HNO3 + Li = H2O + N2O + LiNO3

HNO_3 nitric acid + Li lithium ⟶ H_2O water + N_2O nitrous oxide + LiNO_3 lithium nitrate

Balance the chemical equation algebraically: HNO_3 + Li ⟶ H_2O + N_2O + LiNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Li ⟶ c_3 H_2O + c_4 N_2O + c_5 LiNO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Li: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + c_5 O: | 3 c_1 = c_3 + c_4 + 3 c_5 Li: | c_2 = c_5 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 8 c_3 = 5 c_4 = 1 c_5 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 HNO_3 + 8 Li ⟶ 5 H_2O + N_2O + 8 LiNO_3

+ ⟶ + +

nitric acid + lithium ⟶ water + nitrous oxide + lithium nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Li ⟶ H_2O + N_2O + LiNO_3 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: 10 HNO_3 + 8 Li ⟶ 5 H_2O + N_2O + 8 LiNO_3 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 HNO_3 | 10 | -10 Li | 8 | -8 H_2O | 5 | 5 N_2O | 1 | 1 LiNO_3 | 8 | 8 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 10 | -10 | ([HNO3])^(-10) Li | 8 | -8 | ([Li])^(-8) H_2O | 5 | 5 | ([H2O])^5 N_2O | 1 | 1 | [N2O] LiNO_3 | 8 | 8 | ([LiNO3])^8 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 = ([HNO3])^(-10) ([Li])^(-8) ([H2O])^5 [N2O] ([LiNO3])^8 = (([H2O])^5 [N2O] ([LiNO3])^8)/(([HNO3])^10 ([Li])^8)

Construct the rate of reaction expression for: HNO_3 + Li ⟶ H_2O + N_2O + LiNO_3 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: 10 HNO_3 + 8 Li ⟶ 5 H_2O + N_2O + 8 LiNO_3 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 HNO_3 | 10 | -10 Li | 8 | -8 H_2O | 5 | 5 N_2O | 1 | 1 LiNO_3 | 8 | 8 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 HNO_3 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) Li | 8 | -8 | -1/8 (Δ[Li])/(Δt) H_2O | 5 | 5 | 1/5 (Δ[H2O])/(Δt) N_2O | 1 | 1 | (Δ[N2O])/(Δt) LiNO_3 | 8 | 8 | 1/8 (Δ[LiNO3])/(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 = -1/10 (Δ[HNO3])/(Δt) = -1/8 (Δ[Li])/(Δt) = 1/5 (Δ[H2O])/(Δt) = (Δ[N2O])/(Δt) = 1/8 (Δ[LiNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | lithium | water | nitrous oxide | lithium nitrate formula | HNO_3 | Li | H_2O | N_2O | LiNO_3 name | nitric acid | lithium | water | nitrous oxide | lithium nitrate