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H2O + Cl2 + PH3 = HCl + H3PO4

Input interpretation

H_2O (water) + Cl_2 (chlorine) + PH_3 (phosphine) ⟶ HCl (hydrogen chloride) + H_3PO_4 (phosphoric acid)
H_2O (water) + Cl_2 (chlorine) + PH_3 (phosphine) ⟶ HCl (hydrogen chloride) + H_3PO_4 (phosphoric acid)

Balanced equation

Balance the chemical equation algebraically: H_2O + Cl_2 + PH_3 ⟶ HCl + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Cl_2 + c_3 PH_3 ⟶ c_4 HCl + c_5 H_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Cl and P: H: | 2 c_1 + 3 c_3 = c_4 + 3 c_5 O: | c_1 = 4 c_5 Cl: | 2 c_2 = c_4 P: | c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 4 c_3 = 1 c_4 = 8 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 4 H_2O + 4 Cl_2 + PH_3 ⟶ 8 HCl + H_3PO_4
Balance the chemical equation algebraically: H_2O + Cl_2 + PH_3 ⟶ HCl + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Cl_2 + c_3 PH_3 ⟶ c_4 HCl + c_5 H_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Cl and P: H: | 2 c_1 + 3 c_3 = c_4 + 3 c_5 O: | c_1 = 4 c_5 Cl: | 2 c_2 = c_4 P: | c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 4 c_3 = 1 c_4 = 8 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 H_2O + 4 Cl_2 + PH_3 ⟶ 8 HCl + H_3PO_4

Structures

 + + ⟶ +
+ + ⟶ +

Names

water + chlorine + phosphine ⟶ hydrogen chloride + phosphoric acid
water + chlorine + phosphine ⟶ hydrogen chloride + phosphoric acid

Reaction thermodynamics

Gibbs free energy

 | water | chlorine | phosphine | hydrogen chloride | phosphoric acid molecular free energy | -237.1 kJ/mol | 0 kJ/mol | 13.5 kJ/mol | -95.3 kJ/mol | -1124 kJ/mol total free energy | -948.4 kJ/mol | 0 kJ/mol | 13.5 kJ/mol | -762.4 kJ/mol | -1124 kJ/mol  | G_initial = -934.9 kJ/mol | | | G_final = -1886 kJ/mol |  ΔG_rxn^0 | -1886 kJ/mol - -934.9 kJ/mol = -951.1 kJ/mol (exergonic) | | | |
| water | chlorine | phosphine | hydrogen chloride | phosphoric acid molecular free energy | -237.1 kJ/mol | 0 kJ/mol | 13.5 kJ/mol | -95.3 kJ/mol | -1124 kJ/mol total free energy | -948.4 kJ/mol | 0 kJ/mol | 13.5 kJ/mol | -762.4 kJ/mol | -1124 kJ/mol | G_initial = -934.9 kJ/mol | | | G_final = -1886 kJ/mol | ΔG_rxn^0 | -1886 kJ/mol - -934.9 kJ/mol = -951.1 kJ/mol (exergonic) | | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + Cl_2 + PH_3 ⟶ HCl + 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: 4 H_2O + 4 Cl_2 + PH_3 ⟶ 8 HCl + 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 | 4 | -4 Cl_2 | 4 | -4 PH_3 | 1 | -1 HCl | 8 | 8 H_3PO_4 | 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 | 4 | -4 | ([H2O])^(-4) Cl_2 | 4 | -4 | ([Cl2])^(-4) PH_3 | 1 | -1 | ([PH3])^(-1) HCl | 8 | 8 | ([HCl])^8 H_3PO_4 | 1 | 1 | [H3PO4] 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])^(-4) ([Cl2])^(-4) ([PH3])^(-1) ([HCl])^8 [H3PO4] = (([HCl])^8 [H3PO4])/(([H2O])^4 ([Cl2])^4 [PH3])
Construct the equilibrium constant, K, expression for: H_2O + Cl_2 + PH_3 ⟶ HCl + 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: 4 H_2O + 4 Cl_2 + PH_3 ⟶ 8 HCl + 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 | 4 | -4 Cl_2 | 4 | -4 PH_3 | 1 | -1 HCl | 8 | 8 H_3PO_4 | 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 | 4 | -4 | ([H2O])^(-4) Cl_2 | 4 | -4 | ([Cl2])^(-4) PH_3 | 1 | -1 | ([PH3])^(-1) HCl | 8 | 8 | ([HCl])^8 H_3PO_4 | 1 | 1 | [H3PO4] 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])^(-4) ([Cl2])^(-4) ([PH3])^(-1) ([HCl])^8 [H3PO4] = (([HCl])^8 [H3PO4])/(([H2O])^4 ([Cl2])^4 [PH3])

Rate of reaction

Construct the rate of reaction expression for: H_2O + Cl_2 + PH_3 ⟶ HCl + 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: 4 H_2O + 4 Cl_2 + PH_3 ⟶ 8 HCl + 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 | 4 | -4 Cl_2 | 4 | -4 PH_3 | 1 | -1 HCl | 8 | 8 H_3PO_4 | 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 | 4 | -4 | -1/4 (Δ[H2O])/(Δt) Cl_2 | 4 | -4 | -1/4 (Δ[Cl2])/(Δt) PH_3 | 1 | -1 | -(Δ[PH3])/(Δt) HCl | 8 | 8 | 1/8 (Δ[HCl])/(Δt) H_3PO_4 | 1 | 1 | (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 = -1/4 (Δ[H2O])/(Δt) = -1/4 (Δ[Cl2])/(Δt) = -(Δ[PH3])/(Δt) = 1/8 (Δ[HCl])/(Δt) = (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + Cl_2 + PH_3 ⟶ HCl + 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: 4 H_2O + 4 Cl_2 + PH_3 ⟶ 8 HCl + 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 | 4 | -4 Cl_2 | 4 | -4 PH_3 | 1 | -1 HCl | 8 | 8 H_3PO_4 | 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 | 4 | -4 | -1/4 (Δ[H2O])/(Δt) Cl_2 | 4 | -4 | -1/4 (Δ[Cl2])/(Δt) PH_3 | 1 | -1 | -(Δ[PH3])/(Δt) HCl | 8 | 8 | 1/8 (Δ[HCl])/(Δt) H_3PO_4 | 1 | 1 | (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 = -1/4 (Δ[H2O])/(Δt) = -1/4 (Δ[Cl2])/(Δt) = -(Δ[PH3])/(Δt) = 1/8 (Δ[HCl])/(Δt) = (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | water | chlorine | phosphine | hydrogen chloride | phosphoric acid formula | H_2O | Cl_2 | PH_3 | HCl | H_3PO_4 Hill formula | H_2O | Cl_2 | H_3P | ClH | H_3O_4P name | water | chlorine | phosphine | hydrogen chloride | phosphoric acid IUPAC name | water | molecular chlorine | phosphine | hydrogen chloride | phosphoric acid
| water | chlorine | phosphine | hydrogen chloride | phosphoric acid formula | H_2O | Cl_2 | PH_3 | HCl | H_3PO_4 Hill formula | H_2O | Cl_2 | H_3P | ClH | H_3O_4P name | water | chlorine | phosphine | hydrogen chloride | phosphoric acid IUPAC name | water | molecular chlorine | phosphine | hydrogen chloride | phosphoric acid

Substance properties

 | water | chlorine | phosphine | hydrogen chloride | phosphoric acid molar mass | 18.015 g/mol | 70.9 g/mol | 33.998 g/mol | 36.46 g/mol | 97.994 g/mol phase | liquid (at STP) | gas (at STP) | gas (at STP) | gas (at STP) | liquid (at STP) melting point | 0 °C | -101 °C | -132.8 °C | -114.17 °C | 42.4 °C boiling point | 99.9839 °C | -34 °C | -87.5 °C | -85 °C | 158 °C density | 1 g/cm^3 | 0.003214 g/cm^3 (at 0 °C) | 0.00139 g/cm^3 (at 25 °C) | 0.00149 g/cm^3 (at 25 °C) | 1.685 g/cm^3 solubility in water | | | slightly soluble | miscible | very 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 | chlorine | phosphine | hydrogen chloride | phosphoric acid molar mass | 18.015 g/mol | 70.9 g/mol | 33.998 g/mol | 36.46 g/mol | 97.994 g/mol phase | liquid (at STP) | gas (at STP) | gas (at STP) | gas (at STP) | liquid (at STP) melting point | 0 °C | -101 °C | -132.8 °C | -114.17 °C | 42.4 °C boiling point | 99.9839 °C | -34 °C | -87.5 °C | -85 °C | 158 °C density | 1 g/cm^3 | 0.003214 g/cm^3 (at 0 °C) | 0.00139 g/cm^3 (at 25 °C) | 0.00149 g/cm^3 (at 25 °C) | 1.685 g/cm^3 solubility in water | | | slightly soluble | miscible | very 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