KOH + C2H5Br = H2O + KBr + C2H4

KOH potassium hydroxide + CH_3CH_2Br bromoethane ⟶ H_2O water + KBr potassium bromide + CH_2=CH_2 ethylene

Balance the chemical equation algebraically: KOH + CH_3CH_2Br ⟶ H_2O + KBr + CH_2=CH_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 CH_3CH_2Br ⟶ c_3 H_2O + c_4 KBr + c_5 CH_2=CH_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, Br and C: H: | c_1 + 5 c_2 = 2 c_3 + 4 c_5 K: | c_1 = c_4 O: | c_1 = c_3 Br: | c_2 = c_4 C: | 2 c_2 = 2 c_5 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 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | KOH + CH_3CH_2Br ⟶ H_2O + KBr + CH_2=CH_2

+ ⟶ + +

potassium hydroxide + bromoethane ⟶ water + potassium bromide + ethylene

Construct the equilibrium constant, K, expression for: KOH + CH_3CH_2Br ⟶ H_2O + KBr + CH_2=CH_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: KOH + CH_3CH_2Br ⟶ H_2O + KBr + CH_2=CH_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 KOH | 1 | -1 CH_3CH_2Br | 1 | -1 H_2O | 1 | 1 KBr | 1 | 1 CH_2=CH_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 1 | -1 | ([KOH])^(-1) CH_3CH_2Br | 1 | -1 | ([CH3CH2Br])^(-1) H_2O | 1 | 1 | [H2O] KBr | 1 | 1 | [KBr] CH_2=CH_2 | 1 | 1 | [CH2=CH2] 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 = ([KOH])^(-1) ([CH3CH2Br])^(-1) [H2O] [KBr] [CH2=CH2] = ([H2O] [KBr] [CH2=CH2])/([KOH] [CH3CH2Br])

Construct the rate of reaction expression for: KOH + CH_3CH_2Br ⟶ H_2O + KBr + CH_2=CH_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: KOH + CH_3CH_2Br ⟶ H_2O + KBr + CH_2=CH_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 KOH | 1 | -1 CH_3CH_2Br | 1 | -1 H_2O | 1 | 1 KBr | 1 | 1 CH_2=CH_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 KOH | 1 | -1 | -(Δ[KOH])/(Δt) CH_3CH_2Br | 1 | -1 | -(Δ[CH3CH2Br])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KBr | 1 | 1 | (Δ[KBr])/(Δt) CH_2=CH_2 | 1 | 1 | (Δ[CH2=CH2])/(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[KOH])/(Δt) = -(Δ[CH3CH2Br])/(Δt) = (Δ[H2O])/(Δt) = (Δ[KBr])/(Δt) = (Δ[CH2=CH2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | bromoethane | water | potassium bromide | ethylene formula | KOH | CH_3CH_2Br | H_2O | KBr | CH_2=CH_2 Hill formula | HKO | C_2H_5Br | H_2O | BrK | C_2H_4 name | potassium hydroxide | bromoethane | water | potassium bromide | ethylene