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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11 - Accounting for the rate laws"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E1 - Pg 255"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Max. velocity and Michaelis constant\n",
"#Initialization of variables\n",
"import math\n",
"import numpy as np\n",
"from numpy import linalg\n",
"S=[10, 20, 40, 80, 120, 180, 300]\n",
"v=[0.32, 0.58, 0.9, 1.22, 1.42, 1.58, 1.74]\n",
"#calculations\n",
"n=len(S)\n",
"def fun1(x):\n",
"\tfor i in range(0,len(x)):\n",
"\t\tx[i]=1000./(x[i])\n",
"\treturn x\n",
"\n",
"\n",
"def fun2(x):\n",
"\tfor i in range(0,len(x)):\n",
"\t\tx[i]=1./(x[i])\n",
"\treturn x\n",
"bys=fun1(S)\n",
"byv=fun2(v)\n",
"x=bys\n",
"y=byv\n",
"A = np.vstack([x, np.ones(len(x))]).T\n",
"m1, b1 = np.linalg.lstsq(A, y)[0]\n",
"print '%s' %(\"From graph,\")\n",
"vmax=1/b1\n",
"Km=m1*1000./b1\n",
"#results\n",
"print '%s %.2f %s' %(\"Max. velocity =\",vmax,\" mumol/L s\")\n",
"print '%s %.1f %s' %(\"\\n Michaelis constant =\",Km,\" mumol/L\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From graph,\n",
"Max. velocity = 2.10 mumol/L s\n",
"\n",
" Michaelis constant = 55.0 mumol/L\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E2 - Pg 257"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Equilibrium constant\n",
"#Initialization of variables\n",
"c=1.234\n",
"m=2.044\n",
"#calculations\n",
"Ki=c/m\n",
"#results\n",
"print '%s %.2f' %(\"KI = \",Ki)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"KI = 0.60\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E3 - Pg 265"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the no. of diheptane molecules destroyed \n",
"#Initialization of variables\n",
"import math\n",
"P=50. #J/s\n",
"l=313.*math.pow(10,-9) #m\n",
"h=6.62608*math.pow(10,-34) #Js\n",
"N=6.023*math.pow(10,23)\n",
"c=2.99792*math.pow(10,8) #m/s\n",
"yiel=0.21\n",
"#calculations\n",
"rate=P*l/(h*c)\n",
"Frate=yiel*rate\n",
"molrate=Frate/N\n",
"#results\n",
"print '%s %.1e %s' %(\"No.of diheptane molecules destroyed =\",Frate,\" s^-1\")\n",
"print '%s %.2e %s' %(\"\\n Moles of diheptane molecules destroyed =\",molrate,\"mol s^-1\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"No.of diheptane molecules destroyed = 1.7e+19 s^-1\n",
"\n",
" Moles of diheptane molecules destroyed = 2.75e-05 mol s^-1\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I1 - Pg 243"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Equilibrium constant for dimerization\n",
"#Initialization of variables\n",
"import math\n",
"kf=8.18*math.pow(10,8) #L/mol s\n",
"kb=2*math.pow(10,6) #s^-1\n",
"#calculations\n",
"K=kf/kb\n",
"#results\n",
"print '%s %.1e' %(\"Equilibrium constant for dimerization = \",K)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Equilibrium constant for dimerization = 4.1e+02\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I2 - Pg 248"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate if the reaction step is far from equilibrium and calculate the heat generated\n",
"#Initialization of variables\n",
"import math\n",
"F16bP=1.9*math.pow(10,-5) #mmol/L\n",
"ADP=1.3/1000. #mmol/L\n",
"ATP=11.4/1000. #mmol/L\n",
"F6P=8.9*math.pow(10,-5) #mmol/L\n",
"k=1.2*1000.\n",
"#calculations\n",
"Q=F16bP*ADP/(F6P*ATP)\n",
"if(Q<k):\n",
" print '%s %.3f' %(\"The reaction step is far from equilibrium and Q= \",Q)\n",
"else:\n",
" print '%s %.3f' %(\"The reaction step is at equilibrium and Q= \",Q)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The reaction step is far from equilibrium and Q= 0.024\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
