HBr + Na2O = H2O + NaBr

HBr hydrogen bromide + Na_2O sodium oxide ⟶ H_2O water + NaBr sodium bromide

Balance the chemical equation algebraically: HBr + Na_2O ⟶ H_2O + NaBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 Na_2O ⟶ c_3 H_2O + c_4 NaBr Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H, Na and O: Br: | c_1 = c_4 H: | c_1 = 2 c_3 Na: | 2 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HBr + Na_2O ⟶ H_2O + 2 NaBr

+ ⟶ +

hydrogen bromide + sodium oxide ⟶ water + sodium bromide

Construct the equilibrium constant, K, expression for: HBr + Na_2O ⟶ H_2O + NaBr 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 HBr + Na_2O ⟶ H_2O + 2 NaBr 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 | 2 | -2 Na_2O | 1 | -1 H_2O | 1 | 1 NaBr | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 2 | -2 | ([HBr])^(-2) Na_2O | 1 | -1 | ([Na2O])^(-1) H_2O | 1 | 1 | [H2O] NaBr | 2 | 2 | ([NaBr])^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 = ([HBr])^(-2) ([Na2O])^(-1) [H2O] ([NaBr])^2 = ([H2O] ([NaBr])^2)/(([HBr])^2 [Na2O])

Construct the rate of reaction expression for: HBr + Na_2O ⟶ H_2O + NaBr 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 HBr + Na_2O ⟶ H_2O + 2 NaBr 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 | 2 | -2 Na_2O | 1 | -1 H_2O | 1 | 1 NaBr | 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 HBr | 2 | -2 | -1/2 (Δ[HBr])/(Δt) Na_2O | 1 | -1 | -(Δ[Na2O])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaBr | 2 | 2 | 1/2 (Δ[NaBr])/(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[HBr])/(Δt) = -(Δ[Na2O])/(Δt) = (Δ[H2O])/(Δt) = 1/2 (Δ[NaBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen bromide | sodium oxide | water | sodium bromide formula | HBr | Na_2O | H_2O | NaBr Hill formula | BrH | Na_2O | H_2O | BrNa name | hydrogen bromide | sodium oxide | water | sodium bromide IUPAC name | hydrogen bromide | disodium oxygen(-2) anion | water | sodium bromide

| hydrogen bromide | sodium oxide | water | sodium bromide molar mass | 80.912 g/mol | 61.979 g/mol | 18.015 g/mol | 102.89 g/mol phase | gas (at STP) | | liquid (at STP) | solid (at STP) melting point | -86.8 °C | | 0 °C | 755 °C boiling point | -66.38 °C | | 99.9839 °C | 1396 °C density | 0.003307 g/cm^3 (at 25 °C) | 2.27 g/cm^3 | 1 g/cm^3 | 3.2 g/cm^3 solubility in water | miscible | | | soluble surface tension | 0.0271 N/m | | 0.0728 N/m | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | | | odorless |