I2 + Pb = PbI4

I_2 iodine + Pb lead ⟶ PbI4

Balance the chemical equation algebraically: I_2 + Pb ⟶ PbI4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 I_2 + c_2 Pb ⟶ c_3 PbI4 Set the number of atoms in the reactants equal to the number of atoms in the products for I and Pb: I: | 2 c_1 = 4 c_3 Pb: | c_2 = c_3 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 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 I_2 + Pb ⟶ PbI4

+ ⟶ PbI4

iodine + lead ⟶ PbI4

Construct the equilibrium constant, K, expression for: I_2 + Pb ⟶ PbI4 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 I_2 + Pb ⟶ PbI4 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 I_2 | 2 | -2 Pb | 1 | -1 PbI4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression I_2 | 2 | -2 | ([I2])^(-2) Pb | 1 | -1 | ([Pb])^(-1) PbI4 | 1 | 1 | [PbI4] 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 = ([I2])^(-2) ([Pb])^(-1) [PbI4] = ([PbI4])/(([I2])^2 [Pb])

Construct the rate of reaction expression for: I_2 + Pb ⟶ PbI4 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 I_2 + Pb ⟶ PbI4 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 I_2 | 2 | -2 Pb | 1 | -1 PbI4 | 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 I_2 | 2 | -2 | -1/2 (Δ[I2])/(Δt) Pb | 1 | -1 | -(Δ[Pb])/(Δt) PbI4 | 1 | 1 | (Δ[PbI4])/(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[I2])/(Δt) = -(Δ[Pb])/(Δt) = (Δ[PbI4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iodine | lead | PbI4 formula | I_2 | Pb | PbI4 Hill formula | I_2 | Pb | I4Pb name | iodine | lead | IUPAC name | molecular iodine | lead |

| iodine | lead | PbI4 molar mass | 253.80894 g/mol | 207.2 g/mol | 714.8 g/mol phase | solid (at STP) | solid (at STP) | melting point | 113 °C | 327.4 °C | boiling point | 184 °C | 1740 °C | density | 4.94 g/cm^3 | 11.34 g/cm^3 | solubility in water | | insoluble | dynamic viscosity | 0.00227 Pa s (at 116 °C) | 0.00183 Pa s (at 38 °C) |