=0.00414 \n",
+ "#let us assume that x moles have reacted, so, HI=HIo+2x, H2=0.00623-x, I2=0.00414-x, when we substitute in Kc=(HI)**2/(H2)*(I2) we get 54.3=(2x+0.0224)**2/((0.00623-x)*(0.00414-x)) simplifying we get 50.3x**2-0.654x+8.98*10**-4=0\n",
+ "a=50.3 \n",
+ "b=-0.654 \n",
+ "c=8.98*10**-4 \n",
+ "\n",
+ "#Calculation\n",
+ "x1=(-b+math.sqrt(b**2-4*a*c))/(2*a) \n",
+ "x2=(-b-math.sqrt(b**2-4*a*c))/(2*a) \n",
+ "if(x1>I2o):\n",
+ " x=x2 \n",
+ "else:\n",
+ " x=x1 \n",
+ "H2=0.00623-x \n",
+ "I2=0.00414-x \n",
+ "HI=2*x+0.0224 \n",
+ "\n",
+ "#Result\n",
+ "print\"The equilibrium concentration of H2 is :\",round(H2,5),\"M\"\n",
+ "print\"The equilibrium concentration of I2 is :\",round(I2,5),\"M\"\n",
+ "print\"The equilibrium concentration of HI is :\",round(HI,4),\"M\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The equilibrium concentration of H2 is : 0.00467 M\n",
+ "The equilibrium concentration of I2 is : 0.00258 M\n",
+ "The equilibrium concentration of HI is : 0.0255 M\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:14.11,Page no:639"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Kc=2.37*10**-3 #equilibrium constant for the reaction\n",
+ "N2=0.683 #conc of N2, M\n",
+ "H2=8.8 #conc of H2, M\n",
+ "NH3=3.65 #conc of NH3, M\n",
+ "\n",
+ "#Calculation\n",
+ "Qc=NH3**2/(N2*H2**3) #reaction quotient initial\n",
+ "print\"Qc=%.2e\"%Qc\n",
+ "\n",
+ "#Result\n",
+ "if(Qc==Kc):\n",
+ " d=\"the system is in equilibrium\" \n",
+ "elif(Qc<Kc):\n",
+ " d=\"the system is not in equilibrium and the reaction will move from left to right\" \n",
+ "else:\n",
+ " d=\"the system is not in equilibrium and the reaction will move from right to left\" \n",
+ "print d\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Qc=2.86e-02\n",
+ "the system is not in equilibrium and the reaction will move from right to left\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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