H2 + K = KH4

H_2 hydrogen + K potassium ⟶ KH4

Balance the chemical equation algebraically: H_2 + K ⟶ KH4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 K ⟶ c_3 KH4 Set the number of atoms in the reactants equal to the number of atoms in the products for H and K: H: | 2 c_1 = 4 c_3 K: | c_2 = c_3 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 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2 + K ⟶ KH4

+ ⟶ KH4

hydrogen + potassium ⟶ KH4

Construct the equilibrium constant, K, expression for: H_2 + K ⟶ KH4 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_2 + K ⟶ KH4 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_2 | 2 | -2 K | 1 | -1 KH4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 2 | -2 | ([H2])^(-2) K | 1 | -1 | ([K])^(-1) KH4 | 1 | 1 | [KH4] 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 = ([H2])^(-2) ([K])^(-1) [KH4] = ([KH4])/(([H2])^2 [K])

Construct the rate of reaction expression for: H_2 + K ⟶ KH4 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_2 + K ⟶ KH4 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_2 | 2 | -2 K | 1 | -1 KH4 | 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_2 | 2 | -2 | -1/2 (Δ[H2])/(Δt) K | 1 | -1 | -(Δ[K])/(Δt) KH4 | 1 | 1 | (Δ[KH4])/(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[H2])/(Δt) = -(Δ[K])/(Δt) = (Δ[KH4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | potassium | KH4 formula | H_2 | K | KH4 Hill formula | H_2 | K | H4K name | hydrogen | potassium | IUPAC name | molecular hydrogen | potassium |

| hydrogen | potassium | KH4 molar mass | 2.016 g/mol | 39.0983 g/mol | 43.13 g/mol phase | gas (at STP) | solid (at STP) | melting point | -259.2 °C | 64 °C | boiling point | -252.8 °C | 760 °C | density | 8.99×10^-5 g/cm^3 (at 0 °C) | 0.86 g/cm^3 | solubility in water | | reacts | dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | | odor | odorless | |