Na2S + BaNO3 = BaS + Na2NO3

Na_2S sodium sulfide + BaNO3 ⟶ BaS barium sulfide + Na2NO3

Balance the chemical equation algebraically: Na_2S + BaNO3 ⟶ BaS + Na2NO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Na_2S + c_2 BaNO3 ⟶ c_3 BaS + c_4 Na2NO3 Set the number of atoms in the reactants equal to the number of atoms in the products for Na, S, Ba, N and O: Na: | 2 c_1 = 2 c_4 S: | c_1 = c_3 Ba: | c_2 = c_3 N: | c_2 = c_4 O: | 3 c_2 = 3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Na_2S + BaNO3 ⟶ BaS + Na2NO3

+ BaNO3 ⟶ + Na2NO3

sodium sulfide + BaNO3 ⟶ barium sulfide + Na2NO3

Construct the equilibrium constant, K, expression for: Na_2S + BaNO3 ⟶ BaS + Na2NO3 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: Na_2S + BaNO3 ⟶ BaS + Na2NO3 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 Na_2S | 1 | -1 BaNO3 | 1 | -1 BaS | 1 | 1 Na2NO3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Na_2S | 1 | -1 | ([Na2S])^(-1) BaNO3 | 1 | -1 | ([BaNO3])^(-1) BaS | 1 | 1 | [BaS] Na2NO3 | 1 | 1 | [Na2NO3] 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 = ([Na2S])^(-1) ([BaNO3])^(-1) [BaS] [Na2NO3] = ([BaS] [Na2NO3])/([Na2S] [BaNO3])

Construct the rate of reaction expression for: Na_2S + BaNO3 ⟶ BaS + Na2NO3 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: Na_2S + BaNO3 ⟶ BaS + Na2NO3 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 Na_2S | 1 | -1 BaNO3 | 1 | -1 BaS | 1 | 1 Na2NO3 | 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 Na_2S | 1 | -1 | -(Δ[Na2S])/(Δt) BaNO3 | 1 | -1 | -(Δ[BaNO3])/(Δt) BaS | 1 | 1 | (Δ[BaS])/(Δt) Na2NO3 | 1 | 1 | (Δ[Na2NO3])/(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[Na2S])/(Δt) = -(Δ[BaNO3])/(Δt) = (Δ[BaS])/(Δt) = (Δ[Na2NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium sulfide | BaNO3 | barium sulfide | Na2NO3 formula | Na_2S | BaNO3 | BaS | Na2NO3 Hill formula | Na_2S_1 | BaNO3 | BaS | NNa2O3 name | sodium sulfide | | barium sulfide | IUPAC name | | | thioxobarium |

| sodium sulfide | BaNO3 | barium sulfide | Na2NO3 molar mass | 78.04 g/mol | 199.33 g/mol | 169.39 g/mol | 107.98 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 1172 °C | | 1999.85 °C | density | 1.856 g/cm^3 | | 4.25 g/cm^3 |