Al + Sn(NO3)2 = Sn + Al(NO3)2

Al aluminum + Sn(NO3)2 ⟶ Sn white tin + Al(NO3)2

Balance the chemical equation algebraically: Al + Sn(NO3)2 ⟶ Sn + Al(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 Sn(NO3)2 ⟶ c_3 Sn + c_4 Al(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for Al, Sn, N and O: Al: | c_1 = c_4 Sn: | c_2 = c_3 N: | 2 c_2 = 2 c_4 O: | 6 c_2 = 6 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_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: | | Al + Sn(NO3)2 ⟶ Sn + Al(NO3)2

+ Sn(NO3)2 ⟶ + Al(NO3)2

aluminum + Sn(NO3)2 ⟶ white tin + Al(NO3)2

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

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

| aluminum | Sn(NO3)2 | white tin | Al(NO3)2 formula | Al | Sn(NO3)2 | Sn | Al(NO3)2 Hill formula | Al | N2O6Sn | Sn | AlN2O6 name | aluminum | | white tin | IUPAC name | aluminum | | tin |

| aluminum | Sn(NO3)2 | white tin | Al(NO3)2 molar mass | 26.9815385 g/mol | 242.72 g/mol | 118.71 g/mol | 150.99 g/mol phase | solid (at STP) | | solid (at STP) | melting point | 660.4 °C | | 231.9 °C | boiling point | 2460 °C | | 2602 °C | density | 2.7 g/cm^3 | | 7.31 g/cm^3 | solubility in water | insoluble | | insoluble | surface tension | 0.817 N/m | | | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | | 0.001 Pa s (at 600 °C) | odor | odorless | | odorless |