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KOH + Te = H2O + K2TeO3 + K2Te

Input interpretation

KOH potassium hydroxide + Te tellurium ⟶ H_2O water + K_2TeO_3 potassium tellurite + K2Te
KOH potassium hydroxide + Te tellurium ⟶ H_2O water + K_2TeO_3 potassium tellurite + K2Te

Balanced equation

Balance the chemical equation algebraically: KOH + Te ⟶ H_2O + K_2TeO_3 + K2Te Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Te ⟶ c_3 H_2O + c_4 K_2TeO_3 + c_5 K2Te Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and Te: H: | c_1 = 2 c_3 K: | c_1 = 2 c_4 + 2 c_5 O: | c_1 = c_3 + 3 c_4 Te: | c_2 = c_4 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 3 c_3 = 3 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 KOH + 3 Te ⟶ 3 H_2O + K_2TeO_3 + 2 K2Te
Balance the chemical equation algebraically: KOH + Te ⟶ H_2O + K_2TeO_3 + K2Te Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 Te ⟶ c_3 H_2O + c_4 K_2TeO_3 + c_5 K2Te Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and Te: H: | c_1 = 2 c_3 K: | c_1 = 2 c_4 + 2 c_5 O: | c_1 = c_3 + 3 c_4 Te: | c_2 = c_4 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 3 c_3 = 3 c_4 = 1 c_5 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 KOH + 3 Te ⟶ 3 H_2O + K_2TeO_3 + 2 K2Te

Structures

+ ⟶ + + K2Te
+ ⟶ + + K2Te

Names

potassium hydroxide + tellurium ⟶ water + potassium tellurite + K2Te
potassium hydroxide + tellurium ⟶ water + potassium tellurite + K2Te

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + Te ⟶ H_2O + K_2TeO_3 + K2Te 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 + 3 Te ⟶ 3 H_2O + K_2TeO_3 + 2 K2Te 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 Te | 3 | -3 H_2O | 3 | 3 K_2TeO_3 | 1 | 1 K2Te | 2 | 2 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) Te | 3 | -3 | ([Te])^(-3) H_2O | 3 | 3 | ([H2O])^3 K_2TeO_3 | 1 | 1 | [K2TeO3] K2Te | 2 | 2 | ([K2Te])^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 = ([KOH])^(-6) ([Te])^(-3) ([H2O])^3 [K2TeO3] ([K2Te])^2 = (([H2O])^3 [K2TeO3] ([K2Te])^2)/(([KOH])^6 ([Te])^3)
Construct the equilibrium constant, K, expression for: KOH + Te ⟶ H_2O + K_2TeO_3 + K2Te 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 + 3 Te ⟶ 3 H_2O + K_2TeO_3 + 2 K2Te 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 Te | 3 | -3 H_2O | 3 | 3 K_2TeO_3 | 1 | 1 K2Te | 2 | 2 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) Te | 3 | -3 | ([Te])^(-3) H_2O | 3 | 3 | ([H2O])^3 K_2TeO_3 | 1 | 1 | [K2TeO3] K2Te | 2 | 2 | ([K2Te])^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 = ([KOH])^(-6) ([Te])^(-3) ([H2O])^3 [K2TeO3] ([K2Te])^2 = (([H2O])^3 [K2TeO3] ([K2Te])^2)/(([KOH])^6 ([Te])^3)

Rate of reaction

Construct the rate of reaction expression for: KOH + Te ⟶ H_2O + K_2TeO_3 + K2Te 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 + 3 Te ⟶ 3 H_2O + K_2TeO_3 + 2 K2Te 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 Te | 3 | -3 H_2O | 3 | 3 K_2TeO_3 | 1 | 1 K2Te | 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 KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) Te | 3 | -3 | -1/3 (Δ[Te])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) K_2TeO_3 | 1 | 1 | (Δ[K2TeO3])/(Δt) K2Te | 2 | 2 | 1/2 (Δ[K2Te])/(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) = -1/3 (Δ[Te])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[K2TeO3])/(Δt) = 1/2 (Δ[K2Te])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + Te ⟶ H_2O + K_2TeO_3 + K2Te 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 + 3 Te ⟶ 3 H_2O + K_2TeO_3 + 2 K2Te 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 Te | 3 | -3 H_2O | 3 | 3 K_2TeO_3 | 1 | 1 K2Te | 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 KOH | 6 | -6 | -1/6 (Δ[KOH])/(Δt) Te | 3 | -3 | -1/3 (Δ[Te])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) K_2TeO_3 | 1 | 1 | (Δ[K2TeO3])/(Δt) K2Te | 2 | 2 | 1/2 (Δ[K2Te])/(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) = -1/3 (Δ[Te])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[K2TeO3])/(Δt) = 1/2 (Δ[K2Te])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| potassium hydroxide | tellurium | water | potassium tellurite | K2Te formula | KOH | Te | H_2O | K_2TeO_3 | K2Te Hill formula | HKO | Te | H_2O | K_2O_3Te | K2Te name | potassium hydroxide | tellurium | water | potassium tellurite |
| potassium hydroxide | tellurium | water | potassium tellurite | K2Te formula | KOH | Te | H_2O | K_2TeO_3 | K2Te Hill formula | HKO | Te | H_2O | K_2O_3Te | K2Te name | potassium hydroxide | tellurium | water | potassium tellurite |

Substance properties

| potassium hydroxide | tellurium | water | potassium tellurite | K2Te molar mass | 56.105 g/mol | 127.6 g/mol | 18.015 g/mol | 253.79 g/mol | 205.8 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | 450 °C | 0 °C | 465 °C | boiling point | 1327 °C | 990 °C | 99.9839 °C | | density | 2.044 g/cm^3 | 6.24 g/cm^3 | 1 g/cm^3 | | solubility in water | soluble | insoluble | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |
| potassium hydroxide | tellurium | water | potassium tellurite | K2Te molar mass | 56.105 g/mol | 127.6 g/mol | 18.015 g/mol | 253.79 g/mol | 205.8 g/mol phase | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | 450 °C | 0 °C | 465 °C | boiling point | 1327 °C | 990 °C | 99.9839 °C | | density | 2.044 g/cm^3 | 6.24 g/cm^3 | 1 g/cm^3 | | solubility in water | soluble | insoluble | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | |

Units

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