NH3 + HBr2 = N2 + HBr

NH_3 ammonia + HBr2 ⟶ N_2 nitrogen + HBr hydrogen bromide

Balance the chemical equation algebraically: NH_3 + HBr2 ⟶ N_2 + HBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 HBr2 ⟶ c_3 N_2 + c_4 HBr Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and Br: H: | 3 c_1 + c_2 = c_4 N: | c_1 = 2 c_3 Br: | 2 c_2 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 6 c_3 = 1 c_4 = 12 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NH_3 + 6 HBr2 ⟶ N_2 + 12 HBr

+ HBr2 ⟶ +

ammonia + HBr2 ⟶ nitrogen + hydrogen bromide

Construct the equilibrium constant, K, expression for: NH_3 + HBr2 ⟶ N_2 + HBr 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: 2 NH_3 + 6 HBr2 ⟶ N_2 + 12 HBr 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 NH_3 | 2 | -2 HBr2 | 6 | -6 N_2 | 1 | 1 HBr | 12 | 12 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_3 | 2 | -2 | ([NH3])^(-2) HBr2 | 6 | -6 | ([HBr2])^(-6) N_2 | 1 | 1 | [N2] HBr | 12 | 12 | ([HBr])^12 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 = ([NH3])^(-2) ([HBr2])^(-6) [N2] ([HBr])^12 = ([N2] ([HBr])^12)/(([NH3])^2 ([HBr2])^6)

Construct the rate of reaction expression for: NH_3 + HBr2 ⟶ N_2 + HBr 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: 2 NH_3 + 6 HBr2 ⟶ N_2 + 12 HBr 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 NH_3 | 2 | -2 HBr2 | 6 | -6 N_2 | 1 | 1 HBr | 12 | 12 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 NH_3 | 2 | -2 | -1/2 (Δ[NH3])/(Δt) HBr2 | 6 | -6 | -1/6 (Δ[HBr2])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) HBr | 12 | 12 | 1/12 (Δ[HBr])/(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 = -1/2 (Δ[NH3])/(Δt) = -1/6 (Δ[HBr2])/(Δt) = (Δ[N2])/(Δt) = 1/12 (Δ[HBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ammonia | HBr2 | nitrogen | hydrogen bromide formula | NH_3 | HBr2 | N_2 | HBr Hill formula | H_3N | HBr2 | N_2 | BrH name | ammonia | | nitrogen | hydrogen bromide IUPAC name | ammonia | | molecular nitrogen | hydrogen bromide

| ammonia | HBr2 | nitrogen | hydrogen bromide molar mass | 17.031 g/mol | 160.82 g/mol | 28.014 g/mol | 80.912 g/mol phase | gas (at STP) | | gas (at STP) | gas (at STP) melting point | -77.73 °C | | -210 °C | -86.8 °C boiling point | -33.33 °C | | -195.79 °C | -66.38 °C density | 6.96×10^-4 g/cm^3 (at 25 °C) | | 0.001251 g/cm^3 (at 0 °C) | 0.003307 g/cm^3 (at 25 °C) solubility in water | | | insoluble | miscible surface tension | 0.0234 N/m | | 0.0066 N/m | 0.0271 N/m dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | | 1.78×10^-5 Pa s (at 25 °C) | 8.4×10^-4 Pa s (at -75 °C) odor | | | odorless |