NaOH + Br2 + Cr(NO3)3 = H2O + HNO3 + NaBr + Na2CrO4

NaOH sodium hydroxide + Br_2 bromine + CrN_3O_9 chromium nitrate ⟶ H_2O water + HNO_3 nitric acid + NaBr sodium bromide + Na_2CrO_4 sodium chromate

Balance the chemical equation algebraically: NaOH + Br_2 + CrN_3O_9 ⟶ H_2O + HNO_3 + NaBr + Na_2CrO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 Br_2 + c_3 CrN_3O_9 ⟶ c_4 H_2O + c_5 HNO_3 + c_6 NaBr + c_7 Na_2CrO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, Br, Cr and N: H: | c_1 = 2 c_4 + c_5 Na: | c_1 = c_6 + 2 c_7 O: | c_1 + 9 c_3 = c_4 + 3 c_5 + 4 c_7 Br: | 2 c_2 = c_6 Cr: | c_3 = c_7 N: | 3 c_3 = 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 = 5 c_2 = 3/2 c_3 = 1 c_4 = 1 c_5 = 3 c_6 = 3 c_7 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 10 c_2 = 3 c_3 = 2 c_4 = 2 c_5 = 6 c_6 = 6 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 NaOH + 3 Br_2 + 2 CrN_3O_9 ⟶ 2 H_2O + 6 HNO_3 + 6 NaBr + 2 Na_2CrO_4

+ + ⟶ + + +

sodium hydroxide + bromine + chromium nitrate ⟶ water + nitric acid + sodium bromide + sodium chromate

Construct the equilibrium constant, K, expression for: NaOH + Br_2 + CrN_3O_9 ⟶ H_2O + HNO_3 + NaBr + Na_2CrO_4 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: 10 NaOH + 3 Br_2 + 2 CrN_3O_9 ⟶ 2 H_2O + 6 HNO_3 + 6 NaBr + 2 Na_2CrO_4 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 NaOH | 10 | -10 Br_2 | 3 | -3 CrN_3O_9 | 2 | -2 H_2O | 2 | 2 HNO_3 | 6 | 6 NaBr | 6 | 6 Na_2CrO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 10 | -10 | ([NaOH])^(-10) Br_2 | 3 | -3 | ([Br2])^(-3) CrN_3O_9 | 2 | -2 | ([CrN3O9])^(-2) H_2O | 2 | 2 | ([H2O])^2 HNO_3 | 6 | 6 | ([HNO3])^6 NaBr | 6 | 6 | ([NaBr])^6 Na_2CrO_4 | 2 | 2 | ([Na2CrO4])^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 = ([NaOH])^(-10) ([Br2])^(-3) ([CrN3O9])^(-2) ([H2O])^2 ([HNO3])^6 ([NaBr])^6 ([Na2CrO4])^2 = (([H2O])^2 ([HNO3])^6 ([NaBr])^6 ([Na2CrO4])^2)/(([NaOH])^10 ([Br2])^3 ([CrN3O9])^2)

Construct the rate of reaction expression for: NaOH + Br_2 + CrN_3O_9 ⟶ H_2O + HNO_3 + NaBr + Na_2CrO_4 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: 10 NaOH + 3 Br_2 + 2 CrN_3O_9 ⟶ 2 H_2O + 6 HNO_3 + 6 NaBr + 2 Na_2CrO_4 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 NaOH | 10 | -10 Br_2 | 3 | -3 CrN_3O_9 | 2 | -2 H_2O | 2 | 2 HNO_3 | 6 | 6 NaBr | 6 | 6 Na_2CrO_4 | 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 NaOH | 10 | -10 | -1/10 (Δ[NaOH])/(Δt) Br_2 | 3 | -3 | -1/3 (Δ[Br2])/(Δt) CrN_3O_9 | 2 | -2 | -1/2 (Δ[CrN3O9])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) HNO_3 | 6 | 6 | 1/6 (Δ[HNO3])/(Δt) NaBr | 6 | 6 | 1/6 (Δ[NaBr])/(Δt) Na_2CrO_4 | 2 | 2 | 1/2 (Δ[Na2CrO4])/(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 = -1/10 (Δ[NaOH])/(Δt) = -1/3 (Δ[Br2])/(Δt) = -1/2 (Δ[CrN3O9])/(Δt) = 1/2 (Δ[H2O])/(Δt) = 1/6 (Δ[HNO3])/(Δt) = 1/6 (Δ[NaBr])/(Δt) = 1/2 (Δ[Na2CrO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | bromine | chromium nitrate | water | nitric acid | sodium bromide | sodium chromate formula | NaOH | Br_2 | CrN_3O_9 | H_2O | HNO_3 | NaBr | Na_2CrO_4 Hill formula | HNaO | Br_2 | CrN_3O_9 | H_2O | HNO_3 | BrNa | CrNa_2O_4 name | sodium hydroxide | bromine | chromium nitrate | water | nitric acid | sodium bromide | sodium chromate IUPAC name | sodium hydroxide | molecular bromine | chromium(+3) cation trinitrate | water | nitric acid | sodium bromide | disodium dioxido(dioxo)chromium