HNO3 + Te = NO + H2TeO4

HNO_3 nitric acid + Te tellurium ⟶ NO nitric oxide + H2TeO4

Balance the chemical equation algebraically: HNO_3 + Te ⟶ NO + H2TeO4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Te ⟶ c_3 NO + c_4 H2TeO4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Te: H: | c_1 = 2 c_4 N: | c_1 = c_3 O: | 3 c_1 = c_3 + 4 c_4 Te: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HNO_3 + Te ⟶ 2 NO + H2TeO4

+ ⟶ + H2TeO4

nitric acid + tellurium ⟶ nitric oxide + H2TeO4

Construct the equilibrium constant, K, expression for: HNO_3 + Te ⟶ NO + H2TeO4 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 + Te ⟶ 2 NO + H2TeO4 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 Te | 1 | -1 NO | 2 | 2 H2TeO4 | 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 | 2 | -2 | ([HNO3])^(-2) Te | 1 | -1 | ([Te])^(-1) NO | 2 | 2 | ([NO])^2 H2TeO4 | 1 | 1 | [H2TeO4] 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) ([Te])^(-1) ([NO])^2 [H2TeO4] = (([NO])^2 [H2TeO4])/(([HNO3])^2 [Te])

Construct the rate of reaction expression for: HNO_3 + Te ⟶ NO + H2TeO4 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 + Te ⟶ 2 NO + H2TeO4 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 Te | 1 | -1 NO | 2 | 2 H2TeO4 | 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 | 2 | -2 | -1/2 (Δ[HNO3])/(Δt) Te | 1 | -1 | -(Δ[Te])/(Δt) NO | 2 | 2 | 1/2 (Δ[NO])/(Δt) H2TeO4 | 1 | 1 | (Δ[H2TeO4])/(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) = -(Δ[Te])/(Δt) = 1/2 (Δ[NO])/(Δt) = (Δ[H2TeO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | tellurium | nitric oxide | H2TeO4 formula | HNO_3 | Te | NO | H2TeO4 Hill formula | HNO_3 | Te | NO | H2O4Te name | nitric acid | tellurium | nitric oxide |

| nitric acid | tellurium | nitric oxide | H2TeO4 molar mass | 63.012 g/mol | 127.6 g/mol | 30.006 g/mol | 193.61 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | melting point | -41.6 °C | 450 °C | -163.6 °C | boiling point | 83 °C | 990 °C | -151.7 °C | density | 1.5129 g/cm^3 | 6.24 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 1.911×10^-5 Pa s (at 25 °C) |