Fe + Pb(NO3)2 = Pb + Fe(NO3)2

Fe iron + Pb(NO_3)_2 lead(II) nitrate ⟶ Pb lead + Fe(NO_3)_2 iron(II) nitrate

Balance the chemical equation algebraically: Fe + Pb(NO_3)_2 ⟶ Pb + Fe(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Fe + c_2 Pb(NO_3)_2 ⟶ c_3 Pb + c_4 Fe(NO_3)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Fe, N, O and Pb: Fe: | c_1 = c_4 N: | 2 c_2 = 2 c_4 O: | 6 c_2 = 6 c_4 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Fe + Pb(NO_3)_2 ⟶ Pb + Fe(NO_3)_2

+ ⟶ +

iron + lead(II) nitrate ⟶ lead + iron(II) nitrate

Construct the equilibrium constant, K, expression for: Fe + Pb(NO_3)_2 ⟶ Pb + Fe(NO_3)_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: Fe + Pb(NO_3)_2 ⟶ Pb + Fe(NO_3)_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 Fe | 1 | -1 Pb(NO_3)_2 | 1 | -1 Pb | 1 | 1 Fe(NO_3)_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Fe | 1 | -1 | ([Fe])^(-1) Pb(NO_3)_2 | 1 | -1 | ([Pb(NO3)2])^(-1) Pb | 1 | 1 | [Pb] Fe(NO_3)_2 | 1 | 1 | [Fe(NO3)2] 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 = ([Fe])^(-1) ([Pb(NO3)2])^(-1) [Pb] [Fe(NO3)2] = ([Pb] [Fe(NO3)2])/([Fe] [Pb(NO3)2])

Construct the rate of reaction expression for: Fe + Pb(NO_3)_2 ⟶ Pb + Fe(NO_3)_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: Fe + Pb(NO_3)_2 ⟶ Pb + Fe(NO_3)_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 Fe | 1 | -1 Pb(NO_3)_2 | 1 | -1 Pb | 1 | 1 Fe(NO_3)_2 | 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 Fe | 1 | -1 | -(Δ[Fe])/(Δt) Pb(NO_3)_2 | 1 | -1 | -(Δ[Pb(NO3)2])/(Δt) Pb | 1 | 1 | (Δ[Pb])/(Δt) Fe(NO_3)_2 | 1 | 1 | (Δ[Fe(NO3)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[Fe])/(Δt) = -(Δ[Pb(NO3)2])/(Δt) = (Δ[Pb])/(Δt) = (Δ[Fe(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| iron | lead(II) nitrate | lead | iron(II) nitrate formula | Fe | Pb(NO_3)_2 | Pb | Fe(NO_3)_2 Hill formula | Fe | N_2O_6Pb | Pb | FeN_2O_6 name | iron | lead(II) nitrate | lead | iron(II) nitrate IUPAC name | iron | plumbous dinitrate | lead |

| iron | lead(II) nitrate | lead | iron(II) nitrate molar mass | 55.845 g/mol | 331.2 g/mol | 207.2 g/mol | 179.85 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | melting point | 1535 °C | 470 °C | 327.4 °C | boiling point | 2750 °C | | 1740 °C | density | 7.874 g/cm^3 | | 11.34 g/cm^3 | solubility in water | insoluble | | insoluble | dynamic viscosity | | | 0.00183 Pa s (at 38 °C) | odor | | odorless | |