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H2O + HNO3 + As4O6 = NO + NO2 + H3AsO4

Input interpretation

H_2O water + HNO_3 nitric acid + As4O6 ⟶ NO nitric oxide + NO_2 nitrogen dioxide + H_3AsO_4 arsenic acid, solid
H_2O water + HNO_3 nitric acid + As4O6 ⟶ NO nitric oxide + NO_2 nitrogen dioxide + H_3AsO_4 arsenic acid, solid

Balanced equation

Balance the chemical equation algebraically: H_2O + HNO_3 + As4O6 ⟶ NO + NO_2 + H_3AsO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HNO_3 + c_3 As4O6 ⟶ c_4 NO + c_5 NO_2 + c_6 H_3AsO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and As: H: | 2 c_1 + c_2 = 3 c_6 O: | c_1 + 3 c_2 + 6 c_3 = c_4 + 2 c_5 + 4 c_6 N: | c_2 = c_4 + c_5 As: | 4 c_3 = c_6 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_2 = 12 - 2 c_1 c_3 = 1 c_4 = c_1 - 2 c_5 = 14 - 3 c_1 c_6 = 4 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_2 = 36 - 2 c_1 c_3 = 3 c_4 = c_1 - 6 c_5 = 42 - 3 c_1 c_6 = 12 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 13 and solve for the remaining coefficients: c_1 = 13 c_2 = 10 c_3 = 3 c_4 = 7 c_5 = 3 c_6 = 12 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 13 H_2O + 10 HNO_3 + 3 As4O6 ⟶ 7 NO + 3 NO_2 + 12 H_3AsO_4
Balance the chemical equation algebraically: H_2O + HNO_3 + As4O6 ⟶ NO + NO_2 + H_3AsO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HNO_3 + c_3 As4O6 ⟶ c_4 NO + c_5 NO_2 + c_6 H_3AsO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and As: H: | 2 c_1 + c_2 = 3 c_6 O: | c_1 + 3 c_2 + 6 c_3 = c_4 + 2 c_5 + 4 c_6 N: | c_2 = c_4 + c_5 As: | 4 c_3 = c_6 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_2 = 12 - 2 c_1 c_3 = 1 c_4 = c_1 - 2 c_5 = 14 - 3 c_1 c_6 = 4 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_2 = 36 - 2 c_1 c_3 = 3 c_4 = c_1 - 6 c_5 = 42 - 3 c_1 c_6 = 12 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 13 and solve for the remaining coefficients: c_1 = 13 c_2 = 10 c_3 = 3 c_4 = 7 c_5 = 3 c_6 = 12 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 13 H_2O + 10 HNO_3 + 3 As4O6 ⟶ 7 NO + 3 NO_2 + 12 H_3AsO_4

Structures

+ + As4O6 ⟶ + +
+ + As4O6 ⟶ + +

Names

water + nitric acid + As4O6 ⟶ nitric oxide + nitrogen dioxide + arsenic acid, solid
water + nitric acid + As4O6 ⟶ nitric oxide + nitrogen dioxide + arsenic acid, solid

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + HNO_3 + As4O6 ⟶ NO + NO_2 + H_3AsO_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: 13 H_2O + 10 HNO_3 + 3 As4O6 ⟶ 7 NO + 3 NO_2 + 12 H_3AsO_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 | 13 | -13 HNO_3 | 10 | -10 As4O6 | 3 | -3 NO | 7 | 7 NO_2 | 3 | 3 H_3AsO_4 | 12 | 12 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 13 | -13 | ([H2O])^(-13) HNO_3 | 10 | -10 | ([HNO3])^(-10) As4O6 | 3 | -3 | ([As4O6])^(-3) NO | 7 | 7 | ([NO])^7 NO_2 | 3 | 3 | ([NO2])^3 H_3AsO_4 | 12 | 12 | ([H3AsO4])^12 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])^(-13) ([HNO3])^(-10) ([As4O6])^(-3) ([NO])^7 ([NO2])^3 ([H3AsO4])^12 = (([NO])^7 ([NO2])^3 ([H3AsO4])^12)/(([H2O])^13 ([HNO3])^10 ([As4O6])^3)
Construct the equilibrium constant, K, expression for: H_2O + HNO_3 + As4O6 ⟶ NO + NO_2 + H_3AsO_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: 13 H_2O + 10 HNO_3 + 3 As4O6 ⟶ 7 NO + 3 NO_2 + 12 H_3AsO_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 | 13 | -13 HNO_3 | 10 | -10 As4O6 | 3 | -3 NO | 7 | 7 NO_2 | 3 | 3 H_3AsO_4 | 12 | 12 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 13 | -13 | ([H2O])^(-13) HNO_3 | 10 | -10 | ([HNO3])^(-10) As4O6 | 3 | -3 | ([As4O6])^(-3) NO | 7 | 7 | ([NO])^7 NO_2 | 3 | 3 | ([NO2])^3 H_3AsO_4 | 12 | 12 | ([H3AsO4])^12 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])^(-13) ([HNO3])^(-10) ([As4O6])^(-3) ([NO])^7 ([NO2])^3 ([H3AsO4])^12 = (([NO])^7 ([NO2])^3 ([H3AsO4])^12)/(([H2O])^13 ([HNO3])^10 ([As4O6])^3)

Rate of reaction

Construct the rate of reaction expression for: H_2O + HNO_3 + As4O6 ⟶ NO + NO_2 + H_3AsO_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: 13 H_2O + 10 HNO_3 + 3 As4O6 ⟶ 7 NO + 3 NO_2 + 12 H_3AsO_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 | 13 | -13 HNO_3 | 10 | -10 As4O6 | 3 | -3 NO | 7 | 7 NO_2 | 3 | 3 H_3AsO_4 | 12 | 12 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 | 13 | -13 | -1/13 (Δ[H2O])/(Δt) HNO_3 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) As4O6 | 3 | -3 | -1/3 (Δ[As4O6])/(Δt) NO | 7 | 7 | 1/7 (Δ[NO])/(Δt) NO_2 | 3 | 3 | 1/3 (Δ[NO2])/(Δt) H_3AsO_4 | 12 | 12 | 1/12 (Δ[H3AsO4])/(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 = -1/13 (Δ[H2O])/(Δt) = -1/10 (Δ[HNO3])/(Δt) = -1/3 (Δ[As4O6])/(Δt) = 1/7 (Δ[NO])/(Δt) = 1/3 (Δ[NO2])/(Δt) = 1/12 (Δ[H3AsO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + HNO_3 + As4O6 ⟶ NO + NO_2 + H_3AsO_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: 13 H_2O + 10 HNO_3 + 3 As4O6 ⟶ 7 NO + 3 NO_2 + 12 H_3AsO_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 | 13 | -13 HNO_3 | 10 | -10 As4O6 | 3 | -3 NO | 7 | 7 NO_2 | 3 | 3 H_3AsO_4 | 12 | 12 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 | 13 | -13 | -1/13 (Δ[H2O])/(Δt) HNO_3 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) As4O6 | 3 | -3 | -1/3 (Δ[As4O6])/(Δt) NO | 7 | 7 | 1/7 (Δ[NO])/(Δt) NO_2 | 3 | 3 | 1/3 (Δ[NO2])/(Δt) H_3AsO_4 | 12 | 12 | 1/12 (Δ[H3AsO4])/(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 = -1/13 (Δ[H2O])/(Δt) = -1/10 (Δ[HNO3])/(Δt) = -1/3 (Δ[As4O6])/(Δt) = 1/7 (Δ[NO])/(Δt) = 1/3 (Δ[NO2])/(Δt) = 1/12 (Δ[H3AsO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | nitric acid | As4O6 | nitric oxide | nitrogen dioxide | arsenic acid, solid formula | H_2O | HNO_3 | As4O6 | NO | NO_2 | H_3AsO_4 Hill formula | H_2O | HNO_3 | As4O6 | NO | NO_2 | AsH_3O_4 name | water | nitric acid | | nitric oxide | nitrogen dioxide | arsenic acid, solid IUPAC name | water | nitric acid | | nitric oxide | Nitrogen dioxide | arsoric acid
| water | nitric acid | As4O6 | nitric oxide | nitrogen dioxide | arsenic acid, solid formula | H_2O | HNO_3 | As4O6 | NO | NO_2 | H_3AsO_4 Hill formula | H_2O | HNO_3 | As4O6 | NO | NO_2 | AsH_3O_4 name | water | nitric acid | | nitric oxide | nitrogen dioxide | arsenic acid, solid IUPAC name | water | nitric acid | | nitric oxide | Nitrogen dioxide | arsoric acid

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