HBr + Ba = H2 + BaBr2 (2024)

HBr hydrogen bromide + Ba barium ⟶ H_2 hydrogen + BaBr_2 barium bromide

Balance the chemical equation algebraically: HBr + Ba ⟶ H_2 + BaBr_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 Ba ⟶ c_3 H_2 + c_4 BaBr_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H and Ba: Br: | c_1 = 2 c_4 H: | c_1 = 2 c_3 Ba: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 1 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 HBr + Ba ⟶ H_2 + BaBr_2

+ ⟶ +

hydrogen bromide + barium ⟶ hydrogen + barium bromide

| hydrogen bromide | barium | hydrogen | barium bromide molecular enthalpy | -36.3 kJ/mol | 0 kJ/mol | 0 kJ/mol | -757.3 kJ/mol total enthalpy | -72.6 kJ/mol | 0 kJ/mol | 0 kJ/mol | -757.3 kJ/mol | H_initial = -72.6 kJ/mol | | H_final = -757.3 kJ/mol | ΔH_rxn^0 | -757.3 kJ/mol - -72.6 kJ/mol = -684.7 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: HBr + Ba ⟶ H_2 + BaBr_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 + Ba ⟶ H_2 + BaBr_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 Ba | 1 | -1 H_2 | 1 | 1 BaBr_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) Ba | 1 | -1 | ([Ba])^(-1) H_2 | 1 | 1 | [H2] BaBr_2 | 1 | 1 | [BaBr2] 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) ([Ba])^(-1) [H2] [BaBr2] = ([H2] [BaBr2])/(([HBr])^2 [Ba])

Construct the rate of reaction expression for: HBr + Ba ⟶ H_2 + BaBr_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 + Ba ⟶ H_2 + BaBr_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 Ba | 1 | -1 H_2 | 1 | 1 BaBr_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) Ba | 1 | -1 | -(Δ[Ba])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) BaBr_2 | 1 | 1 | (Δ[BaBr2])/(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[Ba])/(Δt) = (Δ[H2])/(Δt) = (Δ[BaBr2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen bromide | barium | hydrogen | barium bromide formula | HBr | Ba | H_2 | BaBr_2 Hill formula | BrH | Ba | H_2 | BaBr_2 name | hydrogen bromide | barium | hydrogen | barium bromide IUPAC name | hydrogen bromide | barium | molecular hydrogen | barium(+2) cation dibromide

| hydrogen bromide | barium | hydrogen | barium bromide molar mass | 80.912 g/mol | 137.327 g/mol | 2.016 g/mol | 297.13 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | -86.8 °C | 725 °C | -259.2 °C | 860 °C boiling point | -66.38 °C | 1640 °C | -252.8 °C | 1835 °C density | 0.003307 g/cm^3 (at 25 °C) | 3.6 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 4.78 g/cm^3 solubility in water | miscible | insoluble | | surface tension | 0.0271 N/m | 0.224 N/m | | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |

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