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KNO3 = O2 + N2 + K2O

Input interpretation

KNO_3 potassium nitrate ⟶ O_2 oxygen + N_2 nitrogen + K_2O potassium oxide
KNO_3 potassium nitrate ⟶ O_2 oxygen + N_2 nitrogen + K_2O potassium oxide

Balanced equation

Balance the chemical equation algebraically: KNO_3 ⟶ O_2 + N_2 + K_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KNO_3 ⟶ c_2 O_2 + c_3 N_2 + c_4 K_2O Set the number of atoms in the reactants equal to the number of atoms in the products for K, N and O: K: | c_1 = 2 c_4 N: | c_1 = 2 c_3 O: | 3 c_1 = 2 c_2 + 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 = 2 c_2 = 5/2 c_3 = 1 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 5 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 4 KNO_3 ⟶ 5 O_2 + 2 N_2 + 2 K_2O
Balance the chemical equation algebraically: KNO_3 ⟶ O_2 + N_2 + K_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KNO_3 ⟶ c_2 O_2 + c_3 N_2 + c_4 K_2O Set the number of atoms in the reactants equal to the number of atoms in the products for K, N and O: K: | c_1 = 2 c_4 N: | c_1 = 2 c_3 O: | 3 c_1 = 2 c_2 + 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 = 2 c_2 = 5/2 c_3 = 1 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 4 c_2 = 5 c_3 = 2 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 KNO_3 ⟶ 5 O_2 + 2 N_2 + 2 K_2O

Structures

 ⟶ + +
⟶ + +

Names

potassium nitrate ⟶ oxygen + nitrogen + potassium oxide
potassium nitrate ⟶ oxygen + nitrogen + potassium oxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: KNO_3 ⟶ O_2 + N_2 + K_2O 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: 4 KNO_3 ⟶ 5 O_2 + 2 N_2 + 2 K_2O 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 KNO_3 | 4 | -4 O_2 | 5 | 5 N_2 | 2 | 2 K_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KNO_3 | 4 | -4 | ([KNO3])^(-4) O_2 | 5 | 5 | ([O2])^5 N_2 | 2 | 2 | ([N2])^2 K_2O | 2 | 2 | ([K2O])^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 = ([KNO3])^(-4) ([O2])^5 ([N2])^2 ([K2O])^2 = (([O2])^5 ([N2])^2 ([K2O])^2)/([KNO3])^4
Construct the equilibrium constant, K, expression for: KNO_3 ⟶ O_2 + N_2 + K_2O 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: 4 KNO_3 ⟶ 5 O_2 + 2 N_2 + 2 K_2O 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 KNO_3 | 4 | -4 O_2 | 5 | 5 N_2 | 2 | 2 K_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KNO_3 | 4 | -4 | ([KNO3])^(-4) O_2 | 5 | 5 | ([O2])^5 N_2 | 2 | 2 | ([N2])^2 K_2O | 2 | 2 | ([K2O])^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 = ([KNO3])^(-4) ([O2])^5 ([N2])^2 ([K2O])^2 = (([O2])^5 ([N2])^2 ([K2O])^2)/([KNO3])^4

Rate of reaction

Construct the rate of reaction expression for: KNO_3 ⟶ O_2 + N_2 + K_2O 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: 4 KNO_3 ⟶ 5 O_2 + 2 N_2 + 2 K_2O 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 KNO_3 | 4 | -4 O_2 | 5 | 5 N_2 | 2 | 2 K_2O | 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 KNO_3 | 4 | -4 | -1/4 (Δ[KNO3])/(Δt) O_2 | 5 | 5 | 1/5 (Δ[O2])/(Δt) N_2 | 2 | 2 | 1/2 (Δ[N2])/(Δt) K_2O | 2 | 2 | 1/2 (Δ[K2O])/(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 = -1/4 (Δ[KNO3])/(Δt) = 1/5 (Δ[O2])/(Δt) = 1/2 (Δ[N2])/(Δt) = 1/2 (Δ[K2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KNO_3 ⟶ O_2 + N_2 + K_2O 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: 4 KNO_3 ⟶ 5 O_2 + 2 N_2 + 2 K_2O 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 KNO_3 | 4 | -4 O_2 | 5 | 5 N_2 | 2 | 2 K_2O | 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 KNO_3 | 4 | -4 | -1/4 (Δ[KNO3])/(Δt) O_2 | 5 | 5 | 1/5 (Δ[O2])/(Δt) N_2 | 2 | 2 | 1/2 (Δ[N2])/(Δt) K_2O | 2 | 2 | 1/2 (Δ[K2O])/(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 = -1/4 (Δ[KNO3])/(Δt) = 1/5 (Δ[O2])/(Δt) = 1/2 (Δ[N2])/(Δt) = 1/2 (Δ[K2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | potassium nitrate | oxygen | nitrogen | potassium oxide formula | KNO_3 | O_2 | N_2 | K_2O name | potassium nitrate | oxygen | nitrogen | potassium oxide IUPAC name | potassium nitrate | molecular oxygen | molecular nitrogen | dipotassium oxygen(2-)
| potassium nitrate | oxygen | nitrogen | potassium oxide formula | KNO_3 | O_2 | N_2 | K_2O name | potassium nitrate | oxygen | nitrogen | potassium oxide IUPAC name | potassium nitrate | molecular oxygen | molecular nitrogen | dipotassium oxygen(2-)

Substance properties

 | potassium nitrate | oxygen | nitrogen | potassium oxide molar mass | 101.1 g/mol | 31.998 g/mol | 28.014 g/mol | 94.196 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) |  melting point | 334 °C | -218 °C | -210 °C |  boiling point | | -183 °C | -195.79 °C |  density | | 0.001429 g/cm^3 (at 0 °C) | 0.001251 g/cm^3 (at 0 °C) |  solubility in water | soluble | | insoluble |  surface tension | | 0.01347 N/m | 0.0066 N/m |  dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) |  odor | odorless | odorless | odorless |
| potassium nitrate | oxygen | nitrogen | potassium oxide molar mass | 101.1 g/mol | 31.998 g/mol | 28.014 g/mol | 94.196 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | melting point | 334 °C | -218 °C | -210 °C | boiling point | | -183 °C | -195.79 °C | density | | 0.001429 g/cm^3 (at 0 °C) | 0.001251 g/cm^3 (at 0 °C) | solubility in water | soluble | | insoluble | surface tension | | 0.01347 N/m | 0.0066 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | odorless | odorless | odorless |

Units