LiNO3 = O2 + NO2 + Li2O

LiNO_3 lithium nitrate ⟶ O_2 oxygen + NO_2 nitrogen dioxide + Li_2O lithium oxide

Balance the chemical equation algebraically: LiNO_3 ⟶ O_2 + NO_2 + Li_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 LiNO_3 ⟶ c_2 O_2 + c_3 NO_2 + c_4 Li_2O Set the number of atoms in the reactants equal to the number of atoms in the products for Li, N and O: Li: | c_1 = 2 c_4 N: | c_1 = c_3 O: | 3 c_1 = 2 c_2 + 2 c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 4 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 LiNO_3 ⟶ O_2 + 4 NO_2 + 2 Li_2O

⟶ + +

lithium nitrate ⟶ oxygen + nitrogen dioxide + lithium oxide

Construct the equilibrium constant, K, expression for: LiNO_3 ⟶ O_2 + NO_2 + Li_2O 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: 4 LiNO_3 ⟶ O_2 + 4 NO_2 + 2 Li_2O 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 LiNO_3 | 4 | -4 O_2 | 1 | 1 NO_2 | 4 | 4 Li_2O | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression LiNO_3 | 4 | -4 | ([LiNO3])^(-4) O_2 | 1 | 1 | [O2] NO_2 | 4 | 4 | ([NO2])^4 Li_2O | 2 | 2 | ([Li2O])^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 = ([LiNO3])^(-4) [O2] ([NO2])^4 ([Li2O])^2 = ([O2] ([NO2])^4 ([Li2O])^2)/([LiNO3])^4

Construct the rate of reaction expression for: LiNO_3 ⟶ O_2 + NO_2 + Li_2O 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: 4 LiNO_3 ⟶ O_2 + 4 NO_2 + 2 Li_2O 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 LiNO_3 | 4 | -4 O_2 | 1 | 1 NO_2 | 4 | 4 Li_2O | 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 LiNO_3 | 4 | -4 | -1/4 (Δ[LiNO3])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) NO_2 | 4 | 4 | 1/4 (Δ[NO2])/(Δt) Li_2O | 2 | 2 | 1/2 (Δ[Li2O])/(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 = -1/4 (Δ[LiNO3])/(Δt) = (Δ[O2])/(Δt) = 1/4 (Δ[NO2])/(Δt) = 1/2 (Δ[Li2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| lithium nitrate | oxygen | nitrogen dioxide | lithium oxide formula | LiNO_3 | O_2 | NO_2 | Li_2O name | lithium nitrate | oxygen | nitrogen dioxide | lithium oxide IUPAC name | lithium nitrate | molecular oxygen | Nitrogen dioxide | dilithium oxygen(-2) anion

| lithium nitrate | oxygen | nitrogen dioxide | lithium oxide molar mass | 68.94 g/mol | 31.998 g/mol | 46.005 g/mol | 29.9 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | melting point | 264 °C | -218 °C | -11 °C | boiling point | | -183 °C | 21 °C | density | | 0.001429 g/cm^3 (at 0 °C) | 0.00188 g/cm^3 (at 25 °C) | 2.013 g/cm^3 solubility in water | | | reacts | surface tension | | 0.01347 N/m | | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | odor | | odorless | |