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H2O + P4 + Ba(OH)2 = PH3 + Ba(H2PO4)2

Input interpretation

H_2O water + P_4 white phosphorus + Ba(OH)_2 barium hydroxide ⟶ PH_3 phosphine + Ba(H2PO4)2
H_2O water + P_4 white phosphorus + Ba(OH)_2 barium hydroxide ⟶ PH_3 phosphine + Ba(H2PO4)2

Balanced equation

Balance the chemical equation algebraically: H_2O + P_4 + Ba(OH)_2 ⟶ PH_3 + Ba(H2PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P_4 + c_3 Ba(OH)_2 ⟶ c_4 PH_3 + c_5 Ba(H2PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Ba: H: | 2 c_1 + 2 c_3 = 3 c_4 + 4 c_5 O: | c_1 + 2 c_3 = 8 c_5 P: | 4 c_2 = c_4 + 2 c_5 Ba: | 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 = 6 c_2 = 4/3 c_3 = 1 c_4 = 10/3 c_5 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 18 c_2 = 4 c_3 = 3 c_4 = 10 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 18 H_2O + 4 P_4 + 3 Ba(OH)_2 ⟶ 10 PH_3 + 3 Ba(H2PO4)2
Balance the chemical equation algebraically: H_2O + P_4 + Ba(OH)_2 ⟶ PH_3 + Ba(H2PO4)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P_4 + c_3 Ba(OH)_2 ⟶ c_4 PH_3 + c_5 Ba(H2PO4)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Ba: H: | 2 c_1 + 2 c_3 = 3 c_4 + 4 c_5 O: | c_1 + 2 c_3 = 8 c_5 P: | 4 c_2 = c_4 + 2 c_5 Ba: | 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 = 6 c_2 = 4/3 c_3 = 1 c_4 = 10/3 c_5 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 18 c_2 = 4 c_3 = 3 c_4 = 10 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 18 H_2O + 4 P_4 + 3 Ba(OH)_2 ⟶ 10 PH_3 + 3 Ba(H2PO4)2

Structures

+ + ⟶ + Ba(H2PO4)2
+ + ⟶ + Ba(H2PO4)2

Names

water + white phosphorus + barium hydroxide ⟶ phosphine + Ba(H2PO4)2
water + white phosphorus + barium hydroxide ⟶ phosphine + Ba(H2PO4)2

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + P_4 + Ba(OH)_2 ⟶ PH_3 + Ba(H2PO4)2 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: 18 H_2O + 4 P_4 + 3 Ba(OH)_2 ⟶ 10 PH_3 + 3 Ba(H2PO4)2 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 | 18 | -18 P_4 | 4 | -4 Ba(OH)_2 | 3 | -3 PH_3 | 10 | 10 Ba(H2PO4)2 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 18 | -18 | ([H2O])^(-18) P_4 | 4 | -4 | ([P4])^(-4) Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) PH_3 | 10 | 10 | ([PH3])^10 Ba(H2PO4)2 | 3 | 3 | ([Ba(H2PO4)2])^3 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])^(-18) ([P4])^(-4) ([Ba(OH)2])^(-3) ([PH3])^10 ([Ba(H2PO4)2])^3 = (([PH3])^10 ([Ba(H2PO4)2])^3)/(([H2O])^18 ([P4])^4 ([Ba(OH)2])^3)
Construct the equilibrium constant, K, expression for: H_2O + P_4 + Ba(OH)_2 ⟶ PH_3 + Ba(H2PO4)2 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: 18 H_2O + 4 P_4 + 3 Ba(OH)_2 ⟶ 10 PH_3 + 3 Ba(H2PO4)2 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 | 18 | -18 P_4 | 4 | -4 Ba(OH)_2 | 3 | -3 PH_3 | 10 | 10 Ba(H2PO4)2 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 18 | -18 | ([H2O])^(-18) P_4 | 4 | -4 | ([P4])^(-4) Ba(OH)_2 | 3 | -3 | ([Ba(OH)2])^(-3) PH_3 | 10 | 10 | ([PH3])^10 Ba(H2PO4)2 | 3 | 3 | ([Ba(H2PO4)2])^3 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])^(-18) ([P4])^(-4) ([Ba(OH)2])^(-3) ([PH3])^10 ([Ba(H2PO4)2])^3 = (([PH3])^10 ([Ba(H2PO4)2])^3)/(([H2O])^18 ([P4])^4 ([Ba(OH)2])^3)

Rate of reaction

Construct the rate of reaction expression for: H_2O + P_4 + Ba(OH)_2 ⟶ PH_3 + Ba(H2PO4)2 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: 18 H_2O + 4 P_4 + 3 Ba(OH)_2 ⟶ 10 PH_3 + 3 Ba(H2PO4)2 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 | 18 | -18 P_4 | 4 | -4 Ba(OH)_2 | 3 | -3 PH_3 | 10 | 10 Ba(H2PO4)2 | 3 | 3 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 | 18 | -18 | -1/18 (Δ[H2O])/(Δt) P_4 | 4 | -4 | -1/4 (Δ[P4])/(Δt) Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) PH_3 | 10 | 10 | 1/10 (Δ[PH3])/(Δt) Ba(H2PO4)2 | 3 | 3 | 1/3 (Δ[Ba(H2PO4)2])/(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 = -1/18 (Δ[H2O])/(Δt) = -1/4 (Δ[P4])/(Δt) = -1/3 (Δ[Ba(OH)2])/(Δt) = 1/10 (Δ[PH3])/(Δt) = 1/3 (Δ[Ba(H2PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + P_4 + Ba(OH)_2 ⟶ PH_3 + Ba(H2PO4)2 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: 18 H_2O + 4 P_4 + 3 Ba(OH)_2 ⟶ 10 PH_3 + 3 Ba(H2PO4)2 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 | 18 | -18 P_4 | 4 | -4 Ba(OH)_2 | 3 | -3 PH_3 | 10 | 10 Ba(H2PO4)2 | 3 | 3 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 | 18 | -18 | -1/18 (Δ[H2O])/(Δt) P_4 | 4 | -4 | -1/4 (Δ[P4])/(Δt) Ba(OH)_2 | 3 | -3 | -1/3 (Δ[Ba(OH)2])/(Δt) PH_3 | 10 | 10 | 1/10 (Δ[PH3])/(Δt) Ba(H2PO4)2 | 3 | 3 | 1/3 (Δ[Ba(H2PO4)2])/(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 = -1/18 (Δ[H2O])/(Δt) = -1/4 (Δ[P4])/(Δt) = -1/3 (Δ[Ba(OH)2])/(Δt) = 1/10 (Δ[PH3])/(Δt) = 1/3 (Δ[Ba(H2PO4)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | white phosphorus | barium hydroxide | phosphine | Ba(H2PO4)2 formula | H_2O | P_4 | Ba(OH)_2 | PH_3 | Ba(H2PO4)2 Hill formula | H_2O | P_4 | BaH_2O_2 | H_3P | H4BaO8P2 name | water | white phosphorus | barium hydroxide | phosphine | IUPAC name | water | tetraphosphorus | barium(+2) cation dihydroxide | phosphine |
| water | white phosphorus | barium hydroxide | phosphine | Ba(H2PO4)2 formula | H_2O | P_4 | Ba(OH)_2 | PH_3 | Ba(H2PO4)2 Hill formula | H_2O | P_4 | BaH_2O_2 | H_3P | H4BaO8P2 name | water | white phosphorus | barium hydroxide | phosphine | IUPAC name | water | tetraphosphorus | barium(+2) cation dihydroxide | phosphine |

Substance properties

| water | white phosphorus | barium hydroxide | phosphine | Ba(H2PO4)2 molar mass | 18.015 g/mol | 123.89504799 g/mol | 171.34 g/mol | 33.998 g/mol | 331.3 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | 44.15 °C | 300 °C | -132.8 °C | boiling point | 99.9839 °C | 280.5 °C | | -87.5 °C | density | 1 g/cm^3 | 1.823 g/cm^3 | 2.2 g/cm^3 | 0.00139 g/cm^3 (at 25 °C) | solubility in water | | insoluble | | slightly soluble | surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | | 1.1×10^-5 Pa s (at 0 °C) | odor | odorless | odorless | | |
| water | white phosphorus | barium hydroxide | phosphine | Ba(H2PO4)2 molar mass | 18.015 g/mol | 123.89504799 g/mol | 171.34 g/mol | 33.998 g/mol | 331.3 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | 44.15 °C | 300 °C | -132.8 °C | boiling point | 99.9839 °C | 280.5 °C | | -87.5 °C | density | 1 g/cm^3 | 1.823 g/cm^3 | 2.2 g/cm^3 | 0.00139 g/cm^3 (at 25 °C) | solubility in water | | insoluble | | slightly soluble | surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | | 1.1×10^-5 Pa s (at 0 °C) | odor | odorless | odorless | | |

Units

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