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H2O + O2 + NaOH + Na3[IrCl6] = NaCl + Ir(OH)4

Input interpretation

H_2O water + O_2 oxygen + NaOH sodium hydroxide + Na3IrCl6 ⟶ NaCl sodium chloride + Ir(OH)4
H_2O water + O_2 oxygen + NaOH sodium hydroxide + Na3IrCl6 ⟶ NaCl sodium chloride + Ir(OH)4

Balanced equation

Balance the chemical equation algebraically: H_2O + O_2 + NaOH + Na3IrCl6 ⟶ NaCl + Ir(OH)4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 NaOH + c_4 Na3IrCl6 ⟶ c_5 NaCl + c_6 Ir(OH)4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Na, Ir and Cl: H: | 2 c_1 + c_3 = 4 c_6 O: | c_1 + 2 c_2 + c_3 = 4 c_6 Na: | c_3 + 3 c_4 = c_5 Ir: | c_4 = c_6 Cl: | 6 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 12 c_4 = 4 c_5 = 24 c_6 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + O_2 + 12 NaOH + 4 Na3IrCl6 ⟶ 24 NaCl + 4 Ir(OH)4
Balance the chemical equation algebraically: H_2O + O_2 + NaOH + Na3IrCl6 ⟶ NaCl + Ir(OH)4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 O_2 + c_3 NaOH + c_4 Na3IrCl6 ⟶ c_5 NaCl + c_6 Ir(OH)4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, Na, Ir and Cl: H: | 2 c_1 + c_3 = 4 c_6 O: | c_1 + 2 c_2 + c_3 = 4 c_6 Na: | c_3 + 3 c_4 = c_5 Ir: | c_4 = c_6 Cl: | 6 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 12 c_4 = 4 c_5 = 24 c_6 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + O_2 + 12 NaOH + 4 Na3IrCl6 ⟶ 24 NaCl + 4 Ir(OH)4

Structures

+ + + Na3IrCl6 ⟶ + Ir(OH)4
+ + + Na3IrCl6 ⟶ + Ir(OH)4

Names

water + oxygen + sodium hydroxide + Na3IrCl6 ⟶ sodium chloride + Ir(OH)4
water + oxygen + sodium hydroxide + Na3IrCl6 ⟶ sodium chloride + Ir(OH)4

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + O_2 + NaOH + Na3IrCl6 ⟶ NaCl + Ir(OH)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: 2 H_2O + O_2 + 12 NaOH + 4 Na3IrCl6 ⟶ 24 NaCl + 4 Ir(OH)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 H_2O | 2 | -2 O_2 | 1 | -1 NaOH | 12 | -12 Na3IrCl6 | 4 | -4 NaCl | 24 | 24 Ir(OH)4 | 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) NaOH | 12 | -12 | ([NaOH])^(-12) Na3IrCl6 | 4 | -4 | ([Na3IrCl6])^(-4) NaCl | 24 | 24 | ([NaCl])^24 Ir(OH)4 | 4 | 4 | ([Ir(OH)4])^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) ([NaOH])^(-12) ([Na3IrCl6])^(-4) ([NaCl])^24 ([Ir(OH)4])^4 = (([NaCl])^24 ([Ir(OH)4])^4)/(([H2O])^2 [O2] ([NaOH])^12 ([Na3IrCl6])^4)
Construct the equilibrium constant, K, expression for: H_2O + O_2 + NaOH + Na3IrCl6 ⟶ NaCl + Ir(OH)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: 2 H_2O + O_2 + 12 NaOH + 4 Na3IrCl6 ⟶ 24 NaCl + 4 Ir(OH)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 H_2O | 2 | -2 O_2 | 1 | -1 NaOH | 12 | -12 Na3IrCl6 | 4 | -4 NaCl | 24 | 24 Ir(OH)4 | 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) NaOH | 12 | -12 | ([NaOH])^(-12) Na3IrCl6 | 4 | -4 | ([Na3IrCl6])^(-4) NaCl | 24 | 24 | ([NaCl])^24 Ir(OH)4 | 4 | 4 | ([Ir(OH)4])^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) ([NaOH])^(-12) ([Na3IrCl6])^(-4) ([NaCl])^24 ([Ir(OH)4])^4 = (([NaCl])^24 ([Ir(OH)4])^4)/(([H2O])^2 [O2] ([NaOH])^12 ([Na3IrCl6])^4)

Rate of reaction

Construct the rate of reaction expression for: H_2O + O_2 + NaOH + Na3IrCl6 ⟶ NaCl + Ir(OH)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: 2 H_2O + O_2 + 12 NaOH + 4 Na3IrCl6 ⟶ 24 NaCl + 4 Ir(OH)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 H_2O | 2 | -2 O_2 | 1 | -1 NaOH | 12 | -12 Na3IrCl6 | 4 | -4 NaCl | 24 | 24 Ir(OH)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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) NaOH | 12 | -12 | -1/12 (Δ[NaOH])/(Δt) Na3IrCl6 | 4 | -4 | -1/4 (Δ[Na3IrCl6])/(Δt) NaCl | 24 | 24 | 1/24 (Δ[NaCl])/(Δt) Ir(OH)4 | 4 | 4 | 1/4 (Δ[Ir(OH)4])/(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) = -1/12 (Δ[NaOH])/(Δt) = -1/4 (Δ[Na3IrCl6])/(Δt) = 1/24 (Δ[NaCl])/(Δt) = 1/4 (Δ[Ir(OH)4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + O_2 + NaOH + Na3IrCl6 ⟶ NaCl + Ir(OH)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: 2 H_2O + O_2 + 12 NaOH + 4 Na3IrCl6 ⟶ 24 NaCl + 4 Ir(OH)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 H_2O | 2 | -2 O_2 | 1 | -1 NaOH | 12 | -12 Na3IrCl6 | 4 | -4 NaCl | 24 | 24 Ir(OH)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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) O_2 | 1 | -1 | -(Δ[O2])/(Δt) NaOH | 12 | -12 | -1/12 (Δ[NaOH])/(Δt) Na3IrCl6 | 4 | -4 | -1/4 (Δ[Na3IrCl6])/(Δt) NaCl | 24 | 24 | 1/24 (Δ[NaCl])/(Δt) Ir(OH)4 | 4 | 4 | 1/4 (Δ[Ir(OH)4])/(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) = -1/12 (Δ[NaOH])/(Δt) = -1/4 (Δ[Na3IrCl6])/(Δt) = 1/24 (Δ[NaCl])/(Δt) = 1/4 (Δ[Ir(OH)4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | oxygen | sodium hydroxide | Na3IrCl6 | sodium chloride | Ir(OH)4 formula | H_2O | O_2 | NaOH | Na3IrCl6 | NaCl | Ir(OH)4 Hill formula | H_2O | O_2 | HNaO | Cl6IrNa3 | ClNa | H4IrO4 name | water | oxygen | sodium hydroxide | | sodium chloride | IUPAC name | water | molecular oxygen | sodium hydroxide | | sodium chloride |
| water | oxygen | sodium hydroxide | Na3IrCl6 | sodium chloride | Ir(OH)4 formula | H_2O | O_2 | NaOH | Na3IrCl6 | NaCl | Ir(OH)4 Hill formula | H_2O | O_2 | HNaO | Cl6IrNa3 | ClNa | H4IrO4 name | water | oxygen | sodium hydroxide | | sodium chloride | IUPAC name | water | molecular oxygen | sodium hydroxide | | sodium chloride |

Substance properties

| water | oxygen | sodium hydroxide | Na3IrCl6 | sodium chloride | Ir(OH)4 molar mass | 18.015 g/mol | 31.998 g/mol | 39.997 g/mol | 473.9 g/mol | 58.44 g/mol | 260.24 g/mol phase | liquid (at STP) | gas (at STP) | solid (at STP) | | solid (at STP) | melting point | 0 °C | -218 °C | 323 °C | | 801 °C | boiling point | 99.9839 °C | -183 °C | 1390 °C | | 1413 °C | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 2.13 g/cm^3 | | 2.16 g/cm^3 | solubility in water | | | soluble | | soluble | surface tension | 0.0728 N/m | 0.01347 N/m | 0.07435 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 0.004 Pa s (at 350 °C) | | | odor | odorless | odorless | | | odorless |
| water | oxygen | sodium hydroxide | Na3IrCl6 | sodium chloride | Ir(OH)4 molar mass | 18.015 g/mol | 31.998 g/mol | 39.997 g/mol | 473.9 g/mol | 58.44 g/mol | 260.24 g/mol phase | liquid (at STP) | gas (at STP) | solid (at STP) | | solid (at STP) | melting point | 0 °C | -218 °C | 323 °C | | 801 °C | boiling point | 99.9839 °C | -183 °C | 1390 °C | | 1413 °C | density | 1 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 2.13 g/cm^3 | | 2.16 g/cm^3 | solubility in water | | | soluble | | soluble | surface tension | 0.0728 N/m | 0.01347 N/m | 0.07435 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 2.055×10^-5 Pa s (at 25 °C) | 0.004 Pa s (at 350 °C) | | | odor | odorless | odorless | | | odorless |

Units

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