HBr = H2 + Br2

HBr hydrogen bromide ⟶ H_2 hydrogen + Br_2 bromine

Balance the chemical equation algebraically: HBr ⟶ H_2 + Br_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr ⟶ c_2 H_2 + c_3 Br_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br and H: Br: | c_1 = 2 c_3 H: | c_1 = 2 c_2 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 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HBr ⟶ H_2 + Br_2

⟶ +

hydrogen bromide ⟶ hydrogen + bromine

| hydrogen bromide | hydrogen | bromine molecular enthalpy | -36.3 kJ/mol | 0 kJ/mol | 0 kJ/mol total enthalpy | -72.6 kJ/mol | 0 kJ/mol | 0 kJ/mol | H_initial = -72.6 kJ/mol | H_final = 0 kJ/mol | ΔH_rxn^0 | 0 kJ/mol - -72.6 kJ/mol = 72.6 kJ/mol (endothermic) | |

| hydrogen bromide | hydrogen | bromine molecular free energy | -53.4 kJ/mol | 0 kJ/mol | 0 kJ/mol total free energy | -106.8 kJ/mol | 0 kJ/mol | 0 kJ/mol | G_initial = -106.8 kJ/mol | G_final = 0 kJ/mol | ΔG_rxn^0 | 0 kJ/mol - -106.8 kJ/mol = 106.8 kJ/mol (endergonic) | |

| hydrogen bromide | hydrogen | bromine molecular entropy | 199 J/(mol K) | 115 J/(mol K) | 152.2 J/(mol K) total entropy | 398 J/(mol K) | 115 J/(mol K) | 152.2 J/(mol K) | S_initial = 398 J/(mol K) | S_final = 267.2 J/(mol K) | ΔS_rxn^0 | 267.2 J/(mol K) - 398 J/(mol K) = -130.8 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: HBr ⟶ H_2 + Br_2 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 ⟶ H_2 + Br_2 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 H_2 | 1 | 1 Br_2 | 1 | 1 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) H_2 | 1 | 1 | [H2] Br_2 | 1 | 1 | [Br2] 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) [H2] [Br2] = ([H2] [Br2])/([HBr])^2

Construct the rate of reaction expression for: HBr ⟶ H_2 + Br_2 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 ⟶ H_2 + Br_2 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 H_2 | 1 | 1 Br_2 | 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 | 2 | -2 | -1/2 (Δ[HBr])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(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[H2])/(Δt) = (Δ[Br2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen bromide | hydrogen | bromine formula | HBr | H_2 | Br_2 Hill formula | BrH | H_2 | Br_2 name | hydrogen bromide | hydrogen | bromine IUPAC name | hydrogen bromide | molecular hydrogen | molecular bromine

| hydrogen bromide | hydrogen | bromine molar mass | 80.912 g/mol | 2.016 g/mol | 159.81 g/mol phase | gas (at STP) | gas (at STP) | liquid (at STP) melting point | -86.8 °C | -259.2 °C | -7.2 °C boiling point | -66.38 °C | -252.8 °C | 58.8 °C density | 0.003307 g/cm^3 (at 25 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | 3.119 g/cm^3 solubility in water | miscible | | insoluble surface tension | 0.0271 N/m | | 0.0409 N/m dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | 8.9×10^-6 Pa s (at 25 °C) | 9.44×10^-4 Pa s (at 25 °C) odor | | odorless |