HBr + NH4OH = H2O + NH4Br

HBr hydrogen bromide + NH_4OH ammonium hydroxide ⟶ H_2O water + NH_4Br ammonium bromide

Balance the chemical equation algebraically: HBr + NH_4OH ⟶ H_2O + NH_4Br Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 NH_4OH ⟶ c_3 H_2O + c_4 NH_4Br Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, N and O: Br: | c_1 = c_4 H: | c_1 + 5 c_2 = 2 c_3 + 4 c_4 N: | c_2 = c_4 O: | 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: | | HBr + NH_4OH ⟶ H_2O + NH_4Br

+ ⟶ +

hydrogen bromide + ammonium hydroxide ⟶ water + ammonium bromide

Construct the equilibrium constant, K, expression for: HBr + NH_4OH ⟶ H_2O + NH_4Br 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: HBr + NH_4OH ⟶ H_2O + NH_4Br 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 HBr | 1 | -1 NH_4OH | 1 | -1 H_2O | 1 | 1 NH_4Br | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 1 | -1 | ([HBr])^(-1) NH_4OH | 1 | -1 | ([NH4OH])^(-1) H_2O | 1 | 1 | [H2O] NH_4Br | 1 | 1 | [NH4Br] 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 = ([HBr])^(-1) ([NH4OH])^(-1) [H2O] [NH4Br] = ([H2O] [NH4Br])/([HBr] [NH4OH])

Construct the rate of reaction expression for: HBr + NH_4OH ⟶ H_2O + NH_4Br 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: HBr + NH_4OH ⟶ H_2O + NH_4Br 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 HBr | 1 | -1 NH_4OH | 1 | -1 H_2O | 1 | 1 NH_4Br | 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 HBr | 1 | -1 | -(Δ[HBr])/(Δt) NH_4OH | 1 | -1 | -(Δ[NH4OH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NH_4Br | 1 | 1 | (Δ[NH4Br])/(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[HBr])/(Δt) = -(Δ[NH4OH])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NH4Br])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen bromide | ammonium hydroxide | water | ammonium bromide formula | HBr | NH_4OH | H_2O | NH_4Br Hill formula | BrH | H_5NO | H_2O | BrH_4N name | hydrogen bromide | ammonium hydroxide | water | ammonium bromide