H2O + Li = H2 + LiOH

H_2O (water) + Li (lithium) ⟶ H_2 (hydrogen) + LiOH (lithium hydroxide)

Balance the chemical equation algebraically: H_2O + Li ⟶ H_2 + LiOH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Li ⟶ c_3 H_2 + c_4 LiOH Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Li: H: | 2 c_1 = 2 c_3 + c_4 O: | c_1 = c_4 Li: | c_2 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + 2 Li ⟶ H_2 + 2 LiOH

+ ⟶ +

water + lithium ⟶ hydrogen + lithium hydroxide

| water | lithium | hydrogen | lithium hydroxide molecular enthalpy | -285.8 kJ/mol | 0 kJ/mol | 0 kJ/mol | -487.5 kJ/mol total enthalpy | -571.7 kJ/mol | 0 kJ/mol | 0 kJ/mol | -975 kJ/mol | H_initial = -571.7 kJ/mol | | H_final = -975 kJ/mol | ΔH_rxn^0 | -975 kJ/mol - -571.7 kJ/mol = -403.3 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: H_2O + Li ⟶ H_2 + LiOH 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_2O + 2 Li ⟶ H_2 + 2 LiOH 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_2O | 2 | -2 Li | 2 | -2 H_2 | 1 | 1 LiOH | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) Li | 2 | -2 | ([Li])^(-2) H_2 | 1 | 1 | [H2] LiOH | 2 | 2 | ([LiOH])^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 = ([H2O])^(-2) ([Li])^(-2) [H2] ([LiOH])^2 = ([H2] ([LiOH])^2)/(([H2O])^2 ([Li])^2)

Construct the rate of reaction expression for: H_2O + Li ⟶ H_2 + LiOH 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_2O + 2 Li ⟶ H_2 + 2 LiOH 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_2O | 2 | -2 Li | 2 | -2 H_2 | 1 | 1 LiOH | 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_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) Li | 2 | -2 | -1/2 (Δ[Li])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) LiOH | 2 | 2 | 1/2 (Δ[LiOH])/(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[H2O])/(Δt) = -1/2 (Δ[Li])/(Δt) = (Δ[H2])/(Δt) = 1/2 (Δ[LiOH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | lithium | hydrogen | lithium hydroxide formula | H_2O | Li | H_2 | LiOH Hill formula | H_2O | Li | H_2 | HLiO name | water | lithium | hydrogen | lithium hydroxide IUPAC name | water | lithium | molecular hydrogen | lithium hydroxide