LiClO3 = O2 + LiCl

LiClO3 ⟶ O_2 oxygen + LiCl lithium chloride

Balance the chemical equation algebraically: LiClO3 ⟶ O_2 + LiCl Add stoichiometric coefficients, c_i, to the reactants and products: c_1 LiClO3 ⟶ c_2 O_2 + c_3 LiCl Set the number of atoms in the reactants equal to the number of atoms in the products for Li, Cl and O: Li: | c_1 = c_3 Cl: | c_1 = c_3 O: | 3 c_1 = 2 c_2 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 = 3/2 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 LiClO3 ⟶ 3 O_2 + 2 LiCl

LiClO3 ⟶ +

LiClO3 ⟶ oxygen + lithium chloride

Construct the equilibrium constant, K, expression for: LiClO3 ⟶ O_2 + LiCl 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 LiClO3 ⟶ 3 O_2 + 2 LiCl 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 LiClO3 | 2 | -2 O_2 | 3 | 3 LiCl | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression LiClO3 | 2 | -2 | ([LiClO3])^(-2) O_2 | 3 | 3 | ([O2])^3 LiCl | 2 | 2 | ([LiCl])^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 = ([LiClO3])^(-2) ([O2])^3 ([LiCl])^2 = (([O2])^3 ([LiCl])^2)/([LiClO3])^2

Construct the rate of reaction expression for: LiClO3 ⟶ O_2 + LiCl 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 LiClO3 ⟶ 3 O_2 + 2 LiCl 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 LiClO3 | 2 | -2 O_2 | 3 | 3 LiCl | 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 LiClO3 | 2 | -2 | -1/2 (Δ[LiClO3])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) LiCl | 2 | 2 | 1/2 (Δ[LiCl])/(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[LiClO3])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/2 (Δ[LiCl])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| LiClO3 | oxygen | lithium chloride formula | LiClO3 | O_2 | LiCl Hill formula | ClLiO3 | O_2 | ClLi name | | oxygen | lithium chloride IUPAC name | | molecular oxygen | lithium chloride

| LiClO3 | oxygen | lithium chloride molar mass | 90.4 g/mol | 31.998 g/mol | 42.4 g/mol phase | | gas (at STP) | solid (at STP) melting point | | -218 °C | 605 °C boiling point | | -183 °C | 1382 °C density | | 0.001429 g/cm^3 (at 0 °C) | 2.07 g/cm^3 surface tension | | 0.01347 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 0.00525 Pa s (at 20 °C) odor | | odorless |