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O2 + H3P = H2O + P2O5

Input interpretation

O_2 oxygen + PH_3 phosphine ⟶ H_2O water + P2O5
O_2 oxygen + PH_3 phosphine ⟶ H_2O water + P2O5

Balanced equation

Balance the chemical equation algebraically: O_2 + PH_3 ⟶ H_2O + P2O5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 PH_3 ⟶ c_3 H_2O + c_4 P2O5 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H and P: O: | 2 c_1 = c_3 + 5 c_4 H: | 3 c_2 = 2 c_3 P: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 2 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 4 O_2 + 2 PH_3 ⟶ 3 H_2O + P2O5
Balance the chemical equation algebraically: O_2 + PH_3 ⟶ H_2O + P2O5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 PH_3 ⟶ c_3 H_2O + c_4 P2O5 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H and P: O: | 2 c_1 = c_3 + 5 c_4 H: | 3 c_2 = 2 c_3 P: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 2 c_3 = 3 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 O_2 + 2 PH_3 ⟶ 3 H_2O + P2O5

Structures

 + ⟶ + P2O5
+ ⟶ + P2O5

Names

oxygen + phosphine ⟶ water + P2O5
oxygen + phosphine ⟶ water + P2O5

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + PH_3 ⟶ H_2O + P2O5 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 O_2 + 2 PH_3 ⟶ 3 H_2O + P2O5 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 O_2 | 4 | -4 PH_3 | 2 | -2 H_2O | 3 | 3 P2O5 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 4 | -4 | ([O2])^(-4) PH_3 | 2 | -2 | ([PH3])^(-2) H_2O | 3 | 3 | ([H2O])^3 P2O5 | 1 | 1 | [P2O5] 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 = ([O2])^(-4) ([PH3])^(-2) ([H2O])^3 [P2O5] = (([H2O])^3 [P2O5])/(([O2])^4 ([PH3])^2)
Construct the equilibrium constant, K, expression for: O_2 + PH_3 ⟶ H_2O + P2O5 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 O_2 + 2 PH_3 ⟶ 3 H_2O + P2O5 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 O_2 | 4 | -4 PH_3 | 2 | -2 H_2O | 3 | 3 P2O5 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 4 | -4 | ([O2])^(-4) PH_3 | 2 | -2 | ([PH3])^(-2) H_2O | 3 | 3 | ([H2O])^3 P2O5 | 1 | 1 | [P2O5] 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 = ([O2])^(-4) ([PH3])^(-2) ([H2O])^3 [P2O5] = (([H2O])^3 [P2O5])/(([O2])^4 ([PH3])^2)

Rate of reaction

Construct the rate of reaction expression for: O_2 + PH_3 ⟶ H_2O + P2O5 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 O_2 + 2 PH_3 ⟶ 3 H_2O + P2O5 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 O_2 | 4 | -4 PH_3 | 2 | -2 H_2O | 3 | 3 P2O5 | 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 O_2 | 4 | -4 | -1/4 (Δ[O2])/(Δt) PH_3 | 2 | -2 | -1/2 (Δ[PH3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) P2O5 | 1 | 1 | (Δ[P2O5])/(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[O2])/(Δt) = -1/2 (Δ[PH3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[P2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + PH_3 ⟶ H_2O + P2O5 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 O_2 + 2 PH_3 ⟶ 3 H_2O + P2O5 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 O_2 | 4 | -4 PH_3 | 2 | -2 H_2O | 3 | 3 P2O5 | 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 O_2 | 4 | -4 | -1/4 (Δ[O2])/(Δt) PH_3 | 2 | -2 | -1/2 (Δ[PH3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) P2O5 | 1 | 1 | (Δ[P2O5])/(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[O2])/(Δt) = -1/2 (Δ[PH3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[P2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | oxygen | phosphine | water | P2O5 formula | O_2 | PH_3 | H_2O | P2O5 Hill formula | O_2 | H_3P | H_2O | O5P2 name | oxygen | phosphine | water |  IUPAC name | molecular oxygen | phosphine | water |
| oxygen | phosphine | water | P2O5 formula | O_2 | PH_3 | H_2O | P2O5 Hill formula | O_2 | H_3P | H_2O | O5P2 name | oxygen | phosphine | water | IUPAC name | molecular oxygen | phosphine | water |

Substance properties

 | oxygen | phosphine | water | P2O5 molar mass | 31.998 g/mol | 33.998 g/mol | 18.015 g/mol | 141.94 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) |  melting point | -218 °C | -132.8 °C | 0 °C |  boiling point | -183 °C | -87.5 °C | 99.9839 °C |  density | 0.001429 g/cm^3 (at 0 °C) | 0.00139 g/cm^3 (at 25 °C) | 1 g/cm^3 |  solubility in water | | slightly soluble | |  surface tension | 0.01347 N/m | | 0.0728 N/m |  dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.1×10^-5 Pa s (at 0 °C) | 8.9×10^-4 Pa s (at 25 °C) |  odor | odorless | | odorless |
| oxygen | phosphine | water | P2O5 molar mass | 31.998 g/mol | 33.998 g/mol | 18.015 g/mol | 141.94 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) | melting point | -218 °C | -132.8 °C | 0 °C | boiling point | -183 °C | -87.5 °C | 99.9839 °C | density | 0.001429 g/cm^3 (at 0 °C) | 0.00139 g/cm^3 (at 25 °C) | 1 g/cm^3 | solubility in water | | slightly soluble | | surface tension | 0.01347 N/m | | 0.0728 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 1.1×10^-5 Pa s (at 0 °C) | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |

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