Chemical Reactions¶

Making a Reaction¶

class chemlib.chemistry.Reaction(*args: Any, **kwargs: Any)¶

Instantiate a chemlib.Reaction object with a list of reactant Compounds and product Compounds.

>>> from chemlib import Compound, Reaction
>>> N2O5 = Compound("N2O5")
>>> H2O = Compound("H2O")
>>> HNO3 = Compound("HNO3")
>>> r = Reaction([N2O5, H2O], [HNO3])

classmethod chemlib.chemistry.Reaction.by_formula(cls, formula: str)¶

OR Instantiate a chemlib.Reaction object using a string to represent the formula:

In the formula string, place the formulae of reactants separated by + signs on the left side of the arrow -->. On the right side of the arrow, place the formulae of products separated by + signs:

>>> from chemlib import Reaction
>>> r = Reaction.by_formula("N2O5 + H2O --> HNO3")

chemlib.chemistry.Reaction.formula: str¶
>>> r.formula
'1N₂O₅ + 1H₂O₁ --> 1H₁N₁O₃'
chemlib.chemistry.Reaction.is_balanced: boolean¶
>>> r.is_balanced
False
chemlib.chemistry.Reaction.reactant_formulas: list¶
>>> r.reactant_formulas
['N₂O₅', 'H₂O₁']
chemlib.chemistry.Reaction.product_formulas: list¶
>>> r.product_formulas
['H₁N₁O₃']

Combustion Reactions¶

class chemlib.chemistry.Combustion(compound)¶

Inherits from chemlib.chemistry.Reaction

Makes a chemical reaction involving the combustion of one compound. Formula will be balanced.

>>> from chemlib import Compound, Combustion
>>> methane = Compound('CH4')
>>> c = Combustion(methane)
>>> c.formula
'1C₁H₄ + 2O₂ --> 2H₂O₁ + 1C₁O₂'
>>> c.is_balanced
True

Balancing the Equation¶

chemlib.chemistry.Reaction.balance(self) → None¶

Balances the chemical equation using linear algebra. See Applications of Linear Algebra in Chemistry.

>>> r.balance()
>>> r.formula
'1N₂O₅ + 1H₂O₁ --> 2H₁N₁O₃'
>>> r.is_balanced
True

Stoichiometry¶

chemlib.chemistry.Reaction.get_amounts(self, compound_number, **kwargs)¶

Get stoichiometric amounts of ALL compounds in the reaction given the amount of one compound.

Parameters

compound_number (int) – The chosen compound in the reaction by order of appearance.
kwargs – The amount of the chosen compound (grams=, moles=, or molecules=)

Returns

the amounts of each compound in the reaction

Return type

list of dicts

Raises

ValueError – if the compound_number is less than 1 or greater than the number of compounds in the reaction
ValueError – if more than one argument is given under kwargs

Best demonstrated by example:

>>> r.formula
'1N₂O₅ + 1H₂O₁ --> 2H₁N₁O₃'

Get the amounts of ALL compounds in the above reaction given 5 grams of N₂O₅. It is the first compound in the reaction by order of appearance (left to right).

>>> r.get_amounts(1, grams=5)
[{'Compound': 'N₂O₅', 'Grams': 5, 'Moles': 0.0463, 'Molecules': 2.787e+22}, {'Compound': 'H₂O₁', 'Grams': 0.834, 'Moles': 0.0463, 'Molecules': 2.787e+22}, {'Compound': 'H₁N₁O₃', 'Grams': 5.835, 'Moles': 0.0926, 'Molecules': 5.575e+22}]

Get the amounts of ALL compounds in the above reaction given 3.5 moles of HNO₃. It is the third compound in the reaction by order of appearance (left to right).

>>> r.get_amounts(3, moles=3.5)
[{'Compound': 'N₂O₅', 'Grams': 189.018, 'Moles': 1.75, 'Molecules': 1.054e+24}, {'Compound': 'H₂O₁', 'Grams': 31.518, 'Moles': 1.75, 'Molecules': 1.054e+24}, {'Compound': 'H₁N₁O₃', 'Grams': 220.535, 'Moles': 3.5, 'Molecules': 2.107e+24}]

Limiting Reagent¶

chemlib.chemistry.Reaction.limiting_reagent(self, *args, mode='grams')¶

Get the limiting reagent (limiting reactant) in the chemical reaction.

Parameters

args – The amounts of each reactant to use in the chemical reaction.
mode (str) – The units of each amount in args. Default is grams, can also be moles or molecules.

Returns

The limiting reagent of the reaction.

Return type

chemlib.chemistry.Compound

Raises

TypeError – If the number of args doesn’t match the number of reactants in the reaction.
ValueError – If the mode is not grams, moles, or molecules.

Find the limiting reagent of the reaction when using 50 grams of the first reactant (N₂O₅) and 80 grams of the second reactant (H₂O):

>>> lr = r.limiting_reagent(50, 80)
>>> lr.formula
'N₂O₅'

Find the limiting reagent of the reaction when using 3 moles of the first reactant (N₂O₅) and 1 mole of the second reactant (H₂O):

>>> lr = r.limiting_reagent(3, 1, mode = 'moles')
>>> lr.formula
'H₂O₁'