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H2O + I2 + P4 = H3PO4 + HI

Input interpretation

H_2O water + I_2 iodine + P_4 white phosphorus ⟶ H_3PO_4 phosphoric acid + HI hydrogen iodide
H_2O water + I_2 iodine + P_4 white phosphorus ⟶ H_3PO_4 phosphoric acid + HI hydrogen iodide

Balanced equation

Balance the chemical equation algebraically: H_2O + I_2 + P_4 ⟶ H_3PO_4 + HI Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 I_2 + c_3 P_4 ⟶ c_4 H_3PO_4 + c_5 HI 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 + c_5 O: | c_1 = 4 c_4 I: | 2 c_2 = c_5 P: | 4 c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 16 c_2 = 10 c_3 = 1 c_4 = 4 c_5 = 20 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 16 H_2O + 10 I_2 + P_4 ⟶ 4 H_3PO_4 + 20 HI
Balance the chemical equation algebraically: H_2O + I_2 + P_4 ⟶ H_3PO_4 + HI Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 I_2 + c_3 P_4 ⟶ c_4 H_3PO_4 + c_5 HI 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 + c_5 O: | c_1 = 4 c_4 I: | 2 c_2 = c_5 P: | 4 c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 16 c_2 = 10 c_3 = 1 c_4 = 4 c_5 = 20 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 H_2O + 10 I_2 + P_4 ⟶ 4 H_3PO_4 + 20 HI

Structures

 + + ⟶ +
+ + ⟶ +

Names

water + iodine + white phosphorus ⟶ phosphoric acid + hydrogen iodide
water + iodine + white phosphorus ⟶ phosphoric acid + hydrogen iodide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + I_2 + P_4 ⟶ H_3PO_4 + HI 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 + 10 I_2 + P_4 ⟶ 4 H_3PO_4 + 20 HI 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 I_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HI | 20 | 20 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) I_2 | 10 | -10 | ([I2])^(-10) P_4 | 1 | -1 | ([P4])^(-1) H_3PO_4 | 4 | 4 | ([H3PO4])^4 HI | 20 | 20 | ([HI])^20 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) ([I2])^(-10) ([P4])^(-1) ([H3PO4])^4 ([HI])^20 = (([H3PO4])^4 ([HI])^20)/(([H2O])^16 ([I2])^10 [P4])
Construct the equilibrium constant, K, expression for: H_2O + I_2 + P_4 ⟶ H_3PO_4 + HI 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 + 10 I_2 + P_4 ⟶ 4 H_3PO_4 + 20 HI 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 I_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HI | 20 | 20 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) I_2 | 10 | -10 | ([I2])^(-10) P_4 | 1 | -1 | ([P4])^(-1) H_3PO_4 | 4 | 4 | ([H3PO4])^4 HI | 20 | 20 | ([HI])^20 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) ([I2])^(-10) ([P4])^(-1) ([H3PO4])^4 ([HI])^20 = (([H3PO4])^4 ([HI])^20)/(([H2O])^16 ([I2])^10 [P4])

Rate of reaction

Construct the rate of reaction expression for: H_2O + I_2 + P_4 ⟶ H_3PO_4 + HI 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 + 10 I_2 + P_4 ⟶ 4 H_3PO_4 + 20 HI 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 I_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HI | 20 | 20 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) I_2 | 10 | -10 | -1/10 (Δ[I2])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) H_3PO_4 | 4 | 4 | 1/4 (Δ[H3PO4])/(Δt) HI | 20 | 20 | 1/20 (Δ[HI])/(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) = -1/10 (Δ[I2])/(Δt) = -(Δ[P4])/(Δt) = 1/4 (Δ[H3PO4])/(Δt) = 1/20 (Δ[HI])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + I_2 + P_4 ⟶ H_3PO_4 + HI 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 + 10 I_2 + P_4 ⟶ 4 H_3PO_4 + 20 HI 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 I_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HI | 20 | 20 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) I_2 | 10 | -10 | -1/10 (Δ[I2])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) H_3PO_4 | 4 | 4 | 1/4 (Δ[H3PO4])/(Δt) HI | 20 | 20 | 1/20 (Δ[HI])/(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) = -1/10 (Δ[I2])/(Δt) = -(Δ[P4])/(Δt) = 1/4 (Δ[H3PO4])/(Δt) = 1/20 (Δ[HI])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | water | iodine | white phosphorus | phosphoric acid | hydrogen iodide formula | H_2O | I_2 | P_4 | H_3PO_4 | HI Hill formula | H_2O | I_2 | P_4 | H_3O_4P | HI name | water | iodine | white phosphorus | phosphoric acid | hydrogen iodide IUPAC name | water | molecular iodine | tetraphosphorus | phosphoric acid | hydrogen iodide
| water | iodine | white phosphorus | phosphoric acid | hydrogen iodide formula | H_2O | I_2 | P_4 | H_3PO_4 | HI Hill formula | H_2O | I_2 | P_4 | H_3O_4P | HI name | water | iodine | white phosphorus | phosphoric acid | hydrogen iodide IUPAC name | water | molecular iodine | tetraphosphorus | phosphoric acid | hydrogen iodide

Substance properties

 | water | iodine | white phosphorus | phosphoric acid | hydrogen iodide molar mass | 18.015 g/mol | 253.80894 g/mol | 123.89504799 g/mol | 97.994 g/mol | 127.912 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 0 °C | 113 °C | 44.15 °C | 42.4 °C | -50.76 °C boiling point | 99.9839 °C | 184 °C | 280.5 °C | 158 °C | -35.55 °C density | 1 g/cm^3 | 4.94 g/cm^3 | 1.823 g/cm^3 | 1.685 g/cm^3 | 0.005228 g/cm^3 (at 25 °C) solubility in water | | | insoluble | very soluble | 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) | 0.00169 Pa s (at 50 °C) | | 0.001321 Pa s (at -39 °C) odor | odorless | | odorless | odorless |
| water | iodine | white phosphorus | phosphoric acid | hydrogen iodide molar mass | 18.015 g/mol | 253.80894 g/mol | 123.89504799 g/mol | 97.994 g/mol | 127.912 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 0 °C | 113 °C | 44.15 °C | 42.4 °C | -50.76 °C boiling point | 99.9839 °C | 184 °C | 280.5 °C | 158 °C | -35.55 °C density | 1 g/cm^3 | 4.94 g/cm^3 | 1.823 g/cm^3 | 1.685 g/cm^3 | 0.005228 g/cm^3 (at 25 °C) solubility in water | | | insoluble | very soluble | 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) | 0.00169 Pa s (at 50 °C) | | 0.001321 Pa s (at -39 °C) odor | odorless | | odorless | odorless |

Units