Br2 + Sr = SrBr2

Br_2 bromine + Sr strontium ⟶ SrBr_2 strontium bromide

Balance the chemical equation algebraically: Br_2 + Sr ⟶ SrBr_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Br_2 + c_2 Sr ⟶ c_3 SrBr_2 Set the number of atoms in the reactants equal to the number of atoms in the products for Br and Sr: Br: | 2 c_1 = 2 c_3 Sr: | 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 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Br_2 + Sr ⟶ SrBr_2

+ ⟶

bromine + strontium ⟶ strontium bromide

| bromine | strontium | strontium bromide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -717.6 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -717.6 kJ/mol | H_initial = 0 kJ/mol | | H_final = -717.6 kJ/mol ΔH_rxn^0 | -717.6 kJ/mol - 0 kJ/mol = -717.6 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: Br_2 + Sr ⟶ SrBr_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: Br_2 + Sr ⟶ SrBr_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 Br_2 | 1 | -1 Sr | 1 | -1 SrBr_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Br_2 | 1 | -1 | ([Br2])^(-1) Sr | 1 | -1 | ([Sr])^(-1) SrBr_2 | 1 | 1 | [SrBr2] 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 = ([Br2])^(-1) ([Sr])^(-1) [SrBr2] = ([SrBr2])/([Br2] [Sr])

Construct the rate of reaction expression for: Br_2 + Sr ⟶ SrBr_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: Br_2 + Sr ⟶ SrBr_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 Br_2 | 1 | -1 Sr | 1 | -1 SrBr_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 Br_2 | 1 | -1 | -(Δ[Br2])/(Δt) Sr | 1 | -1 | -(Δ[Sr])/(Δt) SrBr_2 | 1 | 1 | (Δ[SrBr2])/(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[Br2])/(Δt) = -(Δ[Sr])/(Δt) = (Δ[SrBr2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| bromine | strontium | strontium bromide formula | Br_2 | Sr | SrBr_2 Hill formula | Br_2 | Sr | Br_2Sr name | bromine | strontium | strontium bromide IUPAC name | molecular bromine | strontium | strontium dibromide

| bromine | strontium | strontium bromide molar mass | 159.81 g/mol | 87.62 g/mol | 247.4 g/mol phase | liquid (at STP) | solid (at STP) | solid (at STP) melting point | -7.2 °C | 757 °C | 643 °C boiling point | 58.8 °C | 1384 °C | 2146 °C density | 3.119 g/cm^3 | 2.6 g/cm^3 | 4.175 g/cm^3 solubility in water | insoluble | | surface tension | 0.0409 N/m | | dynamic viscosity | 9.44×10^-4 Pa s (at 25 °C) | |