H2O + K = H2 + KOH

H_2O (water) + K (potassium) ⟶ H_2 (hydrogen) + KOH (potassium hydroxide)

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

+ ⟶ +

water + potassium ⟶ hydrogen + potassium hydroxide

| water | potassium | hydrogen | potassium hydroxide molecular entropy | 69.91 J/(mol K) | 64 J/(mol K) | 115 J/(mol K) | 79 J/(mol K) total entropy | 139.8 J/(mol K) | 128 J/(mol K) | 115 J/(mol K) | 158 J/(mol K) | S_initial = 267.8 J/(mol K) | | S_final = 273 J/(mol K) | ΔS_rxn^0 | 273 J/(mol K) - 267.8 J/(mol K) = 5.18 J/(mol K) (endoentropic) | | |

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

Construct the rate of reaction expression for: H_2O + K ⟶ H_2 + KOH 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 K ⟶ H_2 + 2 KOH 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 K | 2 | -2 H_2 | 1 | 1 KOH | 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) K | 2 | -2 | -1/2 (Δ[K])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) KOH | 2 | 2 | 1/2 (Δ[KOH])/(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[K])/(Δt) = (Δ[H2])/(Δt) = 1/2 (Δ[KOH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | potassium | hydrogen | potassium hydroxide formula | H_2O | K | H_2 | KOH Hill formula | H_2O | K | H_2 | HKO name | water | potassium | hydrogen | potassium hydroxide IUPAC name | water | potassium | molecular hydrogen | potassium hydroxide

| water | potassium | hydrogen | potassium hydroxide molar mass | 18.015 g/mol | 39.0983 g/mol | 2.016 g/mol | 56.105 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 0 °C | 64 °C | -259.2 °C | 406 °C boiling point | 99.9839 °C | 760 °C | -252.8 °C | 1327 °C density | 1 g/cm^3 | 0.86 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 2.044 g/cm^3 solubility in water | | reacts | | soluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | 0.001 Pa s (at 550 °C) odor | odorless | | odorless |