H2O + HNO3 + Ti = NO + TiO(OH)2

H_2O water + HNO_3 nitric acid + Ti titanium ⟶ NO nitric oxide + TiO(OH)2

Balance the chemical equation algebraically: H_2O + HNO_3 + Ti ⟶ NO + TiO(OH)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HNO_3 + c_3 Ti ⟶ c_4 NO + c_5 TiO(OH)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and Ti: H: | 2 c_1 + c_2 = 2 c_5 O: | c_1 + 3 c_2 = c_4 + 3 c_5 N: | c_2 = c_4 Ti: | c_3 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 4 c_3 = 3 c_4 = 4 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + 4 HNO_3 + 3 Ti ⟶ 4 NO + 3 TiO(OH)2

+ + ⟶ + TiO(OH)2

water + nitric acid + titanium ⟶ nitric oxide + TiO(OH)2

Construct the equilibrium constant, K, expression for: H_2O + HNO_3 + Ti ⟶ NO + TiO(OH)2 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: H_2O + 4 HNO_3 + 3 Ti ⟶ 4 NO + 3 TiO(OH)2 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 H_2O | 1 | -1 HNO_3 | 4 | -4 Ti | 3 | -3 NO | 4 | 4 TiO(OH)2 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 1 | -1 | ([H2O])^(-1) HNO_3 | 4 | -4 | ([HNO3])^(-4) Ti | 3 | -3 | ([Ti])^(-3) NO | 4 | 4 | ([NO])^4 TiO(OH)2 | 3 | 3 | ([TiO(OH)2])^3 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 = ([H2O])^(-1) ([HNO3])^(-4) ([Ti])^(-3) ([NO])^4 ([TiO(OH)2])^3 = (([NO])^4 ([TiO(OH)2])^3)/([H2O] ([HNO3])^4 ([Ti])^3)

Construct the rate of reaction expression for: H_2O + HNO_3 + Ti ⟶ NO + TiO(OH)2 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: H_2O + 4 HNO_3 + 3 Ti ⟶ 4 NO + 3 TiO(OH)2 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 H_2O | 1 | -1 HNO_3 | 4 | -4 Ti | 3 | -3 NO | 4 | 4 TiO(OH)2 | 3 | 3 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 H_2O | 1 | -1 | -(Δ[H2O])/(Δt) HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Ti | 3 | -3 | -1/3 (Δ[Ti])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(Δt) TiO(OH)2 | 3 | 3 | 1/3 (Δ[TiO(OH)2])/(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[H2O])/(Δt) = -1/4 (Δ[HNO3])/(Δt) = -1/3 (Δ[Ti])/(Δt) = 1/4 (Δ[NO])/(Δt) = 1/3 (Δ[TiO(OH)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | nitric acid | titanium | nitric oxide | TiO(OH)2 formula | H_2O | HNO_3 | Ti | NO | TiO(OH)2 Hill formula | H_2O | HNO_3 | Ti | NO | H2O3Ti name | water | nitric acid | titanium | nitric oxide |

| water | nitric acid | titanium | nitric oxide | TiO(OH)2 molar mass | 18.015 g/mol | 63.012 g/mol | 47.867 g/mol | 30.006 g/mol | 97.88 g/mol phase | liquid (at STP) | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | 0 °C | -41.6 °C | 1660 °C | -163.6 °C | boiling point | 99.9839 °C | 83 °C | 3287 °C | -151.7 °C | density | 1 g/cm^3 | 1.5129 g/cm^3 | 4.5 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | | miscible | insoluble | | surface tension | 0.0728 N/m | | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 25 °C) | | 1.911×10^-5 Pa s (at 25 °C) | odor | odorless | | | |