H2O + P4 = H2 + H3PO4

H_2O water + P_4 white phosphorus ⟶ H_2 hydrogen + H_3PO_4 phosphoric acid

Balance the chemical equation algebraically: H_2O + P_4 ⟶ H_2 + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P_4 ⟶ c_3 H_2 + c_4 H_3PO_4 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 = 2 c_3 + 3 c_4 O: | c_1 = 4 c_4 P: | 4 c_2 = c_4 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 = 16 c_2 = 1 c_3 = 10 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 H_2O + P_4 ⟶ 10 H_2 + 4 H_3PO_4

+ ⟶ +

water + white phosphorus ⟶ hydrogen + phosphoric acid

Construct the equilibrium constant, K, expression for: H_2O + P_4 ⟶ H_2 + H_3PO_4 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 16 H_2O + P_4 ⟶ 10 H_2 + 4 H_3PO_4 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i H_2O | 16 | -16 P_4 | 1 | -1 H_2 | 10 | 10 H_3PO_4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 16 | -16 | ([H2O])^(-16) P_4 | 1 | -1 | ([P4])^(-1) H_2 | 10 | 10 | ([H2])^10 H_3PO_4 | 4 | 4 | ([H3PO4])^4 The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([H2O])^(-16) ([P4])^(-1) ([H2])^10 ([H3PO4])^4 = (([H2])^10 ([H3PO4])^4)/(([H2O])^16 [P4])

Construct the rate of reaction expression for: H_2O + P_4 ⟶ H_2 + H_3PO_4 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 16 H_2O + P_4 ⟶ 10 H_2 + 4 H_3PO_4 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i H_2O | 16 | -16 P_4 | 1 | -1 H_2 | 10 | 10 H_3PO_4 | 4 | 4 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term H_2O | 16 | -16 | -1/16 (Δ[H2O])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) H_2 | 10 | 10 | 1/10 (Δ[H2])/(Δt) H_3PO_4 | 4 | 4 | 1/4 (Δ[H3PO4])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/16 (Δ[H2O])/(Δt) = -(Δ[P4])/(Δt) = 1/10 (Δ[H2])/(Δt) = 1/4 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | white phosphorus | hydrogen | phosphoric acid formula | H_2O | P_4 | H_2 | H_3PO_4 Hill formula | H_2O | P_4 | H_2 | H_3O_4P name | water | white phosphorus | hydrogen | phosphoric acid IUPAC name | water | tetraphosphorus | molecular hydrogen | phosphoric acid

| water | white phosphorus | hydrogen | phosphoric acid molar mass | 18.015 g/mol | 123.89504799 g/mol | 2.016 g/mol | 97.994 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | liquid (at STP) melting point | 0 °C | 44.15 °C | -259.2 °C | 42.4 °C boiling point | 99.9839 °C | 280.5 °C | -252.8 °C | 158 °C density | 1 g/cm^3 | 1.823 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 1.685 g/cm^3 solubility in water | | insoluble | | very soluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | odorless | odorless | odorless