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H2O + I2 + P = H3PO4 + PH4I

Input interpretation

H_2O water + I_2 iodine + P red phosphorus ⟶ H_3PO_4 phosphoric acid + PH_4I phosphonium iodide
H_2O water + I_2 iodine + P red phosphorus ⟶ H_3PO_4 phosphoric acid + PH_4I phosphonium iodide

Balanced equation

Balance the chemical equation algebraically: H_2O + I_2 + P ⟶ H_3PO_4 + PH_4I Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 I_2 + c_3 P ⟶ c_4 H_3PO_4 + c_5 PH_4I Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, I and P: H: | 2 c_1 = 3 c_4 + 4 c_5 O: | c_1 = 4 c_4 I: | 2 c_2 = c_5 P: | c_3 = c_4 + c_5 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 = 32/5 c_2 = 1 c_3 = 18/5 c_4 = 8/5 c_5 = 2 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 32 c_2 = 5 c_3 = 18 c_4 = 8 c_5 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 32 H_2O + 5 I_2 + 18 P ⟶ 8 H_3PO_4 + 10 PH_4I
Balance the chemical equation algebraically: H_2O + I_2 + P ⟶ H_3PO_4 + PH_4I Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 I_2 + c_3 P ⟶ c_4 H_3PO_4 + c_5 PH_4I Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, I and P: H: | 2 c_1 = 3 c_4 + 4 c_5 O: | c_1 = 4 c_4 I: | 2 c_2 = c_5 P: | c_3 = c_4 + c_5 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 = 32/5 c_2 = 1 c_3 = 18/5 c_4 = 8/5 c_5 = 2 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 32 c_2 = 5 c_3 = 18 c_4 = 8 c_5 = 10 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 32 H_2O + 5 I_2 + 18 P ⟶ 8 H_3PO_4 + 10 PH_4I

Structures

 + + ⟶ +
+ + ⟶ +

Names

water + iodine + red phosphorus ⟶ phosphoric acid + phosphonium iodide
water + iodine + red phosphorus ⟶ phosphoric acid + phosphonium iodide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + I_2 + P ⟶ H_3PO_4 + PH_4I 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: 32 H_2O + 5 I_2 + 18 P ⟶ 8 H_3PO_4 + 10 PH_4I 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 | 32 | -32 I_2 | 5 | -5 P | 18 | -18 H_3PO_4 | 8 | 8 PH_4I | 10 | 10 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 32 | -32 | ([H2O])^(-32) I_2 | 5 | -5 | ([I2])^(-5) P | 18 | -18 | ([P])^(-18) H_3PO_4 | 8 | 8 | ([H3PO4])^8 PH_4I | 10 | 10 | ([PH4I])^10 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])^(-32) ([I2])^(-5) ([P])^(-18) ([H3PO4])^8 ([PH4I])^10 = (([H3PO4])^8 ([PH4I])^10)/(([H2O])^32 ([I2])^5 ([P])^18)
Construct the equilibrium constant, K, expression for: H_2O + I_2 + P ⟶ H_3PO_4 + PH_4I 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: 32 H_2O + 5 I_2 + 18 P ⟶ 8 H_3PO_4 + 10 PH_4I 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 | 32 | -32 I_2 | 5 | -5 P | 18 | -18 H_3PO_4 | 8 | 8 PH_4I | 10 | 10 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 32 | -32 | ([H2O])^(-32) I_2 | 5 | -5 | ([I2])^(-5) P | 18 | -18 | ([P])^(-18) H_3PO_4 | 8 | 8 | ([H3PO4])^8 PH_4I | 10 | 10 | ([PH4I])^10 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])^(-32) ([I2])^(-5) ([P])^(-18) ([H3PO4])^8 ([PH4I])^10 = (([H3PO4])^8 ([PH4I])^10)/(([H2O])^32 ([I2])^5 ([P])^18)

Rate of reaction

Construct the rate of reaction expression for: H_2O + I_2 + P ⟶ H_3PO_4 + PH_4I 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: 32 H_2O + 5 I_2 + 18 P ⟶ 8 H_3PO_4 + 10 PH_4I 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 | 32 | -32 I_2 | 5 | -5 P | 18 | -18 H_3PO_4 | 8 | 8 PH_4I | 10 | 10 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 | 32 | -32 | -1/32 (Δ[H2O])/(Δt) I_2 | 5 | -5 | -1/5 (Δ[I2])/(Δt) P | 18 | -18 | -1/18 (Δ[P])/(Δt) H_3PO_4 | 8 | 8 | 1/8 (Δ[H3PO4])/(Δt) PH_4I | 10 | 10 | 1/10 (Δ[PH4I])/(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 = -1/32 (Δ[H2O])/(Δt) = -1/5 (Δ[I2])/(Δt) = -1/18 (Δ[P])/(Δt) = 1/8 (Δ[H3PO4])/(Δt) = 1/10 (Δ[PH4I])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + I_2 + P ⟶ H_3PO_4 + PH_4I 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: 32 H_2O + 5 I_2 + 18 P ⟶ 8 H_3PO_4 + 10 PH_4I 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 | 32 | -32 I_2 | 5 | -5 P | 18 | -18 H_3PO_4 | 8 | 8 PH_4I | 10 | 10 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 | 32 | -32 | -1/32 (Δ[H2O])/(Δt) I_2 | 5 | -5 | -1/5 (Δ[I2])/(Δt) P | 18 | -18 | -1/18 (Δ[P])/(Δt) H_3PO_4 | 8 | 8 | 1/8 (Δ[H3PO4])/(Δt) PH_4I | 10 | 10 | 1/10 (Δ[PH4I])/(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 = -1/32 (Δ[H2O])/(Δt) = -1/5 (Δ[I2])/(Δt) = -1/18 (Δ[P])/(Δt) = 1/8 (Δ[H3PO4])/(Δt) = 1/10 (Δ[PH4I])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | water | iodine | red phosphorus | phosphoric acid | phosphonium iodide formula | H_2O | I_2 | P | H_3PO_4 | PH_4I Hill formula | H_2O | I_2 | P | H_3O_4P | H_4IP name | water | iodine | red phosphorus | phosphoric acid | phosphonium iodide IUPAC name | water | molecular iodine | phosphorus | phosphoric acid |
| water | iodine | red phosphorus | phosphoric acid | phosphonium iodide formula | H_2O | I_2 | P | H_3PO_4 | PH_4I Hill formula | H_2O | I_2 | P | H_3O_4P | H_4IP name | water | iodine | red phosphorus | phosphoric acid | phosphonium iodide IUPAC name | water | molecular iodine | phosphorus | phosphoric acid |

Substance properties

 | water | iodine | red phosphorus | phosphoric acid | phosphonium iodide molar mass | 18.015 g/mol | 253.80894 g/mol | 30.973761998 g/mol | 97.994 g/mol | 161.91 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) |  melting point | 0 °C | 113 °C | 579.2 °C | 42.4 °C | 18.5 °C boiling point | 99.9839 °C | 184 °C | | 158 °C | 62.5 °C density | 1 g/cm^3 | 4.94 g/cm^3 | 2.16 g/cm^3 | 1.685 g/cm^3 | 2.86 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) | 0.00227 Pa s (at 116 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | |  odor | odorless | | | odorless |
| water | iodine | red phosphorus | phosphoric acid | phosphonium iodide molar mass | 18.015 g/mol | 253.80894 g/mol | 30.973761998 g/mol | 97.994 g/mol | 161.91 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | melting point | 0 °C | 113 °C | 579.2 °C | 42.4 °C | 18.5 °C boiling point | 99.9839 °C | 184 °C | | 158 °C | 62.5 °C density | 1 g/cm^3 | 4.94 g/cm^3 | 2.16 g/cm^3 | 1.685 g/cm^3 | 2.86 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) | 0.00227 Pa s (at 116 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | | odor | odorless | | | odorless |

Units