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H2O + P + HBrO3 = H3PO4 + HBr

Input interpretation

H_2O water + P red phosphorus + HO_3Br bromic acid ⟶ H_3PO_4 phosphoric acid + HBr hydrogen bromide
H_2O water + P red phosphorus + HO_3Br bromic acid ⟶ H_3PO_4 phosphoric acid + HBr hydrogen bromide

Balanced equation

Balance the chemical equation algebraically: H_2O + P + HO_3Br ⟶ H_3PO_4 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P + c_3 HO_3Br ⟶ 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, P and Br: H: | 2 c_1 + c_3 = 3 c_4 + c_5 O: | c_1 + 3 c_3 = 4 c_4 P: | c_2 = c_4 Br: | 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 = 9/5 c_2 = 6/5 c_3 = 1 c_4 = 6/5 c_5 = 1 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 9 c_2 = 6 c_3 = 5 c_4 = 6 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 H_2O + 6 P + 5 HO_3Br ⟶ 6 H_3PO_4 + 5 HBr
Balance the chemical equation algebraically: H_2O + P + HO_3Br ⟶ H_3PO_4 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 P + c_3 HO_3Br ⟶ 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, P and Br: H: | 2 c_1 + c_3 = 3 c_4 + c_5 O: | c_1 + 3 c_3 = 4 c_4 P: | c_2 = c_4 Br: | 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 = 9/5 c_2 = 6/5 c_3 = 1 c_4 = 6/5 c_5 = 1 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 9 c_2 = 6 c_3 = 5 c_4 = 6 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 H_2O + 6 P + 5 HO_3Br ⟶ 6 H_3PO_4 + 5 HBr

Structures

+ + ⟶ +
+ + ⟶ +

Names

water + red phosphorus + bromic acid ⟶ phosphoric acid + hydrogen bromide
water + red phosphorus + bromic acid ⟶ phosphoric acid + hydrogen bromide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + P + HO_3Br ⟶ 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: 9 H_2O + 6 P + 5 HO_3Br ⟶ 6 H_3PO_4 + 5 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 | 9 | -9 P | 6 | -6 HO_3Br | 5 | -5 H_3PO_4 | 6 | 6 HBr | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 9 | -9 | ([H2O])^(-9) P | 6 | -6 | ([P])^(-6) HO_3Br | 5 | -5 | ([H1O3Br1])^(-5) H_3PO_4 | 6 | 6 | ([H3PO4])^6 HBr | 5 | 5 | ([HBr])^5 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])^(-9) ([P])^(-6) ([H1O3Br1])^(-5) ([H3PO4])^6 ([HBr])^5 = (([H3PO4])^6 ([HBr])^5)/(([H2O])^9 ([P])^6 ([H1O3Br1])^5)
Construct the equilibrium constant, K, expression for: H_2O + P + HO_3Br ⟶ 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: 9 H_2O + 6 P + 5 HO_3Br ⟶ 6 H_3PO_4 + 5 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 | 9 | -9 P | 6 | -6 HO_3Br | 5 | -5 H_3PO_4 | 6 | 6 HBr | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 9 | -9 | ([H2O])^(-9) P | 6 | -6 | ([P])^(-6) HO_3Br | 5 | -5 | ([H1O3Br1])^(-5) H_3PO_4 | 6 | 6 | ([H3PO4])^6 HBr | 5 | 5 | ([HBr])^5 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])^(-9) ([P])^(-6) ([H1O3Br1])^(-5) ([H3PO4])^6 ([HBr])^5 = (([H3PO4])^6 ([HBr])^5)/(([H2O])^9 ([P])^6 ([H1O3Br1])^5)

Rate of reaction

Construct the rate of reaction expression for: H_2O + P + HO_3Br ⟶ 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: 9 H_2O + 6 P + 5 HO_3Br ⟶ 6 H_3PO_4 + 5 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 | 9 | -9 P | 6 | -6 HO_3Br | 5 | -5 H_3PO_4 | 6 | 6 HBr | 5 | 5 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 | 9 | -9 | -1/9 (Δ[H2O])/(Δt) P | 6 | -6 | -1/6 (Δ[P])/(Δt) HO_3Br | 5 | -5 | -1/5 (Δ[H1O3Br1])/(Δt) H_3PO_4 | 6 | 6 | 1/6 (Δ[H3PO4])/(Δt) HBr | 5 | 5 | 1/5 (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 = -1/9 (Δ[H2O])/(Δt) = -1/6 (Δ[P])/(Δt) = -1/5 (Δ[H1O3Br1])/(Δt) = 1/6 (Δ[H3PO4])/(Δt) = 1/5 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + P + HO_3Br ⟶ 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: 9 H_2O + 6 P + 5 HO_3Br ⟶ 6 H_3PO_4 + 5 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 | 9 | -9 P | 6 | -6 HO_3Br | 5 | -5 H_3PO_4 | 6 | 6 HBr | 5 | 5 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 | 9 | -9 | -1/9 (Δ[H2O])/(Δt) P | 6 | -6 | -1/6 (Δ[P])/(Δt) HO_3Br | 5 | -5 | -1/5 (Δ[H1O3Br1])/(Δt) H_3PO_4 | 6 | 6 | 1/6 (Δ[H3PO4])/(Δt) HBr | 5 | 5 | 1/5 (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 = -1/9 (Δ[H2O])/(Δt) = -1/6 (Δ[P])/(Δt) = -1/5 (Δ[H1O3Br1])/(Δt) = 1/6 (Δ[H3PO4])/(Δt) = 1/5 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | red phosphorus | bromic acid | phosphoric acid | hydrogen bromide formula | H_2O | P | HO_3Br | H_3PO_4 | HBr Hill formula | H_2O | P | BrHO_3 | H_3O_4P | BrH name | water | red phosphorus | bromic acid | phosphoric acid | hydrogen bromide IUPAC name | water | phosphorus | bromic acid | phosphoric acid | hydrogen bromide
| water | red phosphorus | bromic acid | phosphoric acid | hydrogen bromide formula | H_2O | P | HO_3Br | H_3PO_4 | HBr Hill formula | H_2O | P | BrHO_3 | H_3O_4P | BrH name | water | red phosphorus | bromic acid | phosphoric acid | hydrogen bromide IUPAC name | water | phosphorus | bromic acid | phosphoric acid | hydrogen bromide

Substance properties

| water | red phosphorus | bromic acid | phosphoric acid | hydrogen bromide molar mass | 18.015 g/mol | 30.973761998 g/mol | 128.91 g/mol | 97.994 g/mol | 80.912 g/mol phase | liquid (at STP) | solid (at STP) | | liquid (at STP) | gas (at STP) melting point | 0 °C | 579.2 °C | | 42.4 °C | -86.8 °C boiling point | 99.9839 °C | | | 158 °C | -66.38 °C density | 1 g/cm^3 | 2.16 g/cm^3 | | 1.685 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) solubility in water | | insoluble | | very soluble | miscible surface tension | 0.0728 N/m | | | | 0.0271 N/m dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | | | 8.4×10^-4 Pa s (at -75 °C) odor | odorless | | | odorless |
| water | red phosphorus | bromic acid | phosphoric acid | hydrogen bromide molar mass | 18.015 g/mol | 30.973761998 g/mol | 128.91 g/mol | 97.994 g/mol | 80.912 g/mol phase | liquid (at STP) | solid (at STP) | | liquid (at STP) | gas (at STP) melting point | 0 °C | 579.2 °C | | 42.4 °C | -86.8 °C boiling point | 99.9839 °C | | | 158 °C | -66.38 °C density | 1 g/cm^3 | 2.16 g/cm^3 | | 1.685 g/cm^3 | 0.003307 g/cm^3 (at 25 °C) solubility in water | | insoluble | | very soluble | miscible surface tension | 0.0728 N/m | | | | 0.0271 N/m dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | | | 8.4×10^-4 Pa s (at -75 °C) odor | odorless | | | odorless |

Units

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