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H2O + P = H2 + H3PO4

Input interpretation

H_2O water + P red phosphorus ⟶ H_2 hydrogen + H_3PO_4 phosphoric acid
H_2O water + P red phosphorus ⟶ H_2 hydrogen + H_3PO_4 phosphoric acid

Balanced equation

Balance the chemical equation algebraically: H_2O + P ⟶ H_2 + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P ⟶ 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: | 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 = 4 c_2 = 1 c_3 = 5/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 2 c_3 = 5 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 8 H_2O + 2 P ⟶ 5 H_2 + 2 H_3PO_4
Balance the chemical equation algebraically: H_2O + P ⟶ H_2 + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P ⟶ 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: | 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 = 4 c_2 = 1 c_3 = 5/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 2 c_3 = 5 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 H_2O + 2 P ⟶ 5 H_2 + 2 H_3PO_4

Structures

 + ⟶ +
+ ⟶ +

Names

water + red phosphorus ⟶ hydrogen + phosphoric acid
water + red phosphorus ⟶ hydrogen + phosphoric acid

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + P ⟶ 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: 8 H_2O + 2 P ⟶ 5 H_2 + 2 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 | 8 | -8 P | 2 | -2 H_2 | 5 | 5 H_3PO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 8 | -8 | ([H2O])^(-8) P | 2 | -2 | ([P])^(-2) H_2 | 5 | 5 | ([H2])^5 H_3PO_4 | 2 | 2 | ([H3PO4])^2 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])^(-8) ([P])^(-2) ([H2])^5 ([H3PO4])^2 = (([H2])^5 ([H3PO4])^2)/(([H2O])^8 ([P])^2)
Construct the equilibrium constant, K, expression for: H_2O + P ⟶ 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: 8 H_2O + 2 P ⟶ 5 H_2 + 2 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 | 8 | -8 P | 2 | -2 H_2 | 5 | 5 H_3PO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 8 | -8 | ([H2O])^(-8) P | 2 | -2 | ([P])^(-2) H_2 | 5 | 5 | ([H2])^5 H_3PO_4 | 2 | 2 | ([H3PO4])^2 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])^(-8) ([P])^(-2) ([H2])^5 ([H3PO4])^2 = (([H2])^5 ([H3PO4])^2)/(([H2O])^8 ([P])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2O + P ⟶ 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: 8 H_2O + 2 P ⟶ 5 H_2 + 2 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 | 8 | -8 P | 2 | -2 H_2 | 5 | 5 H_3PO_4 | 2 | 2 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 | 8 | -8 | -1/8 (Δ[H2O])/(Δt) P | 2 | -2 | -1/2 (Δ[P])/(Δt) H_2 | 5 | 5 | 1/5 (Δ[H2])/(Δt) H_3PO_4 | 2 | 2 | 1/2 (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 = -1/8 (Δ[H2O])/(Δt) = -1/2 (Δ[P])/(Δt) = 1/5 (Δ[H2])/(Δt) = 1/2 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + P ⟶ 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: 8 H_2O + 2 P ⟶ 5 H_2 + 2 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 | 8 | -8 P | 2 | -2 H_2 | 5 | 5 H_3PO_4 | 2 | 2 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 | 8 | -8 | -1/8 (Δ[H2O])/(Δt) P | 2 | -2 | -1/2 (Δ[P])/(Δt) H_2 | 5 | 5 | 1/5 (Δ[H2])/(Δt) H_3PO_4 | 2 | 2 | 1/2 (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 = -1/8 (Δ[H2O])/(Δt) = -1/2 (Δ[P])/(Δt) = 1/5 (Δ[H2])/(Δt) = 1/2 (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | water | red phosphorus | hydrogen | phosphoric acid formula | H_2O | P | H_2 | H_3PO_4 Hill formula | H_2O | P | H_2 | H_3O_4P name | water | red phosphorus | hydrogen | phosphoric acid IUPAC name | water | phosphorus | molecular hydrogen | phosphoric acid
| water | red phosphorus | hydrogen | phosphoric acid formula | H_2O | P | H_2 | H_3PO_4 Hill formula | H_2O | P | H_2 | H_3O_4P name | water | red phosphorus | hydrogen | phosphoric acid IUPAC name | water | phosphorus | molecular hydrogen | phosphoric acid

Substance properties

 | water | red phosphorus | hydrogen | phosphoric acid molar mass | 18.015 g/mol | 30.973761998 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 | 579.2 °C | -259.2 °C | 42.4 °C boiling point | 99.9839 °C | | -252.8 °C | 158 °C density | 1 g/cm^3 | 2.16 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) | 7.6×10^-4 Pa s (at 20.2 °C) | 8.9×10^-6 Pa s (at 25 °C) |  odor | odorless | | odorless | odorless
| water | red phosphorus | hydrogen | phosphoric acid molar mass | 18.015 g/mol | 30.973761998 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 | 579.2 °C | -259.2 °C | 42.4 °C boiling point | 99.9839 °C | | -252.8 °C | 158 °C density | 1 g/cm^3 | 2.16 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) | 7.6×10^-4 Pa s (at 20.2 °C) | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | | odorless | odorless

Units