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KOH + O3 = H2O + O2 + KO3

Input interpretation

KOH potassium hydroxide + O_3 ozone ⟶ H_2O water + O_2 oxygen + KO3
KOH potassium hydroxide + O_3 ozone ⟶ H_2O water + O_2 oxygen + KO3

Balanced equation

Balance the chemical equation algebraically: KOH + O_3 ⟶ H_2O + O_2 + KO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 O_3 ⟶ c_3 H_2O + c_4 O_2 + c_5 KO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K and O: H: | c_1 = 2 c_3 K: | c_1 = c_5 O: | c_1 + 3 c_2 = c_3 + 2 c_4 + 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 = 2 c_3 = 1 c_4 = (3 c_2)/2 - 5/2 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 6 c_3 = 3 c_4 = (3 c_2)/2 - 15/2 c_5 = 6 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_2 = 7 and solve for the remaining coefficients: c_1 = 6 c_2 = 7 c_3 = 3 c_4 = 3 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 KOH + 7 O_3 ⟶ 3 H_2O + 3 O_2 + 6 KO3
Balance the chemical equation algebraically: KOH + O_3 ⟶ H_2O + O_2 + KO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 O_3 ⟶ c_3 H_2O + c_4 O_2 + c_5 KO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K and O: H: | c_1 = 2 c_3 K: | c_1 = c_5 O: | c_1 + 3 c_2 = c_3 + 2 c_4 + 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 = 2 c_3 = 1 c_4 = (3 c_2)/2 - 5/2 c_5 = 2 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 6 c_3 = 3 c_4 = (3 c_2)/2 - 15/2 c_5 = 6 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_2 = 7 and solve for the remaining coefficients: c_1 = 6 c_2 = 7 c_3 = 3 c_4 = 3 c_5 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 KOH + 7 O_3 ⟶ 3 H_2O + 3 O_2 + 6 KO3

Structures

+ ⟶ + + KO3
+ ⟶ + + KO3

Names

potassium hydroxide + ozone ⟶ water + oxygen + KO3
potassium hydroxide + ozone ⟶ water + oxygen + KO3

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + O_3 ⟶ H_2O + O_2 + KO3 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: 6 KOH + 7 O_3 ⟶ 3 H_2O + 3 O_2 + 6 KO3 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 KOH | 6 | -6 O_3 | 7 | -7 H_2O | 3 | 3 O_2 | 3 | 3 KO3 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 6 | -6 | ([KOH])^(-6) O_3 | 7 | -7 | ([O3])^(-7) H_2O | 3 | 3 | ([H2O])^3 O_2 | 3 | 3 | ([O2])^3 KO3 | 6 | 6 | ([KO3])^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 = ([KOH])^(-6) ([O3])^(-7) ([H2O])^3 ([O2])^3 ([KO3])^6 = (([H2O])^3 ([O2])^3 ([KO3])^6)/(([KOH])^6 ([O3])^7)
Construct the equilibrium constant, K, expression for: KOH + O_3 ⟶ H_2O + O_2 + KO3 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: 6 KOH + 7 O_3 ⟶ 3 H_2O + 3 O_2 + 6 KO3 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 KOH | 6 | -6 O_3 | 7 | -7 H_2O | 3 | 3 O_2 | 3 | 3 KO3 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 6 | -6 | ([KOH])^(-6) O_3 | 7 | -7 | ([O3])^(-7) H_2O | 3 | 3 | ([H2O])^3 O_2 | 3 | 3 | ([O2])^3 KO3 | 6 | 6 | ([KO3])^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 = ([KOH])^(-6) ([O3])^(-7) ([H2O])^3 ([O2])^3 ([KO3])^6 = (([H2O])^3 ([O2])^3 ([KO3])^6)/(([KOH])^6 ([O3])^7)

Rate of reaction

Construct the rate of reaction expression for: KOH + O_3 ⟶ H_2O + O_2 + KO3 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: 6 KOH + 7 O_3 ⟶ 3 H_2O + 3 O_2 + 6 KO3 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 KOH | 6 | -6 O_3 | 7 | -7 H_2O | 3 | 3 O_2 | 3 | 3 KO3 | 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 KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) O_3 | 7 | -7 | -1/7 (Δ[O3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) KO3 | 6 | 6 | 1/6 (Δ[KO3])/(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 = -1/6 (Δ[KOH])/(Δt) = -1/7 (Δ[O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/6 (Δ[KO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + O_3 ⟶ H_2O + O_2 + KO3 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: 6 KOH + 7 O_3 ⟶ 3 H_2O + 3 O_2 + 6 KO3 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 KOH | 6 | -6 O_3 | 7 | -7 H_2O | 3 | 3 O_2 | 3 | 3 KO3 | 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 KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) O_3 | 7 | -7 | -1/7 (Δ[O3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) KO3 | 6 | 6 | 1/6 (Δ[KO3])/(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 = -1/6 (Δ[KOH])/(Δt) = -1/7 (Δ[O3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/6 (Δ[KO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | ozone | water | oxygen | KO3 formula | KOH | O_3 | H_2O | O_2 | KO3 Hill formula | HKO | O_3 | H_2O | O_2 | KO3 name | potassium hydroxide | ozone | water | oxygen | IUPAC name | potassium hydroxide | ozone | water | molecular oxygen |
| potassium hydroxide | ozone | water | oxygen | KO3 formula | KOH | O_3 | H_2O | O_2 | KO3 Hill formula | HKO | O_3 | H_2O | O_2 | KO3 name | potassium hydroxide | ozone | water | oxygen | IUPAC name | potassium hydroxide | ozone | water | molecular oxygen |

Substance properties

| potassium hydroxide | ozone | water | oxygen | KO3 molar mass | 56.105 g/mol | 47.997 g/mol | 18.015 g/mol | 31.998 g/mol | 87.095 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) | melting point | 406 °C | -192.2 °C | 0 °C | -218 °C | boiling point | 1327 °C | -111.9 °C | 99.9839 °C | -183 °C | density | 2.044 g/cm^3 | 0.001962 g/cm^3 (at 25 °C) | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | solubility in water | soluble | | | | surface tension | | | 0.0728 N/m | 0.01347 N/m | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | odor | | | odorless | odorless |
| potassium hydroxide | ozone | water | oxygen | KO3 molar mass | 56.105 g/mol | 47.997 g/mol | 18.015 g/mol | 31.998 g/mol | 87.095 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) | melting point | 406 °C | -192.2 °C | 0 °C | -218 °C | boiling point | 1327 °C | -111.9 °C | 99.9839 °C | -183 °C | density | 2.044 g/cm^3 | 0.001962 g/cm^3 (at 25 °C) | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | solubility in water | soluble | | | | surface tension | | | 0.0728 N/m | 0.01347 N/m | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | odor | | | odorless | odorless |

Units

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