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H2O + Br2 + P4 = H3PO4 + HBr

Input interpretation

H_2O water + Br_2 bromine + P_4 white phosphorus ⟶ H_3PO_4 phosphoric acid + HBr hydrogen bromide
H_2O water + Br_2 bromine + P_4 white phosphorus ⟶ H_3PO_4 phosphoric acid + HBr hydrogen bromide

Balanced equation

Balance the chemical equation algebraically: H_2O + Br_2 + P_4 ⟶ H_3PO_4 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Br_2 + c_3 P_4 ⟶ c_4 H_3PO_4 + c_5 HBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and P: H: | 2 c_1 = 3 c_4 + c_5 O: | c_1 = 4 c_4 Br: | 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 Br_2 + P_4 ⟶ 4 H_3PO_4 + 20 HBr
Balance the chemical equation algebraically: H_2O + Br_2 + P_4 ⟶ H_3PO_4 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Br_2 + c_3 P_4 ⟶ c_4 H_3PO_4 + c_5 HBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Br and P: H: | 2 c_1 = 3 c_4 + c_5 O: | c_1 = 4 c_4 Br: | 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 Br_2 + P_4 ⟶ 4 H_3PO_4 + 20 HBr

Structures

+ + ⟶ +
+ + ⟶ +

Names

water + bromine + white phosphorus ⟶ phosphoric acid + hydrogen bromide
water + bromine + white phosphorus ⟶ phosphoric acid + hydrogen bromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + Br_2 + P_4 ⟶ H_3PO_4 + HBr 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 Br_2 + P_4 ⟶ 4 H_3PO_4 + 20 HBr 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 Br_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HBr | 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) Br_2 | 10 | -10 | ([Br2])^(-10) P_4 | 1 | -1 | ([P4])^(-1) H_3PO_4 | 4 | 4 | ([H3PO4])^4 HBr | 20 | 20 | ([HBr])^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) ([Br2])^(-10) ([P4])^(-1) ([H3PO4])^4 ([HBr])^20 = (([H3PO4])^4 ([HBr])^20)/(([H2O])^16 ([Br2])^10 [P4])
Construct the equilibrium constant, K, expression for: H_2O + Br_2 + P_4 ⟶ H_3PO_4 + HBr 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 Br_2 + P_4 ⟶ 4 H_3PO_4 + 20 HBr 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 Br_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HBr | 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) Br_2 | 10 | -10 | ([Br2])^(-10) P_4 | 1 | -1 | ([P4])^(-1) H_3PO_4 | 4 | 4 | ([H3PO4])^4 HBr | 20 | 20 | ([HBr])^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) ([Br2])^(-10) ([P4])^(-1) ([H3PO4])^4 ([HBr])^20 = (([H3PO4])^4 ([HBr])^20)/(([H2O])^16 ([Br2])^10 [P4])

Rate of reaction

Construct the rate of reaction expression for: H_2O + Br_2 + P_4 ⟶ H_3PO_4 + HBr 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 Br_2 + P_4 ⟶ 4 H_3PO_4 + 20 HBr 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 Br_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HBr | 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) Br_2 | 10 | -10 | -1/10 (Δ[Br2])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) H_3PO_4 | 4 | 4 | 1/4 (Δ[H3PO4])/(Δt) HBr | 20 | 20 | 1/20 (Δ[HBr])/(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[Br2])/(Δt) = -(Δ[P4])/(Δt) = 1/4 (Δ[H3PO4])/(Δt) = 1/20 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + Br_2 + P_4 ⟶ H_3PO_4 + HBr 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 Br_2 + P_4 ⟶ 4 H_3PO_4 + 20 HBr 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 Br_2 | 10 | -10 P_4 | 1 | -1 H_3PO_4 | 4 | 4 HBr | 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) Br_2 | 10 | -10 | -1/10 (Δ[Br2])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) H_3PO_4 | 4 | 4 | 1/4 (Δ[H3PO4])/(Δt) HBr | 20 | 20 | 1/20 (Δ[HBr])/(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[Br2])/(Δt) = -(Δ[P4])/(Δt) = 1/4 (Δ[H3PO4])/(Δt) = 1/20 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | bromine | white phosphorus | phosphoric acid | hydrogen bromide formula | H_2O | Br_2 | P_4 | H_3PO_4 | HBr Hill formula | H_2O | Br_2 | P_4 | H_3O_4P | BrH name | water | bromine | white phosphorus | phosphoric acid | hydrogen bromide IUPAC name | water | molecular bromine | tetraphosphorus | phosphoric acid | hydrogen bromide
| water | bromine | white phosphorus | phosphoric acid | hydrogen bromide formula | H_2O | Br_2 | P_4 | H_3PO_4 | HBr Hill formula | H_2O | Br_2 | P_4 | H_3O_4P | BrH name | water | bromine | white phosphorus | phosphoric acid | hydrogen bromide IUPAC name | water | molecular bromine | tetraphosphorus | phosphoric acid | hydrogen bromide

Substance properties

| water | bromine | white phosphorus | phosphoric acid | hydrogen bromide molar mass | 18.015 g/mol | 159.81 g/mol | 123.89504799 g/mol | 97.994 g/mol | 80.912 g/mol phase | liquid (at STP) | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 0 °C | -7.2 °C | 44.15 °C | 42.4 °C | -86.8 °C boiling point | 99.9839 °C | 58.8 °C | 280.5 °C | 158 °C | -66.38 °C density | 1 g/cm^3 | 3.119 g/cm^3 | 1.823 g/cm^3 | 1.685 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) solubility in water | | insoluble | insoluble | very soluble | miscible surface tension | 0.0728 N/m | 0.0409 N/m | | | 0.0271 N/m dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | | 8.4×10^-4 Pa s (at -75 °C) odor | odorless | | odorless | odorless |
| water | bromine | white phosphorus | phosphoric acid | hydrogen bromide molar mass | 18.015 g/mol | 159.81 g/mol | 123.89504799 g/mol | 97.994 g/mol | 80.912 g/mol phase | liquid (at STP) | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 0 °C | -7.2 °C | 44.15 °C | 42.4 °C | -86.8 °C boiling point | 99.9839 °C | 58.8 °C | 280.5 °C | 158 °C | -66.38 °C density | 1 g/cm^3 | 3.119 g/cm^3 | 1.823 g/cm^3 | 1.685 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) solubility in water | | insoluble | insoluble | very soluble | miscible surface tension | 0.0728 N/m | 0.0409 N/m | | | 0.0271 N/m dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | | 8.4×10^-4 Pa s (at -75 °C) odor | odorless | | odorless | odorless |

Units

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