H2O + HNO3 + NH3 + FeSO4 = H2SO4 + Fe(OH)3 + NH4NO3

H_2O water + HNO_3 nitric acid + NH_3 ammonia + FeSO_4 duretter ⟶ H_2SO_4 sulfuric acid + Fe(OH)_3 iron(III) hydroxide + NH_4NO_3 ammonium nitrate

Balance the chemical equation algebraically: H_2O + HNO_3 + NH_3 + FeSO_4 ⟶ H_2SO_4 + Fe(OH)_3 + NH_4NO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 HNO_3 + c_3 NH_3 + c_4 FeSO_4 ⟶ c_5 H_2SO_4 + c_6 Fe(OH)_3 + c_7 NH_4NO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N, Fe and S: H: | 2 c_1 + c_2 + 3 c_3 = 2 c_5 + 3 c_6 + 4 c_7 O: | c_1 + 3 c_2 + 4 c_4 = 4 c_5 + 3 c_6 + 3 c_7 N: | c_2 + c_3 = 2 c_7 Fe: | c_4 = c_6 S: | c_4 = 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_2 = (2 c_1)/21 + 1 c_3 = 1 c_4 = (8 c_1)/21 c_5 = (8 c_1)/21 c_6 = (8 c_1)/21 c_7 = c_1/21 + 1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 21 and solve for the remaining coefficients: c_1 = 21 c_2 = 3 c_3 = 1 c_4 = 8 c_5 = 8 c_6 = 8 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 21 H_2O + 3 HNO_3 + NH_3 + 8 FeSO_4 ⟶ 8 H_2SO_4 + 8 Fe(OH)_3 + 2 NH_4NO_3

+ + + ⟶ + +

water + nitric acid + ammonia + duretter ⟶ sulfuric acid + iron(III) hydroxide + ammonium nitrate

Construct the equilibrium constant, K, expression for: H_2O + HNO_3 + NH_3 + FeSO_4 ⟶ H_2SO_4 + Fe(OH)_3 + NH_4NO_3 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: 21 H_2O + 3 HNO_3 + NH_3 + 8 FeSO_4 ⟶ 8 H_2SO_4 + 8 Fe(OH)_3 + 2 NH_4NO_3 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 H_2O | 21 | -21 HNO_3 | 3 | -3 NH_3 | 1 | -1 FeSO_4 | 8 | -8 H_2SO_4 | 8 | 8 Fe(OH)_3 | 8 | 8 NH_4NO_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 21 | -21 | ([H2O])^(-21) HNO_3 | 3 | -3 | ([HNO3])^(-3) NH_3 | 1 | -1 | ([NH3])^(-1) FeSO_4 | 8 | -8 | ([FeSO4])^(-8) H_2SO_4 | 8 | 8 | ([H2SO4])^8 Fe(OH)_3 | 8 | 8 | ([Fe(OH)3])^8 NH_4NO_3 | 2 | 2 | ([NH4NO3])^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 = ([H2O])^(-21) ([HNO3])^(-3) ([NH3])^(-1) ([FeSO4])^(-8) ([H2SO4])^8 ([Fe(OH)3])^8 ([NH4NO3])^2 = (([H2SO4])^8 ([Fe(OH)3])^8 ([NH4NO3])^2)/(([H2O])^21 ([HNO3])^3 [NH3] ([FeSO4])^8)

Construct the rate of reaction expression for: H_2O + HNO_3 + NH_3 + FeSO_4 ⟶ H_2SO_4 + Fe(OH)_3 + NH_4NO_3 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: 21 H_2O + 3 HNO_3 + NH_3 + 8 FeSO_4 ⟶ 8 H_2SO_4 + 8 Fe(OH)_3 + 2 NH_4NO_3 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 H_2O | 21 | -21 HNO_3 | 3 | -3 NH_3 | 1 | -1 FeSO_4 | 8 | -8 H_2SO_4 | 8 | 8 Fe(OH)_3 | 8 | 8 NH_4NO_3 | 2 | 2 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 H_2O | 21 | -21 | -1/21 (Δ[H2O])/(Δt) HNO_3 | 3 | -3 | -1/3 (Δ[HNO3])/(Δt) NH_3 | 1 | -1 | -(Δ[NH3])/(Δt) FeSO_4 | 8 | -8 | -1/8 (Δ[FeSO4])/(Δt) H_2SO_4 | 8 | 8 | 1/8 (Δ[H2SO4])/(Δt) Fe(OH)_3 | 8 | 8 | 1/8 (Δ[Fe(OH)3])/(Δt) NH_4NO_3 | 2 | 2 | 1/2 (Δ[NH4NO3])/(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 = -1/21 (Δ[H2O])/(Δt) = -1/3 (Δ[HNO3])/(Δt) = -(Δ[NH3])/(Δt) = -1/8 (Δ[FeSO4])/(Δt) = 1/8 (Δ[H2SO4])/(Δt) = 1/8 (Δ[Fe(OH)3])/(Δt) = 1/2 (Δ[NH4NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | nitric acid | ammonia | duretter | sulfuric acid | iron(III) hydroxide | ammonium nitrate formula | H_2O | HNO_3 | NH_3 | FeSO_4 | H_2SO_4 | Fe(OH)_3 | NH_4NO_3 Hill formula | H_2O | HNO_3 | H_3N | FeO_4S | H_2O_4S | FeH_3O_3 | H_4N_2O_3 name | water | nitric acid | ammonia | duretter | sulfuric acid | iron(III) hydroxide | ammonium nitrate IUPAC name | water | nitric acid | ammonia | iron(+2) cation sulfate | sulfuric acid | ferric trihydroxide |

| water | nitric acid | ammonia | duretter | sulfuric acid | iron(III) hydroxide | ammonium nitrate molar mass | 18.015 g/mol | 63.012 g/mol | 17.031 g/mol | 151.9 g/mol | 98.07 g/mol | 106.87 g/mol | 80.04 g/mol phase | liquid (at STP) | liquid (at STP) | gas (at STP) | | liquid (at STP) | | solid (at STP) melting point | 0 °C | -41.6 °C | -77.73 °C | | 10.371 °C | | 169 °C boiling point | 99.9839 °C | 83 °C | -33.33 °C | | 279.6 °C | | 210 °C density | 1 g/cm^3 | 1.5129 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | 2.841 g/cm^3 | 1.8305 g/cm^3 | | 1.73 g/cm^3 solubility in water | | miscible | | | very soluble | | surface tension | 0.0728 N/m | | 0.0234 N/m | | 0.0735 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | | 0.021 Pa s (at 25 °C) | | odor | odorless | | | | odorless | | odorless