HNO3 + LiOH = H2O + LiNO3

HNO_3 nitric acid + LiOH lithium hydroxide ⟶ H_2O water + LiNO_3 lithium nitrate

Balance the chemical equation algebraically: HNO_3 + LiOH ⟶ H_2O + LiNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 LiOH ⟶ c_3 H_2O + 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 + c_2 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 + c_2 = c_3 + 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | HNO_3 + LiOH ⟶ H_2O + LiNO_3

+ ⟶ +

nitric acid + lithium hydroxide ⟶ water + lithium nitrate

| nitric acid | lithium hydroxide | water | lithium nitrate molecular free energy | -80.7 kJ/mol | -441.5 kJ/mol | -237.1 kJ/mol | -381.1 kJ/mol total free energy | -80.7 kJ/mol | -441.5 kJ/mol | -237.1 kJ/mol | -381.1 kJ/mol | G_initial = -522.2 kJ/mol | | G_final = -618.2 kJ/mol | ΔG_rxn^0 | -618.2 kJ/mol - -522.2 kJ/mol = -96 kJ/mol (exergonic) | | |

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

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

| nitric acid | lithium hydroxide | water | lithium nitrate formula | HNO_3 | LiOH | H_2O | LiNO_3 Hill formula | HNO_3 | HLiO | H_2O | LiNO_3 name | nitric acid | lithium hydroxide | water | lithium nitrate

| nitric acid | lithium hydroxide | water | lithium nitrate molar mass | 63.012 g/mol | 23.95 g/mol | 18.015 g/mol | 68.94 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -41.6 °C | 462 °C | 0 °C | 264 °C boiling point | 83 °C | | 99.9839 °C | density | 1.5129 g/cm^3 | 1.46 g/cm^3 | 1 g/cm^3 | solubility in water | miscible | | | surface tension | | | 0.0728 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | odorless | odorless |