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H3PO4 + K = H2 + KPO4

Input interpretation

H_3PO_4 phosphoric acid + K potassium ⟶ H_2 hydrogen + KPO4
H_3PO_4 phosphoric acid + K potassium ⟶ H_2 hydrogen + KPO4

Balanced equation

Balance the chemical equation algebraically: H_3PO_4 + K ⟶ H_2 + KPO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 K ⟶ c_3 H_2 + c_4 KPO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and K: H: | 3 c_1 = 2 c_3 O: | 4 c_1 = 4 c_4 P: | c_1 = c_4 K: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 2 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 H_3PO_4 + 2 K ⟶ 3 H_2 + 2 KPO4
Balance the chemical equation algebraically: H_3PO_4 + K ⟶ H_2 + KPO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 K ⟶ c_3 H_2 + c_4 KPO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and K: H: | 3 c_1 = 2 c_3 O: | 4 c_1 = 4 c_4 P: | c_1 = c_4 K: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 2 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + 2 K ⟶ 3 H_2 + 2 KPO4

Structures

 + ⟶ + KPO4
+ ⟶ + KPO4

Names

phosphoric acid + potassium ⟶ hydrogen + KPO4
phosphoric acid + potassium ⟶ hydrogen + KPO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_3PO_4 + K ⟶ H_2 + KPO4 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: 2 H_3PO_4 + 2 K ⟶ 3 H_2 + 2 KPO4 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_3PO_4 | 2 | -2 K | 2 | -2 H_2 | 3 | 3 KPO4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_3PO_4 | 2 | -2 | ([H3PO4])^(-2) K | 2 | -2 | ([K])^(-2) H_2 | 3 | 3 | ([H2])^3 KPO4 | 2 | 2 | ([KPO4])^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 = ([H3PO4])^(-2) ([K])^(-2) ([H2])^3 ([KPO4])^2 = (([H2])^3 ([KPO4])^2)/(([H3PO4])^2 ([K])^2)
Construct the equilibrium constant, K, expression for: H_3PO_4 + K ⟶ H_2 + KPO4 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: 2 H_3PO_4 + 2 K ⟶ 3 H_2 + 2 KPO4 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_3PO_4 | 2 | -2 K | 2 | -2 H_2 | 3 | 3 KPO4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_3PO_4 | 2 | -2 | ([H3PO4])^(-2) K | 2 | -2 | ([K])^(-2) H_2 | 3 | 3 | ([H2])^3 KPO4 | 2 | 2 | ([KPO4])^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 = ([H3PO4])^(-2) ([K])^(-2) ([H2])^3 ([KPO4])^2 = (([H2])^3 ([KPO4])^2)/(([H3PO4])^2 ([K])^2)

Rate of reaction

Construct the rate of reaction expression for: H_3PO_4 + K ⟶ H_2 + KPO4 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: 2 H_3PO_4 + 2 K ⟶ 3 H_2 + 2 KPO4 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_3PO_4 | 2 | -2 K | 2 | -2 H_2 | 3 | 3 KPO4 | 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_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) K | 2 | -2 | -1/2 (Δ[K])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(Δt) KPO4 | 2 | 2 | 1/2 (Δ[KPO4])/(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 = -1/2 (Δ[H3PO4])/(Δt) = -1/2 (Δ[K])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[KPO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_3PO_4 + K ⟶ H_2 + KPO4 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: 2 H_3PO_4 + 2 K ⟶ 3 H_2 + 2 KPO4 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_3PO_4 | 2 | -2 K | 2 | -2 H_2 | 3 | 3 KPO4 | 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_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) K | 2 | -2 | -1/2 (Δ[K])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(Δt) KPO4 | 2 | 2 | 1/2 (Δ[KPO4])/(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 = -1/2 (Δ[H3PO4])/(Δt) = -1/2 (Δ[K])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[KPO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | phosphoric acid | potassium | hydrogen | KPO4 formula | H_3PO_4 | K | H_2 | KPO4 Hill formula | H_3O_4P | K | H_2 | KO4P name | phosphoric acid | potassium | hydrogen |  IUPAC name | phosphoric acid | potassium | molecular hydrogen |
| phosphoric acid | potassium | hydrogen | KPO4 formula | H_3PO_4 | K | H_2 | KPO4 Hill formula | H_3O_4P | K | H_2 | KO4P name | phosphoric acid | potassium | hydrogen | IUPAC name | phosphoric acid | potassium | molecular hydrogen |

Substance properties

 | phosphoric acid | potassium | hydrogen | KPO4 molar mass | 97.994 g/mol | 39.0983 g/mol | 2.016 g/mol | 134.07 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) |  melting point | 42.4 °C | 64 °C | -259.2 °C |  boiling point | 158 °C | 760 °C | -252.8 °C |  density | 1.685 g/cm^3 | 0.86 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) |  solubility in water | very soluble | reacts | |  dynamic viscosity | | | 8.9×10^-6 Pa s (at 25 °C) |  odor | odorless | | odorless |
| phosphoric acid | potassium | hydrogen | KPO4 molar mass | 97.994 g/mol | 39.0983 g/mol | 2.016 g/mol | 134.07 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 42.4 °C | 64 °C | -259.2 °C | boiling point | 158 °C | 760 °C | -252.8 °C | density | 1.685 g/cm^3 | 0.86 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | very soluble | reacts | | dynamic viscosity | | | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | | odorless |

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