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Diffstat (limited to 'Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch15.ipynb')
| -rw-r--r-- | Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch15.ipynb | 65 |
1 files changed, 3 insertions, 62 deletions
diff --git a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch15.ipynb b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch15.ipynb index 3eab43ea..a8bdca38 100644 --- a/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch15.ipynb +++ b/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch15.ipynb @@ -1,6 +1,7 @@ { "metadata": { - "name": "" + "name": "", + "signature": "sha256:e78a9af8f6b64f20f90bc773dd4c5b922ed19d33b6608600d787b8b7d2cf559b" }, "nbformat": 3, "nbformat_minor": 0, @@ -27,12 +28,10 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Calculate independent relations among these four species\n", "\n", "import math \n", "\n", "\n", - "# The system contains four species\n", "print \" In this system there are four identifiable chemical species%(which are C,O2,CO2 and CO. The balanced equations we can write among them are\"\n", "\n", "print \" C + 0.5O2 = CO\"\n", @@ -40,20 +39,11 @@ "print \" CO + 0.5O2 = CO2\"\n", "print \" CO2 + C = 2CO\"\n", "\n", - "# Let we call these equations A, B, C and D respectively\n", - "# These relations are not independent.\n", - "# If we add A and C and cancel like terms, we obtain B.\n", - "# So, If we want independent chemical equilibria we must remove equation C\n", "\n", - "# Now, if we reverse the direction of B and add it to A, we see that D is also not independent.\n", - "# Thus, there are only two independent relations among these four species and \n", "print \" There are only two independent relations among these four species and\"\n", "\n", - "# V = C + 2 - P\n", - "# and we have\n", "V = 2# No of the variable\n", "P = 2# No of the phases\n", - "# So\n", "C = V + P - 2\n", "print \" C = V + P - 2\"\n", "print \" C = 4 - 2 = 2\"\n", @@ -92,33 +82,21 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Define Phase rule\n", "\n", "import math \n", "\n", - "#***Data***#\n", - "# This contains three species.\n", "print \" The three species in this system are H2 N2 and NH3\"\n", "N = 3\n", "print \" There is only one balanced chemical reaction among these species\"\n", "Q = 1\n", "\n", - "# 2NH3 = N2 + 3H2\n", "C = N - Q\n", "print \" C = N - Q = %0.0f\"%(C)\n", - "# Now let us we made the system by starting with pure ammonia.\n", - "# Assuming that all the species are in the gas phase, ammonia dissociates in H2 and N2 in the ratio of 3:1.\n", "print \" Let we start with pure ammonia in the system then ammonia will dissociate in H2 and N2 in the ratio of 3:1.\"\n", "\n", - "# We can write an equation among their mole fractions, viz\n", - "# y_H2 = 3*y_N2\n", "print \" And the relation between their mole fraction is y_H2 = 3*y_N2\"\n", "\n", - "# We might modify the phase rule to put in another symbol for stoichiometric restrictions, but the common usage is to write that \n", - "# Components = species - (independent reactions) - (stoichiometric restriction)\n", - "# and stoichiometric restriction SR is \n", "SR = 1\n", - "# so\n", "c = N-Q-SR\n", "print \" We have the modified phase rule as Components = species - independent reactions - stoichiometric restriction\"\n", "print \" C = N - Q - SR = %0.0f\"%(c)\n" @@ -154,24 +132,14 @@ "cell_type": "code", "collapsed": false, "input": [ - "# Find number of the components present in the test tube\n", "\n", "import math \n", "\n", - "#***Data***#\n", - "# We have been given the reaction \n", - "# CaCO3(s) = CaO(s) + CO2(g)\n", "\n", - "# Here we have three species and one balanced chemical reaction between them\n", - "# So\n", "N = 3# No of species\n", "Q = 1 # no of reaction\n", "\n", - "# Since CO2 will mostly be in the gas phase and CaCO3 and CaO will each form separate solid phases, \n", - "# there is no equation we can write among the mole fractions in any of the phases.\n", - "# Hence, there is no stoichiometric restriction i.e.\n", "SR = 0\n", - "# and the number of the components is\n", "C = N - Q - SR\n", "\n", "print \"Number of the components presents in the test tube are %0.0f\"%(C)\n" @@ -201,29 +169,14 @@ "cell_type": "code", "collapsed": false, "input": [ - "'''\n", - "find\n", - "(a) How many phases are present?\n", - "(b) How many degrees of freedom are there?\n", - "(c) If we place a sample of pure CaCO 3 in an evacuated container and heat it, will we find a unique P-T curve?\n", - "'''\n", + "\n", "import math \n", "\n", - "#***Data***#\n", - "# We have been given the reaction \n", - "# CaCO3(s) = CaO(s) + CO2(g)\n", - "# The CaCO3 and CaO form separate solid phases, so we have three phases, two solid and one gas. \n", - "# So\n", "P = 3\n", - "# This is a two component system, so\n", "C = 2\n", "\n", - "# From the phase rule\n", "V = C + 2 - P\n", "\n", - "# If there is only one degree of freedom, then the system should have a unique P-T curve.\n", - "# Reference [ 2, page 214 ] as reported in the book, shows the data to draw such a curve, which can be well represented by\n", - "# math.log(p/torr) = 23.6193 - 19827/T\n", "\n", "print \" The no. of phases present in the system are %0.0f \"%(P)\n", "print \" Total no of degrees of freedom is %0.0f \"%(V)\n", @@ -258,30 +211,18 @@ "cell_type": "code", "collapsed": false, "input": [ - "# find number of components\n", "\n", "import math \n", "\n", - "#***Data***#\n", - "# The system consists of five species.\n", "print \" The five species present in the system are H2O%( HCl%( H+%( OH- and Cl-. \"\n", - "# So\n", "N = 5 # Number of the species \n", "print \" Here we have two chemical relations:\"\n", "print \" H2O = H+ + OH- \"\n", "print \" HCl = H+ + Cl- \"\n", "\n", - "# so\n", "Q = 2 # No of the reactions\n", "\n", - "# In addition we have electroneutrality, which says that at equilibrium the total no of positive ions in the solution must be the same as the total no of nagative ions,or\n", - "# [H+] = [OH-] + [Cl-]\n", - "# To maintain electroneutrality number of positive and negative ion should be same.\n", - "# Here [H+] smath.radians(numpy.arcmath.tan(s for the molality of hydrogen ion. This is convertible to a relation among the 'mu's' hence,\n", - "# it is an additional restriction, so\n", "SR = 1 \n", - "# So\n", - "# The number of components is\n", "C = N - Q - SR\n", "\n", "print \" Number of the components present in the system are C = N - Q - SR = %0.0f\"%(C)\n", |