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KOH + P + KIO3 = H2O + KI + K3PO4

Input interpretation

KOH potassium hydroxide + P red phosphorus + KIO_3 potassium iodate ⟶ H_2O water + KI potassium iodide + K3PO4
KOH potassium hydroxide + P red phosphorus + KIO_3 potassium iodate ⟶ H_2O water + KI potassium iodide + K3PO4

Balanced equation

Balance the chemical equation algebraically: KOH + P + KIO_3 ⟶ H_2O + KI + K3PO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 P + c_3 KIO_3 ⟶ c_4 H_2O + c_5 KI + c_6 K3PO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, P and I: H: | c_1 = 2 c_4 K: | c_1 + c_3 = c_5 + 3 c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_6 P: | c_2 = c_6 I: | 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 18/5 c_2 = 6/5 c_3 = 1 c_4 = 9/5 c_5 = 1 c_6 = 6/5 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 18 c_2 = 6 c_3 = 5 c_4 = 9 c_5 = 5 c_6 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 18 KOH + 6 P + 5 KIO_3 ⟶ 9 H_2O + 5 KI + 6 K3PO4
Balance the chemical equation algebraically: KOH + P + KIO_3 ⟶ H_2O + KI + K3PO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 P + c_3 KIO_3 ⟶ c_4 H_2O + c_5 KI + c_6 K3PO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, P and I: H: | c_1 = 2 c_4 K: | c_1 + c_3 = c_5 + 3 c_6 O: | c_1 + 3 c_3 = c_4 + 4 c_6 P: | c_2 = c_6 I: | 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 18/5 c_2 = 6/5 c_3 = 1 c_4 = 9/5 c_5 = 1 c_6 = 6/5 Multiply by the least common denominator, 5, to eliminate fractional coefficients: c_1 = 18 c_2 = 6 c_3 = 5 c_4 = 9 c_5 = 5 c_6 = 6 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 18 KOH + 6 P + 5 KIO_3 ⟶ 9 H_2O + 5 KI + 6 K3PO4

Structures

 + + ⟶ + + K3PO4
+ + ⟶ + + K3PO4

Names

potassium hydroxide + red phosphorus + potassium iodate ⟶ water + potassium iodide + K3PO4
potassium hydroxide + red phosphorus + potassium iodate ⟶ water + potassium iodide + K3PO4

Equilibrium constant

Construct the equilibrium constant, K, expression for: KOH + P + KIO_3 ⟶ H_2O + KI + K3PO4 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: 18 KOH + 6 P + 5 KIO_3 ⟶ 9 H_2O + 5 KI + 6 K3PO4 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 | 18 | -18 P | 6 | -6 KIO_3 | 5 | -5 H_2O | 9 | 9 KI | 5 | 5 K3PO4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 18 | -18 | ([KOH])^(-18) P | 6 | -6 | ([P])^(-6) KIO_3 | 5 | -5 | ([KIO3])^(-5) H_2O | 9 | 9 | ([H2O])^9 KI | 5 | 5 | ([KI])^5 K3PO4 | 6 | 6 | ([K3PO4])^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])^(-18) ([P])^(-6) ([KIO3])^(-5) ([H2O])^9 ([KI])^5 ([K3PO4])^6 = (([H2O])^9 ([KI])^5 ([K3PO4])^6)/(([KOH])^18 ([P])^6 ([KIO3])^5)
Construct the equilibrium constant, K, expression for: KOH + P + KIO_3 ⟶ H_2O + KI + K3PO4 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: 18 KOH + 6 P + 5 KIO_3 ⟶ 9 H_2O + 5 KI + 6 K3PO4 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 | 18 | -18 P | 6 | -6 KIO_3 | 5 | -5 H_2O | 9 | 9 KI | 5 | 5 K3PO4 | 6 | 6 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 18 | -18 | ([KOH])^(-18) P | 6 | -6 | ([P])^(-6) KIO_3 | 5 | -5 | ([KIO3])^(-5) H_2O | 9 | 9 | ([H2O])^9 KI | 5 | 5 | ([KI])^5 K3PO4 | 6 | 6 | ([K3PO4])^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])^(-18) ([P])^(-6) ([KIO3])^(-5) ([H2O])^9 ([KI])^5 ([K3PO4])^6 = (([H2O])^9 ([KI])^5 ([K3PO4])^6)/(([KOH])^18 ([P])^6 ([KIO3])^5)

Rate of reaction

Construct the rate of reaction expression for: KOH + P + KIO_3 ⟶ H_2O + KI + K3PO4 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: 18 KOH + 6 P + 5 KIO_3 ⟶ 9 H_2O + 5 KI + 6 K3PO4 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 | 18 | -18 P | 6 | -6 KIO_3 | 5 | -5 H_2O | 9 | 9 KI | 5 | 5 K3PO4 | 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 | 18 | -18 | -1/18 (Δ[KOH])/(Δt) P | 6 | -6 | -1/6 (Δ[P])/(Δt) KIO_3 | 5 | -5 | -1/5 (Δ[KIO3])/(Δt) H_2O | 9 | 9 | 1/9 (Δ[H2O])/(Δt) KI | 5 | 5 | 1/5 (Δ[KI])/(Δt) K3PO4 | 6 | 6 | 1/6 (Δ[K3PO4])/(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 = -1/18 (Δ[KOH])/(Δt) = -1/6 (Δ[P])/(Δt) = -1/5 (Δ[KIO3])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/5 (Δ[KI])/(Δt) = 1/6 (Δ[K3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: KOH + P + KIO_3 ⟶ H_2O + KI + K3PO4 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: 18 KOH + 6 P + 5 KIO_3 ⟶ 9 H_2O + 5 KI + 6 K3PO4 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 | 18 | -18 P | 6 | -6 KIO_3 | 5 | -5 H_2O | 9 | 9 KI | 5 | 5 K3PO4 | 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 | 18 | -18 | -1/18 (Δ[KOH])/(Δt) P | 6 | -6 | -1/6 (Δ[P])/(Δt) KIO_3 | 5 | -5 | -1/5 (Δ[KIO3])/(Δt) H_2O | 9 | 9 | 1/9 (Δ[H2O])/(Δt) KI | 5 | 5 | 1/5 (Δ[KI])/(Δt) K3PO4 | 6 | 6 | 1/6 (Δ[K3PO4])/(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 = -1/18 (Δ[KOH])/(Δt) = -1/6 (Δ[P])/(Δt) = -1/5 (Δ[KIO3])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/5 (Δ[KI])/(Δt) = 1/6 (Δ[K3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | potassium hydroxide | red phosphorus | potassium iodate | water | potassium iodide | K3PO4 formula | KOH | P | KIO_3 | H_2O | KI | K3PO4 Hill formula | HKO | P | IKO_3 | H_2O | IK | K3O4P name | potassium hydroxide | red phosphorus | potassium iodate | water | potassium iodide |  IUPAC name | potassium hydroxide | phosphorus | potassium iodate | water | potassium iodide |
| potassium hydroxide | red phosphorus | potassium iodate | water | potassium iodide | K3PO4 formula | KOH | P | KIO_3 | H_2O | KI | K3PO4 Hill formula | HKO | P | IKO_3 | H_2O | IK | K3O4P name | potassium hydroxide | red phosphorus | potassium iodate | water | potassium iodide | IUPAC name | potassium hydroxide | phosphorus | potassium iodate | water | potassium iodide |

Substance properties

 | potassium hydroxide | red phosphorus | potassium iodate | water | potassium iodide | K3PO4 molar mass | 56.105 g/mol | 30.973761998 g/mol | 214 g/mol | 18.015 g/mol | 166.0028 g/mol | 212.26 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) |  melting point | 406 °C | 579.2 °C | 560 °C | 0 °C | 681 °C |  boiling point | 1327 °C | | | 99.9839 °C | 1330 °C |  density | 2.044 g/cm^3 | 2.16 g/cm^3 | 1.005 g/cm^3 | 1 g/cm^3 | 3.123 g/cm^3 |  solubility in water | soluble | insoluble | | | |  surface tension | | | | 0.0728 N/m | |  dynamic viscosity | 0.001 Pa s (at 550 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) |  odor | | | | odorless | |
| potassium hydroxide | red phosphorus | potassium iodate | water | potassium iodide | K3PO4 molar mass | 56.105 g/mol | 30.973761998 g/mol | 214 g/mol | 18.015 g/mol | 166.0028 g/mol | 212.26 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | 406 °C | 579.2 °C | 560 °C | 0 °C | 681 °C | boiling point | 1327 °C | | | 99.9839 °C | 1330 °C | density | 2.044 g/cm^3 | 2.16 g/cm^3 | 1.005 g/cm^3 | 1 g/cm^3 | 3.123 g/cm^3 | solubility in water | soluble | insoluble | | | | surface tension | | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 7.6×10^-4 Pa s (at 20.2 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) | odor | | | | odorless | |

Units