HBr + C3H6 = C3H7Br

HBr hydrogen bromide + C_3H_6 cyclopropane ⟶ (CH_3)_2CHBr 2-bromopropane

Balance the chemical equation algebraically: HBr + C_3H_6 ⟶ (CH_3)_2CHBr Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HBr + c_2 C_3H_6 ⟶ c_3 (CH_3)_2CHBr 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 + 6 c_2 = 7 c_3 C: | 3 c_2 = 3 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 + C_3H_6 ⟶ (CH_3)_2CHBr

+ ⟶

hydrogen bromide + cyclopropane ⟶ 2-bromopropane

Construct the equilibrium constant, K, expression for: HBr + C_3H_6 ⟶ (CH_3)_2CHBr 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 + C_3H_6 ⟶ (CH_3)_2CHBr 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 C_3H_6 | 1 | -1 (CH_3)_2CHBr | 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) C_3H_6 | 1 | -1 | ([C3H6])^(-1) (CH_3)_2CHBr | 1 | 1 | [(CH3)2CHBr] 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) ([C3H6])^(-1) [(CH3)2CHBr] = ([(CH3)2CHBr])/([HBr] [C3H6])

Construct the rate of reaction expression for: HBr + C_3H_6 ⟶ (CH_3)_2CHBr 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 + C_3H_6 ⟶ (CH_3)_2CHBr 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 C_3H_6 | 1 | -1 (CH_3)_2CHBr | 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) C_3H_6 | 1 | -1 | -(Δ[C3H6])/(Δt) (CH_3)_2CHBr | 1 | 1 | (Δ[(CH3)2CHBr])/(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[C3H6])/(Δt) = (Δ[(CH3)2CHBr])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen bromide | cyclopropane | 2-bromopropane formula | HBr | C_3H_6 | (CH_3)_2CHBr Hill formula | BrH | C_3H_6 | C_3H_7Br name | hydrogen bromide | cyclopropane | 2-bromopropane