#### Input interpretation

water + phosphorus pentoxide ⟶ phosphoric acid

#### Balanced equation

Balance the chemical equation algebraically: + ⟶ Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and P: H: | 2 c_1 = 3 c_3 O: | c_1 + 10 c_2 = 4 c_3 P: | 4 c_2 = c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 + ⟶ 4

#### Structures

+ ⟶

#### Names

water + phosphorus pentoxide ⟶ phosphoric acid

#### Reaction thermodynamics

Gibbs free energy

| water | phosphorus pentoxide | phosphoric acid molecular free energy | -237.1 kJ/mol | -2698 kJ/mol | -1124 kJ/mol total free energy | -1423 kJ/mol | -2698 kJ/mol | -4494 kJ/mol | G_initial = -4121 kJ/mol | | G_final = -4494 kJ/mol ΔG_rxn^0 | -4494 kJ/mol - -4121 kJ/mol = -373.8 kJ/mol (exergonic) | |