HNO3 + Li = H2 + LiNO3

HNO_3 nitric acid + Li lithium ⟶ H_2 hydrogen + LiNO_3 lithium nitrate

Balance the chemical equation algebraically: HNO_3 + Li ⟶ H_2 + LiNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Li ⟶ c_3 H_2 + c_4 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 = c_4 O: | 3 c_1 = 3 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 HNO_3 + 2 Li ⟶ H_2 + 2 LiNO_3

+ ⟶ +

nitric acid + lithium ⟶ hydrogen + lithium nitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Li ⟶ H_2 + 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: 2 HNO_3 + 2 Li ⟶ H_2 + 2 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 | 2 | -2 Li | 2 | -2 H_2 | 1 | 1 LiNO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 2 | -2 | ([HNO3])^(-2) Li | 2 | -2 | ([Li])^(-2) H_2 | 1 | 1 | [H2] LiNO_3 | 2 | 2 | ([LiNO3])^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 = ([HNO3])^(-2) ([Li])^(-2) [H2] ([LiNO3])^2 = ([H2] ([LiNO3])^2)/(([HNO3])^2 ([Li])^2)

Construct the rate of reaction expression for: HNO_3 + Li ⟶ H_2 + 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: 2 HNO_3 + 2 Li ⟶ H_2 + 2 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 | 2 | -2 Li | 2 | -2 H_2 | 1 | 1 LiNO_3 | 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 HNO_3 | 2 | -2 | -1/2 (Δ[HNO3])/(Δt) Li | 2 | -2 | -1/2 (Δ[Li])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) LiNO_3 | 2 | 2 | 1/2 (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 = -1/2 (Δ[HNO3])/(Δt) = -1/2 (Δ[Li])/(Δt) = (Δ[H2])/(Δt) = 1/2 (Δ[LiNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | lithium | hydrogen | lithium nitrate formula | HNO_3 | Li | H_2 | LiNO_3 name | nitric acid | lithium | hydrogen | lithium nitrate IUPAC name | nitric acid | lithium | molecular hydrogen | lithium nitrate

| nitric acid | lithium | hydrogen | lithium nitrate molar mass | 63.012 g/mol | 6.94 g/mol | 2.016 g/mol | 68.94 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | 180 °C | -259.2 °C | 264 °C boiling point | 83 °C | 1342 °C | -252.8 °C | density | 1.5129 g/cm^3 | 0.534 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | miscible | decomposes | | surface tension | | 0.3975 N/m | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |