Search

H2O + KI = H2 + KOH + I2

Input interpretation

H_2O water + KI potassium iodide ⟶ H_2 hydrogen + KOH potassium hydroxide + I_2 iodine
H_2O water + KI potassium iodide ⟶ H_2 hydrogen + KOH potassium hydroxide + I_2 iodine

Balanced equation

Balance the chemical equation algebraically: H_2O + KI ⟶ H_2 + KOH + I_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 KI ⟶ c_3 H_2 + c_4 KOH + c_5 I_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, I and K: H: | 2 c_1 = 2 c_3 + c_4 O: | c_1 = c_4 I: | c_2 = 2 c_5 K: | c_2 = c_4 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 = 2 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 H_2O + 2 KI ⟶ H_2 + 2 KOH + I_2
Balance the chemical equation algebraically: H_2O + KI ⟶ H_2 + KOH + I_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 KI ⟶ c_3 H_2 + c_4 KOH + c_5 I_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, I and K: H: | 2 c_1 = 2 c_3 + c_4 O: | c_1 = c_4 I: | c_2 = 2 c_5 K: | c_2 = c_4 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 = 2 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 2 KI ⟶ H_2 + 2 KOH + I_2

Structures

 + ⟶ + +
+ ⟶ + +

Names

water + potassium iodide ⟶ hydrogen + potassium hydroxide + iodine
water + potassium iodide ⟶ hydrogen + potassium hydroxide + iodine

Reaction thermodynamics

Enthalpy

 | water | potassium iodide | hydrogen | potassium hydroxide | iodine molecular enthalpy | -285.8 kJ/mol | -327.9 kJ/mol | 0 kJ/mol | -424.6 kJ/mol | 0 kJ/mol total enthalpy | -571.7 kJ/mol | -655.8 kJ/mol | 0 kJ/mol | -849.2 kJ/mol | 0 kJ/mol  | H_initial = -1227 kJ/mol | | H_final = -849.2 kJ/mol | |  ΔH_rxn^0 | -849.2 kJ/mol - -1227 kJ/mol = 378.3 kJ/mol (endothermic) | | | |
| water | potassium iodide | hydrogen | potassium hydroxide | iodine molecular enthalpy | -285.8 kJ/mol | -327.9 kJ/mol | 0 kJ/mol | -424.6 kJ/mol | 0 kJ/mol total enthalpy | -571.7 kJ/mol | -655.8 kJ/mol | 0 kJ/mol | -849.2 kJ/mol | 0 kJ/mol | H_initial = -1227 kJ/mol | | H_final = -849.2 kJ/mol | | ΔH_rxn^0 | -849.2 kJ/mol - -1227 kJ/mol = 378.3 kJ/mol (endothermic) | | | |

Gibbs free energy

 | water | potassium iodide | hydrogen | potassium hydroxide | iodine molecular free energy | -237.1 kJ/mol | -324.9 kJ/mol | 0 kJ/mol | -379.4 kJ/mol | 0 kJ/mol total free energy | -474.2 kJ/mol | -649.8 kJ/mol | 0 kJ/mol | -758.8 kJ/mol | 0 kJ/mol  | G_initial = -1124 kJ/mol | | G_final = -758.8 kJ/mol | |  ΔG_rxn^0 | -758.8 kJ/mol - -1124 kJ/mol = 365.2 kJ/mol (endergonic) | | | |
| water | potassium iodide | hydrogen | potassium hydroxide | iodine molecular free energy | -237.1 kJ/mol | -324.9 kJ/mol | 0 kJ/mol | -379.4 kJ/mol | 0 kJ/mol total free energy | -474.2 kJ/mol | -649.8 kJ/mol | 0 kJ/mol | -758.8 kJ/mol | 0 kJ/mol | G_initial = -1124 kJ/mol | | G_final = -758.8 kJ/mol | | ΔG_rxn^0 | -758.8 kJ/mol - -1124 kJ/mol = 365.2 kJ/mol (endergonic) | | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + KI ⟶ H_2 + KOH + I_2 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 + 2 KI ⟶ H_2 + 2 KOH + I_2 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 KI | 2 | -2 H_2 | 1 | 1 KOH | 2 | 2 I_2 | 1 | 1 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) KI | 2 | -2 | ([KI])^(-2) H_2 | 1 | 1 | [H2] KOH | 2 | 2 | ([KOH])^2 I_2 | 1 | 1 | [I2] 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) ([KI])^(-2) [H2] ([KOH])^2 [I2] = ([H2] ([KOH])^2 [I2])/(([H2O])^2 ([KI])^2)
Construct the equilibrium constant, K, expression for: H_2O + KI ⟶ H_2 + KOH + I_2 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 + 2 KI ⟶ H_2 + 2 KOH + I_2 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 KI | 2 | -2 H_2 | 1 | 1 KOH | 2 | 2 I_2 | 1 | 1 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) KI | 2 | -2 | ([KI])^(-2) H_2 | 1 | 1 | [H2] KOH | 2 | 2 | ([KOH])^2 I_2 | 1 | 1 | [I2] 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) ([KI])^(-2) [H2] ([KOH])^2 [I2] = ([H2] ([KOH])^2 [I2])/(([H2O])^2 ([KI])^2)

Rate of reaction

Construct the rate of reaction expression for: H_2O + KI ⟶ H_2 + KOH + I_2 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 + 2 KI ⟶ H_2 + 2 KOH + I_2 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 KI | 2 | -2 H_2 | 1 | 1 KOH | 2 | 2 I_2 | 1 | 1 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) KI | 2 | -2 | -1/2 (Δ[KI])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) KOH | 2 | 2 | 1/2 (Δ[KOH])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(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) = -1/2 (Δ[KI])/(Δt) = (Δ[H2])/(Δt) = 1/2 (Δ[KOH])/(Δt) = (Δ[I2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + KI ⟶ H_2 + KOH + I_2 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 + 2 KI ⟶ H_2 + 2 KOH + I_2 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 KI | 2 | -2 H_2 | 1 | 1 KOH | 2 | 2 I_2 | 1 | 1 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) KI | 2 | -2 | -1/2 (Δ[KI])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) KOH | 2 | 2 | 1/2 (Δ[KOH])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(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) = -1/2 (Δ[KI])/(Δt) = (Δ[H2])/(Δt) = 1/2 (Δ[KOH])/(Δt) = (Δ[I2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | water | potassium iodide | hydrogen | potassium hydroxide | iodine formula | H_2O | KI | H_2 | KOH | I_2 Hill formula | H_2O | IK | H_2 | HKO | I_2 name | water | potassium iodide | hydrogen | potassium hydroxide | iodine IUPAC name | water | potassium iodide | molecular hydrogen | potassium hydroxide | molecular iodine
| water | potassium iodide | hydrogen | potassium hydroxide | iodine formula | H_2O | KI | H_2 | KOH | I_2 Hill formula | H_2O | IK | H_2 | HKO | I_2 name | water | potassium iodide | hydrogen | potassium hydroxide | iodine IUPAC name | water | potassium iodide | molecular hydrogen | potassium hydroxide | molecular iodine

Substance properties

 | water | potassium iodide | hydrogen | potassium hydroxide | iodine molar mass | 18.015 g/mol | 166.0028 g/mol | 2.016 g/mol | 56.105 g/mol | 253.80894 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) | solid (at STP) melting point | 0 °C | 681 °C | -259.2 °C | 406 °C | 113 °C boiling point | 99.9839 °C | 1330 °C | -252.8 °C | 1327 °C | 184 °C density | 1 g/cm^3 | 3.123 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 2.044 g/cm^3 | 4.94 g/cm^3 solubility in water | | | | soluble |  surface tension | 0.0728 N/m | | | |  dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) | 8.9×10^-6 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) | 0.00227 Pa s (at 116 °C) odor | odorless | | odorless | |
| water | potassium iodide | hydrogen | potassium hydroxide | iodine molar mass | 18.015 g/mol | 166.0028 g/mol | 2.016 g/mol | 56.105 g/mol | 253.80894 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) | solid (at STP) melting point | 0 °C | 681 °C | -259.2 °C | 406 °C | 113 °C boiling point | 99.9839 °C | 1330 °C | -252.8 °C | 1327 °C | 184 °C density | 1 g/cm^3 | 3.123 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 2.044 g/cm^3 | 4.94 g/cm^3 solubility in water | | | | soluble | surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 0.0010227 Pa s (at 732.9 °C) | 8.9×10^-6 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) | 0.00227 Pa s (at 116 °C) odor | odorless | | odorless | |

Units