HI + MoO3 = H2O + I2 + Mo2O5

HI hydrogen iodide + MoO_3 molybdenum trioxide ⟶ H_2O water + I_2 iodine + Mo2O5

Balance the chemical equation algebraically: HI + MoO_3 ⟶ H_2O + I_2 + Mo2O5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HI + c_2 MoO_3 ⟶ c_3 H_2O + c_4 I_2 + c_5 Mo2O5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, I, Mo and O: H: | c_1 = 2 c_3 I: | c_1 = 2 c_4 Mo: | c_2 = 2 c_5 O: | 3 c_2 = c_3 + 5 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 2 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HI + 2 MoO_3 ⟶ H_2O + I_2 + Mo2O5

+ ⟶ + + Mo2O5

hydrogen iodide + molybdenum trioxide ⟶ water + iodine + Mo2O5

Construct the equilibrium constant, K, expression for: HI + MoO_3 ⟶ H_2O + I_2 + Mo2O5 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 HI + 2 MoO_3 ⟶ H_2O + I_2 + Mo2O5 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 HI | 2 | -2 MoO_3 | 2 | -2 H_2O | 1 | 1 I_2 | 1 | 1 Mo2O5 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HI | 2 | -2 | ([HI])^(-2) MoO_3 | 2 | -2 | ([MoO3])^(-2) H_2O | 1 | 1 | [H2O] I_2 | 1 | 1 | [I2] Mo2O5 | 1 | 1 | [Mo2O5] 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 = ([HI])^(-2) ([MoO3])^(-2) [H2O] [I2] [Mo2O5] = ([H2O] [I2] [Mo2O5])/(([HI])^2 ([MoO3])^2)

Construct the rate of reaction expression for: HI + MoO_3 ⟶ H_2O + I_2 + Mo2O5 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 HI + 2 MoO_3 ⟶ H_2O + I_2 + Mo2O5 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 HI | 2 | -2 MoO_3 | 2 | -2 H_2O | 1 | 1 I_2 | 1 | 1 Mo2O5 | 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 HI | 2 | -2 | -1/2 (Δ[HI])/(Δt) MoO_3 | 2 | -2 | -1/2 (Δ[MoO3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) I_2 | 1 | 1 | (Δ[I2])/(Δt) Mo2O5 | 1 | 1 | (Δ[Mo2O5])/(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[HI])/(Δt) = -1/2 (Δ[MoO3])/(Δt) = (Δ[H2O])/(Δt) = (Δ[I2])/(Δt) = (Δ[Mo2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen iodide | molybdenum trioxide | water | iodine | Mo2O5 formula | HI | MoO_3 | H_2O | I_2 | Mo2O5 name | hydrogen iodide | molybdenum trioxide | water | iodine | IUPAC name | hydrogen iodide | trioxomolybdenum | water | molecular iodine |

| hydrogen iodide | molybdenum trioxide | water | iodine | Mo2O5 molar mass | 127.912 g/mol | 143.9 g/mol | 18.015 g/mol | 253.80894 g/mol | 271.9 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) | melting point | -50.76 °C | 795 °C | 0 °C | 113 °C | boiling point | -35.55 °C | 1155 °C | 99.9839 °C | 184 °C | density | 0.005228 g/cm^3 (at 25 °C) | 4.69 g/cm^3 | 1 g/cm^3 | 4.94 g/cm^3 | solubility in water | very soluble | | | | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.001321 Pa s (at -39 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.00227 Pa s (at 116 °C) | odor | | odorless | odorless | |