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H2O + P4O6 = H3PO4 + PH3

Input interpretation

H_2O water + O_6P_4 tetraphosphorus(III) hexoxide ⟶ H_3PO_4 phosphoric acid + PH_3 phosphine
H_2O water + O_6P_4 tetraphosphorus(III) hexoxide ⟶ H_3PO_4 phosphoric acid + PH_3 phosphine

Balanced equation

Balance the chemical equation algebraically: H_2O + O_6P_4 ⟶ H_3PO_4 + PH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_6P_4 ⟶ c_3 H_3PO_4 + c_4 PH_3 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 = 3 c_3 + 3 c_4 O: | c_1 + 6 c_2 = 4 c_3 P: | 4 c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + O_6P_4 ⟶ 3 H_3PO_4 + PH_3
Balance the chemical equation algebraically: H_2O + O_6P_4 ⟶ H_3PO_4 + PH_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_6P_4 ⟶ c_3 H_3PO_4 + c_4 PH_3 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 = 3 c_3 + 3 c_4 O: | c_1 + 6 c_2 = 4 c_3 P: | 4 c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 H_2O + O_6P_4 ⟶ 3 H_3PO_4 + PH_3

Structures

+ ⟶ +
+ ⟶ +

Names

water + tetraphosphorus(III) hexoxide ⟶ phosphoric acid + phosphine
water + tetraphosphorus(III) hexoxide ⟶ phosphoric acid + phosphine

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + O_6P_4 ⟶ H_3PO_4 + PH_3 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: 6 H_2O + O_6P_4 ⟶ 3 H_3PO_4 + PH_3 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 | 6 | -6 O_6P_4 | 1 | -1 H_3PO_4 | 3 | 3 PH_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 6 | -6 | ([H2O])^(-6) O_6P_4 | 1 | -1 | ([O6P4])^(-1) H_3PO_4 | 3 | 3 | ([H3PO4])^3 PH_3 | 1 | 1 | [PH3] 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])^(-6) ([O6P4])^(-1) ([H3PO4])^3 [PH3] = (([H3PO4])^3 [PH3])/(([H2O])^6 [O6P4])
Construct the equilibrium constant, K, expression for: H_2O + O_6P_4 ⟶ H_3PO_4 + PH_3 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: 6 H_2O + O_6P_4 ⟶ 3 H_3PO_4 + PH_3 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 | 6 | -6 O_6P_4 | 1 | -1 H_3PO_4 | 3 | 3 PH_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 6 | -6 | ([H2O])^(-6) O_6P_4 | 1 | -1 | ([O6P4])^(-1) H_3PO_4 | 3 | 3 | ([H3PO4])^3 PH_3 | 1 | 1 | [PH3] 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])^(-6) ([O6P4])^(-1) ([H3PO4])^3 [PH3] = (([H3PO4])^3 [PH3])/(([H2O])^6 [O6P4])

Rate of reaction

Construct the rate of reaction expression for: H_2O + O_6P_4 ⟶ H_3PO_4 + PH_3 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: 6 H_2O + O_6P_4 ⟶ 3 H_3PO_4 + PH_3 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 | 6 | -6 O_6P_4 | 1 | -1 H_3PO_4 | 3 | 3 PH_3 | 1 | 1 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 | 6 | -6 | -1/6 (Δ[H2O])/(Δt) O_6P_4 | 1 | -1 | -(Δ[O6P4])/(Δt) H_3PO_4 | 3 | 3 | 1/3 (Δ[H3PO4])/(Δt) PH_3 | 1 | 1 | (Δ[PH3])/(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 = -1/6 (Δ[H2O])/(Δt) = -(Δ[O6P4])/(Δt) = 1/3 (Δ[H3PO4])/(Δt) = (Δ[PH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + O_6P_4 ⟶ H_3PO_4 + PH_3 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: 6 H_2O + O_6P_4 ⟶ 3 H_3PO_4 + PH_3 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 | 6 | -6 O_6P_4 | 1 | -1 H_3PO_4 | 3 | 3 PH_3 | 1 | 1 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 | 6 | -6 | -1/6 (Δ[H2O])/(Δt) O_6P_4 | 1 | -1 | -(Δ[O6P4])/(Δt) H_3PO_4 | 3 | 3 | 1/3 (Δ[H3PO4])/(Δt) PH_3 | 1 | 1 | (Δ[PH3])/(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 = -1/6 (Δ[H2O])/(Δt) = -(Δ[O6P4])/(Δt) = 1/3 (Δ[H3PO4])/(Δt) = (Δ[PH3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | tetraphosphorus(III) hexoxide | phosphoric acid | phosphine formula | H_2O | O_6P_4 | H_3PO_4 | PH_3 Hill formula | H_2O | O_6P_4 | H_3O_4P | H_3P name | water | tetraphosphorus(III) hexoxide | phosphoric acid | phosphine
| water | tetraphosphorus(III) hexoxide | phosphoric acid | phosphine formula | H_2O | O_6P_4 | H_3PO_4 | PH_3 Hill formula | H_2O | O_6P_4 | H_3O_4P | H_3P name | water | tetraphosphorus(III) hexoxide | phosphoric acid | phosphine

Substance properties

| water | tetraphosphorus(III) hexoxide | phosphoric acid | phosphine molar mass | 18.015 g/mol | 219.89 g/mol | 97.994 g/mol | 33.998 g/mol phase | liquid (at STP) | | liquid (at STP) | gas (at STP) melting point | 0 °C | | 42.4 °C | -132.8 °C boiling point | 99.9839 °C | | 158 °C | -87.5 °C density | 1 g/cm^3 | | 1.685 g/cm^3 | 0.00139 g/cm^3 (at 25 °C) solubility in water | | | very soluble | slightly soluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | | 1.1×10^-5 Pa s (at 0 °C) odor | odorless | | odorless |
| water | tetraphosphorus(III) hexoxide | phosphoric acid | phosphine molar mass | 18.015 g/mol | 219.89 g/mol | 97.994 g/mol | 33.998 g/mol phase | liquid (at STP) | | liquid (at STP) | gas (at STP) melting point | 0 °C | | 42.4 °C | -132.8 °C boiling point | 99.9839 °C | | 158 °C | -87.5 °C density | 1 g/cm^3 | | 1.685 g/cm^3 | 0.00139 g/cm^3 (at 25 °C) solubility in water | | | very soluble | slightly soluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | | 1.1×10^-5 Pa s (at 0 °C) odor | odorless | | odorless |

Units

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