HNO3 + Te3N2 = H2O + NO + H2TeO4 + Te(NO3)4

HNO_3 nitric acid + Te3N2 ⟶ H_2O water + NO nitric oxide + H2TeO4 + N_4O_12Te tellurium tetranitrate

Balance the chemical equation algebraically: HNO_3 + Te3N2 ⟶ H_2O + NO + H2TeO4 + N_4O_12Te Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Te3N2 ⟶ c_3 H_2O + c_4 NO + c_5 H2TeO4 + c_6 N_4O_12Te 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_3 + 2 c_5 N: | c_1 + 2 c_2 = c_4 + 4 c_6 O: | 3 c_1 = c_3 + c_4 + 4 c_5 + 12 c_6 Te: | 3 c_2 = c_5 + c_6 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_2 = 1 c_3 = (4 c_1)/5 - 26/5 c_4 = 54/5 - c_1/5 c_5 = 26/5 - (3 c_1)/10 c_6 = (3 c_1)/10 - 11/5 Multiply by the least common denominator, 6, to eliminate fractional coefficients: c_2 = 6 c_3 = (4 c_1)/5 - 156/5 c_4 = 324/5 - c_1/5 c_5 = 156/5 - (3 c_1)/10 c_6 = (3 c_1)/10 - 66/5 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 74 and solve for the remaining coefficients: c_1 = 74 c_2 = 6 c_3 = 28 c_4 = 50 c_5 = 9 c_6 = 9 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 74 HNO_3 + 6 Te3N2 ⟶ 28 H_2O + 50 NO + 9 H2TeO4 + 9 N_4O_12Te

+ Te3N2 ⟶ + + H2TeO4 +

nitric acid + Te3N2 ⟶ water + nitric oxide + H2TeO4 + tellurium tetranitrate

Construct the equilibrium constant, K, expression for: HNO_3 + Te3N2 ⟶ H_2O + NO + H2TeO4 + N_4O_12Te 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: 74 HNO_3 + 6 Te3N2 ⟶ 28 H_2O + 50 NO + 9 H2TeO4 + 9 N_4O_12Te 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 | 74 | -74 Te3N2 | 6 | -6 H_2O | 28 | 28 NO | 50 | 50 H2TeO4 | 9 | 9 N_4O_12Te | 9 | 9 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 74 | -74 | ([HNO3])^(-74) Te3N2 | 6 | -6 | ([Te3N2])^(-6) H_2O | 28 | 28 | ([H2O])^28 NO | 50 | 50 | ([NO])^50 H2TeO4 | 9 | 9 | ([H2TeO4])^9 N_4O_12Te | 9 | 9 | ([N4O12Te])^9 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])^(-74) ([Te3N2])^(-6) ([H2O])^28 ([NO])^50 ([H2TeO4])^9 ([N4O12Te])^9 = (([H2O])^28 ([NO])^50 ([H2TeO4])^9 ([N4O12Te])^9)/(([HNO3])^74 ([Te3N2])^6)

Construct the rate of reaction expression for: HNO_3 + Te3N2 ⟶ H_2O + NO + H2TeO4 + N_4O_12Te 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: 74 HNO_3 + 6 Te3N2 ⟶ 28 H_2O + 50 NO + 9 H2TeO4 + 9 N_4O_12Te 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 | 74 | -74 Te3N2 | 6 | -6 H_2O | 28 | 28 NO | 50 | 50 H2TeO4 | 9 | 9 N_4O_12Te | 9 | 9 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 | 74 | -74 | -1/74 (Δ[HNO3])/(Δt) Te3N2 | 6 | -6 | -1/6 (Δ[Te3N2])/(Δt) H_2O | 28 | 28 | 1/28 (Δ[H2O])/(Δt) NO | 50 | 50 | 1/50 (Δ[NO])/(Δt) H2TeO4 | 9 | 9 | 1/9 (Δ[H2TeO4])/(Δt) N_4O_12Te | 9 | 9 | 1/9 (Δ[N4O12Te])/(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 = -1/74 (Δ[HNO3])/(Δt) = -1/6 (Δ[Te3N2])/(Δt) = 1/28 (Δ[H2O])/(Δt) = 1/50 (Δ[NO])/(Δt) = 1/9 (Δ[H2TeO4])/(Δt) = 1/9 (Δ[N4O12Te])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | Te3N2 | water | nitric oxide | H2TeO4 | tellurium tetranitrate formula | HNO_3 | Te3N2 | H_2O | NO | H2TeO4 | N_4O_12Te Hill formula | HNO_3 | N2Te3 | H_2O | NO | H2O4Te | N_4O_12Te name | nitric acid | | water | nitric oxide | | tellurium tetranitrate

| nitric acid | Te3N2 | water | nitric oxide | H2TeO4 | tellurium tetranitrate molar mass | 63.012 g/mol | 410.81 g/mol | 18.015 g/mol | 30.006 g/mol | 193.61 g/mol | 375.62 g/mol phase | liquid (at STP) | | liquid (at STP) | gas (at STP) | | melting point | -41.6 °C | | 0 °C | -163.6 °C | | boiling point | 83 °C | | 99.9839 °C | -151.7 °C | | density | 1.5129 g/cm^3 | | 1 g/cm^3 | 0.001226 g/cm^3 (at 25 °C) | | 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) | 1.911×10^-5 Pa s (at 25 °C) | | odor | | | odorless | | |