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O2 + KNO2 = KNO3

Input interpretation

O_2 oxygen + KNO_2 potassium nitrite ⟶ KNO_3 potassium nitrate
O_2 oxygen + KNO_2 potassium nitrite ⟶ KNO_3 potassium nitrate

Balanced equation

Balance the chemical equation algebraically: O_2 + KNO_2 ⟶ KNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 KNO_2 ⟶ c_3 KNO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, K and N: O: | 2 c_1 + 2 c_2 = 3 c_3 K: | c_2 = c_3 N: | c_2 = c_3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | O_2 + 2 KNO_2 ⟶ 2 KNO_3
Balance the chemical equation algebraically: O_2 + KNO_2 ⟶ KNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 KNO_2 ⟶ c_3 KNO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for O, K and N: O: | 2 c_1 + 2 c_2 = 3 c_3 K: | c_2 = c_3 N: | c_2 = c_3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + 2 KNO_2 ⟶ 2 KNO_3

Structures

 + ⟶
+ ⟶

Names

oxygen + potassium nitrite ⟶ potassium nitrate
oxygen + potassium nitrite ⟶ potassium nitrate

Reaction thermodynamics

Enthalpy

 | oxygen | potassium nitrite | potassium nitrate molecular enthalpy | 0 kJ/mol | -369.8 kJ/mol | -494.6 kJ/mol total enthalpy | 0 kJ/mol | -739.6 kJ/mol | -989.2 kJ/mol  | H_initial = -739.6 kJ/mol | | H_final = -989.2 kJ/mol ΔH_rxn^0 | -989.2 kJ/mol - -739.6 kJ/mol = -249.6 kJ/mol (exothermic) | |
| oxygen | potassium nitrite | potassium nitrate molecular enthalpy | 0 kJ/mol | -369.8 kJ/mol | -494.6 kJ/mol total enthalpy | 0 kJ/mol | -739.6 kJ/mol | -989.2 kJ/mol | H_initial = -739.6 kJ/mol | | H_final = -989.2 kJ/mol ΔH_rxn^0 | -989.2 kJ/mol - -739.6 kJ/mol = -249.6 kJ/mol (exothermic) | |

Gibbs free energy

 | oxygen | potassium nitrite | potassium nitrate molecular free energy | 231.7 kJ/mol | -306.6 kJ/mol | -394.9 kJ/mol total free energy | 231.7 kJ/mol | -613.2 kJ/mol | -789.8 kJ/mol  | G_initial = -381.5 kJ/mol | | G_final = -789.8 kJ/mol ΔG_rxn^0 | -789.8 kJ/mol - -381.5 kJ/mol = -408.3 kJ/mol (exergonic) | |
| oxygen | potassium nitrite | potassium nitrate molecular free energy | 231.7 kJ/mol | -306.6 kJ/mol | -394.9 kJ/mol total free energy | 231.7 kJ/mol | -613.2 kJ/mol | -789.8 kJ/mol | G_initial = -381.5 kJ/mol | | G_final = -789.8 kJ/mol ΔG_rxn^0 | -789.8 kJ/mol - -381.5 kJ/mol = -408.3 kJ/mol (exergonic) | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + KNO_2 ⟶ KNO_3 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: O_2 + 2 KNO_2 ⟶ 2 KNO_3 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 O_2 | 1 | -1 KNO_2 | 2 | -2 KNO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) KNO_2 | 2 | -2 | ([KNO2])^(-2) KNO_3 | 2 | 2 | ([KNO3])^2 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 = ([O2])^(-1) ([KNO2])^(-2) ([KNO3])^2 = ([KNO3])^2/([O2] ([KNO2])^2)
Construct the equilibrium constant, K, expression for: O_2 + KNO_2 ⟶ KNO_3 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: O_2 + 2 KNO_2 ⟶ 2 KNO_3 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 O_2 | 1 | -1 KNO_2 | 2 | -2 KNO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 1 | -1 | ([O2])^(-1) KNO_2 | 2 | -2 | ([KNO2])^(-2) KNO_3 | 2 | 2 | ([KNO3])^2 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 = ([O2])^(-1) ([KNO2])^(-2) ([KNO3])^2 = ([KNO3])^2/([O2] ([KNO2])^2)

Rate of reaction

Construct the rate of reaction expression for: O_2 + KNO_2 ⟶ KNO_3 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: O_2 + 2 KNO_2 ⟶ 2 KNO_3 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 O_2 | 1 | -1 KNO_2 | 2 | -2 KNO_3 | 2 | 2 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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) KNO_2 | 2 | -2 | -1/2 (Δ[KNO2])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(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 = -(Δ[O2])/(Δt) = -1/2 (Δ[KNO2])/(Δt) = 1/2 (Δ[KNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + KNO_2 ⟶ KNO_3 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: O_2 + 2 KNO_2 ⟶ 2 KNO_3 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 O_2 | 1 | -1 KNO_2 | 2 | -2 KNO_3 | 2 | 2 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 O_2 | 1 | -1 | -(Δ[O2])/(Δt) KNO_2 | 2 | -2 | -1/2 (Δ[KNO2])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(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 = -(Δ[O2])/(Δt) = -1/2 (Δ[KNO2])/(Δt) = 1/2 (Δ[KNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | oxygen | potassium nitrite | potassium nitrate formula | O_2 | KNO_2 | KNO_3 name | oxygen | potassium nitrite | potassium nitrate IUPAC name | molecular oxygen | potassium nitrite | potassium nitrate
| oxygen | potassium nitrite | potassium nitrate formula | O_2 | KNO_2 | KNO_3 name | oxygen | potassium nitrite | potassium nitrate IUPAC name | molecular oxygen | potassium nitrite | potassium nitrate

Substance properties

 | oxygen | potassium nitrite | potassium nitrate molar mass | 31.998 g/mol | 85.103 g/mol | 101.1 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 350 °C | 334 °C boiling point | -183 °C | |  density | 0.001429 g/cm^3 (at 0 °C) | 1.915 g/cm^3 |  solubility in water | | | soluble surface tension | 0.01347 N/m | |  dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | |  odor | odorless | | odorless
| oxygen | potassium nitrite | potassium nitrate molar mass | 31.998 g/mol | 85.103 g/mol | 101.1 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 350 °C | 334 °C boiling point | -183 °C | | density | 0.001429 g/cm^3 (at 0 °C) | 1.915 g/cm^3 | solubility in water | | | soluble surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | | odorless

Units