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{
"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": {}
}
]
}
|
