H2 + Ba = BaH

H_2 hydrogen + Ba barium ⟶ BaH

Balance the chemical equation algebraically: H_2 + Ba ⟶ BaH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 Ba ⟶ c_3 BaH Set the number of atoms in the reactants equal to the number of atoms in the products for H and Ba: H: | 2 c_1 = c_3 Ba: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2 + 2 Ba ⟶ 2 BaH

+ ⟶ BaH

hydrogen + barium ⟶ BaH

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

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

| hydrogen | barium | BaH formula | H_2 | Ba | BaH Hill formula | H_2 | Ba | HBa name | hydrogen | barium | IUPAC name | molecular hydrogen | barium |

| hydrogen | barium | BaH molar mass | 2.016 g/mol | 137.327 g/mol | 138.335 g/mol phase | gas (at STP) | solid (at STP) | melting point | -259.2 °C | 725 °C | boiling point | -252.8 °C | 1640 °C | density | 8.99×10^-5 g/cm^3 (at 0 °C) | 3.6 g/cm^3 | solubility in water | | insoluble | surface tension | | 0.224 N/m | dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | | odor | odorless | |