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HNO3 + I2 = H2O + NO + HIO4

Input interpretation

HNO_3 nitric acid + I_2 iodine ⟶ H_2O water + NO nitric oxide + HIO4
HNO_3 nitric acid + I_2 iodine ⟶ H_2O water + NO nitric oxide + HIO4

Balanced equation

Balance the chemical equation algebraically: HNO_3 + I_2 ⟶ H_2O + NO + HIO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 I_2 ⟶ c_3 H_2O + c_4 NO + c_5 HIO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and I: H: | c_1 = 2 c_3 + c_5 N: | c_1 = c_4 O: | 3 c_1 = c_3 + c_4 + 4 c_5 I: | 2 c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 14/3 c_2 = 1 c_3 = 4/3 c_4 = 14/3 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 14 c_2 = 3 c_3 = 4 c_4 = 14 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + 3 I_2 ⟶ 4 H_2O + 14 NO + 6 HIO4
Balance the chemical equation algebraically: HNO_3 + I_2 ⟶ H_2O + NO + HIO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 I_2 ⟶ c_3 H_2O + c_4 NO + c_5 HIO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and I: H: | c_1 = 2 c_3 + c_5 N: | c_1 = c_4 O: | 3 c_1 = c_3 + c_4 + 4 c_5 I: | 2 c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 14/3 c_2 = 1 c_3 = 4/3 c_4 = 14/3 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 14 c_2 = 3 c_3 = 4 c_4 = 14 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 14 HNO_3 + 3 I_2 ⟶ 4 H_2O + 14 NO + 6 HIO4

Structures

+ ⟶ + + HIO4
+ ⟶ + + HIO4

Names

nitric acid + iodine ⟶ water + nitric oxide + HIO4
nitric acid + iodine ⟶ water + nitric oxide + HIO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + I_2 ⟶ H_2O + NO + HIO4 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: 14 HNO_3 + 3 I_2 ⟶ 4 H_2O + 14 NO + 6 HIO4 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 HNO_3 | 14 | -14 I_2 | 3 | -3 H_2O | 4 | 4 NO | 14 | 14 HIO4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 14 | -14 | ([HNO3])^(-14) I_2 | 3 | -3 | ([I2])^(-3) H_2O | 4 | 4 | ([H2O])^4 NO | 14 | 14 | ([NO])^14 HIO4 | 6 | 6 | ([HIO4])^6 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 = ([HNO3])^(-14) ([I2])^(-3) ([H2O])^4 ([NO])^14 ([HIO4])^6 = (([H2O])^4 ([NO])^14 ([HIO4])^6)/(([HNO3])^14 ([I2])^3)
Construct the equilibrium constant, K, expression for: HNO_3 + I_2 ⟶ H_2O + NO + HIO4 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: 14 HNO_3 + 3 I_2 ⟶ 4 H_2O + 14 NO + 6 HIO4 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 HNO_3 | 14 | -14 I_2 | 3 | -3 H_2O | 4 | 4 NO | 14 | 14 HIO4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 14 | -14 | ([HNO3])^(-14) I_2 | 3 | -3 | ([I2])^(-3) H_2O | 4 | 4 | ([H2O])^4 NO | 14 | 14 | ([NO])^14 HIO4 | 6 | 6 | ([HIO4])^6 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 = ([HNO3])^(-14) ([I2])^(-3) ([H2O])^4 ([NO])^14 ([HIO4])^6 = (([H2O])^4 ([NO])^14 ([HIO4])^6)/(([HNO3])^14 ([I2])^3)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + I_2 ⟶ H_2O + NO + HIO4 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: 14 HNO_3 + 3 I_2 ⟶ 4 H_2O + 14 NO + 6 HIO4 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 HNO_3 | 14 | -14 I_2 | 3 | -3 H_2O | 4 | 4 NO | 14 | 14 HIO4 | 6 | 6 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 HNO_3 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) I_2 | 3 | -3 | -1/3 (Δ[I2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) NO | 14 | 14 | 1/14 (Δ[NO])/(Δt) HIO4 | 6 | 6 | 1/6 (Δ[HIO4])/(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 = -1/14 (Δ[HNO3])/(Δt) = -1/3 (Δ[I2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/14 (Δ[NO])/(Δt) = 1/6 (Δ[HIO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + I_2 ⟶ H_2O + NO + HIO4 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: 14 HNO_3 + 3 I_2 ⟶ 4 H_2O + 14 NO + 6 HIO4 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 HNO_3 | 14 | -14 I_2 | 3 | -3 H_2O | 4 | 4 NO | 14 | 14 HIO4 | 6 | 6 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 HNO_3 | 14 | -14 | -1/14 (Δ[HNO3])/(Δt) I_2 | 3 | -3 | -1/3 (Δ[I2])/(Δt) H_2O | 4 | 4 | 1/4 (Δ[H2O])/(Δt) NO | 14 | 14 | 1/14 (Δ[NO])/(Δt) HIO4 | 6 | 6 | 1/6 (Δ[HIO4])/(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 = -1/14 (Δ[HNO3])/(Δt) = -1/3 (Δ[I2])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/14 (Δ[NO])/(Δt) = 1/6 (Δ[HIO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | iodine | water | nitric oxide | HIO4 formula | HNO_3 | I_2 | H_2O | NO | HIO4 name | nitric acid | iodine | water | nitric oxide | IUPAC name | nitric acid | molecular iodine | water | nitric oxide |
| nitric acid | iodine | water | nitric oxide | HIO4 formula | HNO_3 | I_2 | H_2O | NO | HIO4 name | nitric acid | iodine | water | nitric oxide | IUPAC name | nitric acid | molecular iodine | water | nitric oxide |

Substance properties

| nitric acid | iodine | water | nitric oxide | HIO4 molar mass | 63.012 g/mol | 253.80894 g/mol | 18.015 g/mol | 30.006 g/mol | 191.91 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 113 °C | 0 °C | -163.6 °C | boiling point | 83 °C | 184 °C | 99.9839 °C | -151.7 °C | density | 1.5129 g/cm^3 | 4.94 g/cm^3 | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.00227 Pa s (at 116 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | | odorless | |
| nitric acid | iodine | water | nitric oxide | HIO4 molar mass | 63.012 g/mol | 253.80894 g/mol | 18.015 g/mol | 30.006 g/mol | 191.91 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 113 °C | 0 °C | -163.6 °C | boiling point | 83 °C | 184 °C | 99.9839 °C | -151.7 °C | density | 1.5129 g/cm^3 | 4.94 g/cm^3 | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.00227 Pa s (at 116 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.911×10^-5 Pa s (at 25 °C) | odor | | | odorless | |

Units

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