HBr + C2H4 = C2H5Br

HBr hydrogen bromide + CH_2=CH_2 ethylene ⟶ CH_3CH_2Br bromoethane

Balance the chemical equation algebraically: HBr + CH_2=CH_2 ⟶ CH_3CH_2Br Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 CH_2=CH_2 ⟶ c_3 CH_3CH_2Br Set the number of atoms in the reactants equal to the number of atoms in the products for Br, H and C: Br: | c_1 = c_3 H: | c_1 + 4 c_2 = 5 c_3 C: | 2 c_2 = 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: | | HBr + CH_2=CH_2 ⟶ CH_3CH_2Br

+ ⟶

hydrogen bromide + ethylene ⟶ bromoethane

Construct the equilibrium constant, K, expression for: HBr + CH_2=CH_2 ⟶ CH_3CH_2Br 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: HBr + CH_2=CH_2 ⟶ CH_3CH_2Br 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 | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_2Br | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HBr | 1 | -1 | ([HBr])^(-1) CH_2=CH_2 | 1 | -1 | ([CH2=CH2])^(-1) CH_3CH_2Br | 1 | 1 | [CH3CH2Br] 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])^(-1) ([CH2=CH2])^(-1) [CH3CH2Br] = ([CH3CH2Br])/([HBr] [CH2=CH2])

Construct the rate of reaction expression for: HBr + CH_2=CH_2 ⟶ CH_3CH_2Br 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: HBr + CH_2=CH_2 ⟶ CH_3CH_2Br 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 | 1 | -1 CH_2=CH_2 | 1 | -1 CH_3CH_2Br | 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 | 1 | -1 | -(Δ[HBr])/(Δt) CH_2=CH_2 | 1 | -1 | -(Δ[CH2=CH2])/(Δt) CH_3CH_2Br | 1 | 1 | (Δ[CH3CH2Br])/(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[HBr])/(Δt) = -(Δ[CH2=CH2])/(Δt) = (Δ[CH3CH2Br])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen bromide | ethylene | bromoethane formula | HBr | CH_2=CH_2 | CH_3CH_2Br Hill formula | BrH | C_2H_4 | C_2H_5Br name | hydrogen bromide | ethylene | bromoethane

| hydrogen bromide | ethylene | bromoethane molar mass | 80.912 g/mol | 28.054 g/mol | 108.97 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) melting point | -86.8 °C | -169 °C | -139.54 °C boiling point | -66.38 °C | -104 °C | 15.8 °C density | 0.003307 g/cm^3 (at 25 °C) | 1.153 g/cm^3 (at 25 °C) | 1.4933 g/cm^3 (at 20 °C) solubility in water | miscible | insoluble | surface tension | 0.0271 N/m | 0.0181 N/m | dynamic viscosity | 8.4×10^-4 Pa s (at -75 °C) | 1.034×10^-5 Pa s (at 25 °C) | 3.74×10^-4 Pa s (at 25 °C)