H2S + NH3 = (NH4)2S

H_2S hydrogen sulfide + NH_3 ammonia ⟶ (NH_4)_2S diammonium sulfide

Balance the chemical equation algebraically: H_2S + NH_3 ⟶ (NH_4)_2S Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2S + c_2 NH_3 ⟶ c_3 (NH_4)_2S Set the number of atoms in the reactants equal to the number of atoms in the products for H, S and N: H: | 2 c_1 + 3 c_2 = 8 c_3 S: | c_1 = c_3 N: | c_2 = 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 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2S + 2 NH_3 ⟶ (NH_4)_2S

+ ⟶

hydrogen sulfide + ammonia ⟶ diammonium sulfide

Construct the equilibrium constant, K, expression for: H_2S + NH_3 ⟶ (NH_4)_2S 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_2S + 2 NH_3 ⟶ (NH_4)_2S 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_2S | 1 | -1 NH_3 | 2 | -2 (NH_4)_2S | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2S | 1 | -1 | ([H2S])^(-1) NH_3 | 2 | -2 | ([NH3])^(-2) (NH_4)_2S | 1 | 1 | [(NH4)2S] 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 = ([H2S])^(-1) ([NH3])^(-2) [(NH4)2S] = ([(NH4)2S])/([H2S] ([NH3])^2)

Construct the rate of reaction expression for: H_2S + NH_3 ⟶ (NH_4)_2S 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_2S + 2 NH_3 ⟶ (NH_4)_2S 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_2S | 1 | -1 NH_3 | 2 | -2 (NH_4)_2S | 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_2S | 1 | -1 | -(Δ[H2S])/(Δt) NH_3 | 2 | -2 | -1/2 (Δ[NH3])/(Δt) (NH_4)_2S | 1 | 1 | (Δ[(NH4)2S])/(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[H2S])/(Δt) = -1/2 (Δ[NH3])/(Δt) = (Δ[(NH4)2S])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen sulfide | ammonia | diammonium sulfide formula | H_2S | NH_3 | (NH_4)_2S Hill formula | H_2S | H_3N | H_8N_2S name | hydrogen sulfide | ammonia | diammonium sulfide

| hydrogen sulfide | ammonia | diammonium sulfide molar mass | 34.08 g/mol | 17.031 g/mol | 68.14 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) melting point | -85 °C | -77.73 °C | -18 °C boiling point | -60 °C | -33.33 °C | density | 0.001393 g/cm^3 (at 25 °C) | 6.96×10^-4 g/cm^3 (at 25 °C) | 0.997 g/cm^3 solubility in water | | | very soluble surface tension | | 0.0234 N/m | dynamic viscosity | 1.239×10^-5 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) |