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O2 + KOH + Cr(OH)3 = H2O + K2CrO4

Input interpretation

O_2 oxygen + KOH potassium hydroxide + Cr(OH)3 ⟶ H_2O water + K_2CrO_4 potassium chromate
O_2 oxygen + KOH potassium hydroxide + Cr(OH)3 ⟶ H_2O water + K_2CrO_4 potassium chromate

Balanced equation

Balance the chemical equation algebraically: O_2 + KOH + Cr(OH)3 ⟶ H_2O + K_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 KOH + c_3 Cr(OH)3 ⟶ c_4 H_2O + c_5 K_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H, K and Cr: O: | 2 c_1 + c_2 + 3 c_3 = c_4 + 4 c_5 H: | c_2 + 3 c_3 = 2 c_4 K: | c_2 = 2 c_5 Cr: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 8/3 c_3 = 4/3 c_4 = 10/3 c_5 = 4/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 8 c_3 = 4 c_4 = 10 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 8 KOH + 4 Cr(OH)3 ⟶ 10 H_2O + 4 K_2CrO_4
Balance the chemical equation algebraically: O_2 + KOH + Cr(OH)3 ⟶ H_2O + K_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 KOH + c_3 Cr(OH)3 ⟶ c_4 H_2O + c_5 K_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for O, H, K and Cr: O: | 2 c_1 + c_2 + 3 c_3 = c_4 + 4 c_5 H: | c_2 + 3 c_3 = 2 c_4 K: | c_2 = 2 c_5 Cr: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 8/3 c_3 = 4/3 c_4 = 10/3 c_5 = 4/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 8 c_3 = 4 c_4 = 10 c_5 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 O_2 + 8 KOH + 4 Cr(OH)3 ⟶ 10 H_2O + 4 K_2CrO_4

Structures

+ + Cr(OH)3 ⟶ +
+ + Cr(OH)3 ⟶ +

Names

oxygen + potassium hydroxide + Cr(OH)3 ⟶ water + potassium chromate
oxygen + potassium hydroxide + Cr(OH)3 ⟶ water + potassium chromate

Equilibrium constant

Construct the equilibrium constant, K, expression for: O_2 + KOH + Cr(OH)3 ⟶ H_2O + K_2CrO_4 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: 3 O_2 + 8 KOH + 4 Cr(OH)3 ⟶ 10 H_2O + 4 K_2CrO_4 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 | 3 | -3 KOH | 8 | -8 Cr(OH)3 | 4 | -4 H_2O | 10 | 10 K_2CrO_4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) KOH | 8 | -8 | ([KOH])^(-8) Cr(OH)3 | 4 | -4 | ([Cr(OH)3])^(-4) H_2O | 10 | 10 | ([H2O])^10 K_2CrO_4 | 4 | 4 | ([K2CrO4])^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 = ([O2])^(-3) ([KOH])^(-8) ([Cr(OH)3])^(-4) ([H2O])^10 ([K2CrO4])^4 = (([H2O])^10 ([K2CrO4])^4)/(([O2])^3 ([KOH])^8 ([Cr(OH)3])^4)
Construct the equilibrium constant, K, expression for: O_2 + KOH + Cr(OH)3 ⟶ H_2O + K_2CrO_4 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: 3 O_2 + 8 KOH + 4 Cr(OH)3 ⟶ 10 H_2O + 4 K_2CrO_4 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 | 3 | -3 KOH | 8 | -8 Cr(OH)3 | 4 | -4 H_2O | 10 | 10 K_2CrO_4 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 3 | -3 | ([O2])^(-3) KOH | 8 | -8 | ([KOH])^(-8) Cr(OH)3 | 4 | -4 | ([Cr(OH)3])^(-4) H_2O | 10 | 10 | ([H2O])^10 K_2CrO_4 | 4 | 4 | ([K2CrO4])^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 = ([O2])^(-3) ([KOH])^(-8) ([Cr(OH)3])^(-4) ([H2O])^10 ([K2CrO4])^4 = (([H2O])^10 ([K2CrO4])^4)/(([O2])^3 ([KOH])^8 ([Cr(OH)3])^4)

Rate of reaction

Construct the rate of reaction expression for: O_2 + KOH + Cr(OH)3 ⟶ H_2O + K_2CrO_4 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: 3 O_2 + 8 KOH + 4 Cr(OH)3 ⟶ 10 H_2O + 4 K_2CrO_4 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 | 3 | -3 KOH | 8 | -8 Cr(OH)3 | 4 | -4 H_2O | 10 | 10 K_2CrO_4 | 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 O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) KOH | 8 | -8 | -1/8 (Δ[KOH])/(Δt) Cr(OH)3 | 4 | -4 | -1/4 (Δ[Cr(OH)3])/(Δt) H_2O | 10 | 10 | 1/10 (Δ[H2O])/(Δt) K_2CrO_4 | 4 | 4 | 1/4 (Δ[K2CrO4])/(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 = -1/3 (Δ[O2])/(Δt) = -1/8 (Δ[KOH])/(Δt) = -1/4 (Δ[Cr(OH)3])/(Δt) = 1/10 (Δ[H2O])/(Δt) = 1/4 (Δ[K2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: O_2 + KOH + Cr(OH)3 ⟶ H_2O + K_2CrO_4 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: 3 O_2 + 8 KOH + 4 Cr(OH)3 ⟶ 10 H_2O + 4 K_2CrO_4 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 | 3 | -3 KOH | 8 | -8 Cr(OH)3 | 4 | -4 H_2O | 10 | 10 K_2CrO_4 | 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 O_2 | 3 | -3 | -1/3 (Δ[O2])/(Δt) KOH | 8 | -8 | -1/8 (Δ[KOH])/(Δt) Cr(OH)3 | 4 | -4 | -1/4 (Δ[Cr(OH)3])/(Δt) H_2O | 10 | 10 | 1/10 (Δ[H2O])/(Δt) K_2CrO_4 | 4 | 4 | 1/4 (Δ[K2CrO4])/(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 = -1/3 (Δ[O2])/(Δt) = -1/8 (Δ[KOH])/(Δt) = -1/4 (Δ[Cr(OH)3])/(Δt) = 1/10 (Δ[H2O])/(Δt) = 1/4 (Δ[K2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| oxygen | potassium hydroxide | Cr(OH)3 | water | potassium chromate formula | O_2 | KOH | Cr(OH)3 | H_2O | K_2CrO_4 Hill formula | O_2 | HKO | H3CrO3 | H_2O | CrK_2O_4 name | oxygen | potassium hydroxide | | water | potassium chromate IUPAC name | molecular oxygen | potassium hydroxide | | water | dipotassium dioxido-dioxochromium
| oxygen | potassium hydroxide | Cr(OH)3 | water | potassium chromate formula | O_2 | KOH | Cr(OH)3 | H_2O | K_2CrO_4 Hill formula | O_2 | HKO | H3CrO3 | H_2O | CrK_2O_4 name | oxygen | potassium hydroxide | | water | potassium chromate IUPAC name | molecular oxygen | potassium hydroxide | | water | dipotassium dioxido-dioxochromium

Substance properties

| oxygen | potassium hydroxide | Cr(OH)3 | water | potassium chromate molar mass | 31.998 g/mol | 56.105 g/mol | 103.02 g/mol | 18.015 g/mol | 194.19 g/mol phase | gas (at STP) | solid (at STP) | | liquid (at STP) | solid (at STP) melting point | -218 °C | 406 °C | | 0 °C | 971 °C boiling point | -183 °C | 1327 °C | | 99.9839 °C | density | 0.001429 g/cm^3 (at 0 °C) | 2.044 g/cm^3 | | 1 g/cm^3 | 2.73 g/cm^3 solubility in water | | soluble | | | soluble surface tension | 0.01347 N/m | | | 0.0728 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | | odorless | odorless
| oxygen | potassium hydroxide | Cr(OH)3 | water | potassium chromate molar mass | 31.998 g/mol | 56.105 g/mol | 103.02 g/mol | 18.015 g/mol | 194.19 g/mol phase | gas (at STP) | solid (at STP) | | liquid (at STP) | solid (at STP) melting point | -218 °C | 406 °C | | 0 °C | 971 °C boiling point | -183 °C | 1327 °C | | 99.9839 °C | density | 0.001429 g/cm^3 (at 0 °C) | 2.044 g/cm^3 | | 1 g/cm^3 | 2.73 g/cm^3 solubility in water | | soluble | | | soluble surface tension | 0.01347 N/m | | | 0.0728 N/m | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | | odorless | odorless

Units

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