H2S + LiOH = H2O + Li2S

H_2S hydrogen sulfide + LiOH lithium hydroxide ⟶ H_2O water + Li_2S lithium sulfide

Balance the chemical equation algebraically: H_2S + LiOH ⟶ H_2O + Li_2S Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2S + c_2 LiOH ⟶ c_3 H_2O + c_4 Li_2S Set the number of atoms in the reactants equal to the number of atoms in the products for H, S, Li and O: H: | 2 c_1 + c_2 = 2 c_3 S: | c_1 = c_4 Li: | c_2 = 2 c_4 O: | 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 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2S + 2 LiOH ⟶ 2 H_2O + Li_2S

+ ⟶ +

hydrogen sulfide + lithium hydroxide ⟶ water + lithium sulfide

Construct the equilibrium constant, K, expression for: H_2S + LiOH ⟶ H_2O + Li_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 LiOH ⟶ 2 H_2O + Li_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 LiOH | 2 | -2 H_2O | 2 | 2 Li_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) LiOH | 2 | -2 | ([LiOH])^(-2) H_2O | 2 | 2 | ([H2O])^2 Li_2S | 1 | 1 | [Li2S] 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) ([LiOH])^(-2) ([H2O])^2 [Li2S] = (([H2O])^2 [Li2S])/([H2S] ([LiOH])^2)

Construct the rate of reaction expression for: H_2S + LiOH ⟶ H_2O + Li_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 LiOH ⟶ 2 H_2O + Li_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 LiOH | 2 | -2 H_2O | 2 | 2 Li_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) LiOH | 2 | -2 | -1/2 (Δ[LiOH])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Li_2S | 1 | 1 | (Δ[Li2S])/(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[LiOH])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Li2S])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen sulfide | lithium hydroxide | water | lithium sulfide formula | H_2S | LiOH | H_2O | Li_2S Hill formula | H_2S | HLiO | H_2O | Li_2S name | hydrogen sulfide | lithium hydroxide | water | lithium sulfide IUPAC name | hydrogen sulfide | lithium hydroxide | water | dilithium sulfide