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RxnEqn

This is code to support the computation of stoichiometric equations for the Thermodynamic Electon Equivalents Model (TEEM) developed by Rittman and McCarty.

Installation

To install:

git clone https://github.com/djinnome/rxneqn.git
cd rxneqn
pip install .

Usage

ChemicalFormula

>>> from rxneqn import ChemicalFormula
>>> chemical_formula = 'C6H12O5.5N'
>>> cf = ChemicalFormula(chemical_formula)
>>> cf.get_chemical_composition()
C            6
Charge       0
H           12
N            1
O         11/2
dtype: object

Mixture

>>> from rxneqn import Mixture
>>> mix = Mixture('3 C6H12O6 + 4 NO2-')
>>> mix.subtract_from_mixture(Mixture('5 NO2- + H+'))
3 C6H12O6

Reaction

>>> from rxneqn import Reaction
>>> rxn = Reaction('1/4 CO2 + 1/12NH3 + H+  ==> 1/12 CH3CHNH2COOH + 1/3 H2O + OH-')
>>> rxn -  Reaction('CO2 + H2O ==> 4 HCO3')
H+ + 4 HCO3 + 1/12 NH3 ==> 1/12 CH3CHNH2COOH + 3/4 CO2 + 4/3 H2O + OH-

rxn.normalize()
3 CO2 + 12 H+ + NH3 ==> CH3CHNH2COOH + 4 H2O + 12 OH-

HalfReactionBalancer

>>> from rxneqn import HalfReactionBalancer
>>> alanine = 'CH3CHNH2COOH'
>>> carbon_dioxide = 'CO2'
>>> nitrogen_source = 'NH3'
>>> half_rxn = HalfReactionBalancer() #carbon_dioxide, alanine, 'C', nitrogen_source)
>>> truth_eqn = """1/24 C6H12O6 + 1/4 H2O ==> 1/4 CO2 + E- + H+
E- + 6/5 H+ + 1/5 NO3- ==> 3/5 H2O + 1/10 N2
1/24 C6H12O6 + 1/5 H+ + 1/5 NO3- ==> 1/4 CO2 + 7/20 H2O + 1/10 N2
1/4 CO2 + E- + H+ + 1/12 NH3 ==> 1/12 CH3CHNH2COOH + 1/3 H2O
1/5 CO2 + E- + H+ + 1/20 HCO3- + 1/20 NH4+ ==> 1/20 C5H7O2N + 9/20 H2O
1/8 CO2 + E- + H+ ==> 1/8 CH4 + 1/4 H2O
1/6 CO2 + E- + H+ + 1/12 HCO3- + 1/12 NH4+ ==> 1/12 CH3CHNH2COOH + 5/12 H2O
1/12 CH3CHNH2COOH + 1/6 H2O ==> 1/8 CH4 + 1/24 CO2 + 1/12 HCO3- + 1/12 NH4+
1/5 CO2 + E- + H+ + 1/10 HCO3- ==> 1/10 CH3COCOO- + 2/5 H2O
1/4 CO2 + E- + H+ + 1/12 NH3 ==> 1/12 CH3CHNH2COOH + 1/3 H2O
1/6 CO2 + E- + H+ + 1/12 HCO3- + 1/12 NH4+ ==> 1/12 CH3CHNH2COOH + 5/12 H2O
1/12 CO2 + E- + H+ + 1/6 HCO3- + 1/12 NH4+ ==> 1/12 CH3CHNH2COO- + 1/2 H2O""".split('\n')
>>> eqn = {}
>>> eqn[13] = half_rxn.balance_half_reaction('C6H12O6', 'CO2')
>>> eqn[14] = half_rxn.balance_half_reaction('NO3-', 'N2',nitrogen_source='NO3-')
>>> eqn[15] = eqn[13] +  eqn[14] 
>>> eqn[16] = half_rxn.balance_half_reaction(carbon_dioxide, alanine, nitrogen_source)
>>> eqn[17] = half_rxn.balance_half_reaction('CO2', 'C5H7O2N','NH4+')
>>> table3 = {'O-12': '1/8 CO2 + H+ + E- ==> 1/8 CH4 + 1/4 H2O',
         'O-2': '1/6 CO2 + 1/12 NH4+ + 1/12 HCO3- + H+ + E- ==> 1/12 CH3CHNH2COOH + 5/12 H2O',
         'O-3': str(half_rxn.balance_half_reaction('CO2','C6H5COO-'))}

>>> eqn[18] = Reaction(table3['O-12'])
>>> eqn[19] = Reaction(table3['O-2'])
>>> eqn[20] = eqn[18] - eqn[19]
>>> eqn[21] = half_rxn.balance_half_reaction('CO2','CH3COCOO-')
>>> eqn[22] = half_rxn.balance_half_reaction('CO2','CH3CHNH2COOH','NH3')
>>> eqn[23] = half_rxn.balance_half_reaction('CO2','CH3CHNH2COOH','NH4+')
>>> eqn[24] = half_rxn.balance_half_reaction('CO2','CH3CHNH2COO-','NH4+')

>>> k = 0

>>> for i in eqn:
    	assert( str(eqn[i]) == truth_eqn[k] )
    	k += 1
>>> print("{} out of {} tests passed".format(k, len(eqn))
12 out of 12 tests passed