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H2O + O2 + K2TcO4 = KOH + KTcO4

Input interpretation

H_2O water + O_2 oxygen + K2TcO4 ⟶ KOH potassium hydroxide + KTcO4
H_2O water + O_2 oxygen + K2TcO4 ⟶ KOH potassium hydroxide + KTcO4

Balanced equation

Balance the chemical equation algebraically: H_2O + O_2 + K2TcO4 ⟶ KOH + KTcO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 K2TcO4 ⟶ c_4 KOH + c_5 KTcO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K and Tc: H: | 2 c_1 = c_4 O: | c_1 + 2 c_2 + 4 c_3 = c_4 + 4 c_5 K: | 2 c_3 = c_4 + c_5 Tc: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 4 c_4 = 4 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + O_2 + 4 K2TcO4 ⟶ 4 KOH + 4 KTcO4
Balance the chemical equation algebraically: H_2O + O_2 + K2TcO4 ⟶ KOH + KTcO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 K2TcO4 ⟶ c_4 KOH + c_5 KTcO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, K and Tc: H: | 2 c_1 = c_4 O: | c_1 + 2 c_2 + 4 c_3 = c_4 + 4 c_5 K: | 2 c_3 = c_4 + c_5 Tc: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 4 c_4 = 4 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + O_2 + 4 K2TcO4 ⟶ 4 KOH + 4 KTcO4

Structures

+ + K2TcO4 ⟶ + KTcO4
+ + K2TcO4 ⟶ + KTcO4

Names

water + oxygen + K2TcO4 ⟶ potassium hydroxide + KTcO4
water + oxygen + K2TcO4 ⟶ potassium hydroxide + KTcO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + O_2 + K2TcO4 ⟶ KOH + KTcO4 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: 2 H_2O + O_2 + 4 K2TcO4 ⟶ 4 KOH + 4 KTcO4 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 | 2 | -2 O_2 | 1 | -1 K2TcO4 | 4 | -4 KOH | 4 | 4 KTcO4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) O_2 | 1 | -1 | ([O2])^(-1) K2TcO4 | 4 | -4 | ([K2TcO4])^(-4) KOH | 4 | 4 | ([KOH])^4 KTcO4 | 4 | 4 | ([KTcO4])^4 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])^(-2) ([O2])^(-1) ([K2TcO4])^(-4) ([KOH])^4 ([KTcO4])^4 = (([KOH])^4 ([KTcO4])^4)/(([H2O])^2 [O2] ([K2TcO4])^4)
Construct the equilibrium constant, K, expression for: H_2O + O_2 + K2TcO4 ⟶ KOH + KTcO4 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: 2 H_2O + O_2 + 4 K2TcO4 ⟶ 4 KOH + 4 KTcO4 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 | 2 | -2 O_2 | 1 | -1 K2TcO4 | 4 | -4 KOH | 4 | 4 KTcO4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) O_2 | 1 | -1 | ([O2])^(-1) K2TcO4 | 4 | -4 | ([K2TcO4])^(-4) KOH | 4 | 4 | ([KOH])^4 KTcO4 | 4 | 4 | ([KTcO4])^4 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])^(-2) ([O2])^(-1) ([K2TcO4])^(-4) ([KOH])^4 ([KTcO4])^4 = (([KOH])^4 ([KTcO4])^4)/(([H2O])^2 [O2] ([K2TcO4])^4)

Rate of reaction

Construct the rate of reaction expression for: H_2O + O_2 + K2TcO4 ⟶ KOH + KTcO4 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: 2 H_2O + O_2 + 4 K2TcO4 ⟶ 4 KOH + 4 KTcO4 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 | 2 | -2 O_2 | 1 | -1 K2TcO4 | 4 | -4 KOH | 4 | 4 KTcO4 | 4 | 4 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 | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) K2TcO4 | 4 | -4 | -1/4 (Δ[K2TcO4])/(Δt) KOH | 4 | 4 | 1/4 (Δ[KOH])/(Δt) KTcO4 | 4 | 4 | 1/4 (Δ[KTcO4])/(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 = -1/2 (Δ[H2O])/(Δt) = -(Δ[O2])/(Δt) = -1/4 (Δ[K2TcO4])/(Δt) = 1/4 (Δ[KOH])/(Δt) = 1/4 (Δ[KTcO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + O_2 + K2TcO4 ⟶ KOH + KTcO4 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: 2 H_2O + O_2 + 4 K2TcO4 ⟶ 4 KOH + 4 KTcO4 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 | 2 | -2 O_2 | 1 | -1 K2TcO4 | 4 | -4 KOH | 4 | 4 KTcO4 | 4 | 4 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 | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) K2TcO4 | 4 | -4 | -1/4 (Δ[K2TcO4])/(Δt) KOH | 4 | 4 | 1/4 (Δ[KOH])/(Δt) KTcO4 | 4 | 4 | 1/4 (Δ[KTcO4])/(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 = -1/2 (Δ[H2O])/(Δt) = -(Δ[O2])/(Δt) = -1/4 (Δ[K2TcO4])/(Δt) = 1/4 (Δ[KOH])/(Δt) = 1/4 (Δ[KTcO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | oxygen | K2TcO4 | potassium hydroxide | KTcO4 formula | H_2O | O_2 | K2TcO4 | KOH | KTcO4 Hill formula | H_2O | O_2 | K2O4Tc | HKO | KO4Tc name | water | oxygen | | potassium hydroxide | IUPAC name | water | molecular oxygen | | potassium hydroxide |
| water | oxygen | K2TcO4 | potassium hydroxide | KTcO4 formula | H_2O | O_2 | K2TcO4 | KOH | KTcO4 Hill formula | H_2O | O_2 | K2O4Tc | HKO | KO4Tc name | water | oxygen | | potassium hydroxide | IUPAC name | water | molecular oxygen | | potassium hydroxide |

Substance properties

| water | oxygen | K2TcO4 | potassium hydroxide | KTcO4 molar mass | 18.015 g/mol | 31.998 g/mol | 240 g/mol | 56.105 g/mol | 201 g/mol phase | liquid (at STP) | gas (at STP) | | solid (at STP) | melting point | 0 °C | -218 °C | | 406 °C | boiling point | 99.9839 °C | -183 °C | | 1327 °C | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | | 2.044 g/cm^3 | solubility in water | | | | soluble | surface tension | 0.0728 N/m | 0.01347 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | | 0.001 Pa s (at 550 °C) | odor | odorless | odorless | | |
| water | oxygen | K2TcO4 | potassium hydroxide | KTcO4 molar mass | 18.015 g/mol | 31.998 g/mol | 240 g/mol | 56.105 g/mol | 201 g/mol phase | liquid (at STP) | gas (at STP) | | solid (at STP) | melting point | 0 °C | -218 °C | | 406 °C | boiling point | 99.9839 °C | -183 °C | | 1327 °C | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | | 2.044 g/cm^3 | solubility in water | | | | soluble | surface tension | 0.0728 N/m | 0.01347 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | | 0.001 Pa s (at 550 °C) | odor | odorless | odorless | | |

Units

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