KOH + I2 + C3H6O = H2O + KI + CHI3 + CH3COOK

KOH potassium hydroxide + I_2 iodine + CH_3COCH_3 acetone ⟶ H_2O water + KI potassium iodide + CHI_3 iodoform + CH_3COOK potassium acetate

Balance the chemical equation algebraically: KOH + I_2 + CH_3COCH_3 ⟶ H_2O + KI + CHI_3 + CH_3COOK Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 I_2 + c_3 CH_3COCH_3 ⟶ c_4 H_2O + c_5 KI + c_6 CHI_3 + c_7 CH_3COOK Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O, I and C: H: | c_1 + 6 c_3 = 2 c_4 + c_6 + 3 c_7 K: | c_1 = c_5 + c_7 O: | c_1 + c_3 = c_4 + 2 c_7 I: | 2 c_2 = c_5 + 3 c_6 C: | 3 c_3 = c_6 + 2 c_7 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_3 = 1 and solve the system of equations for the remaining coefficients: c_2 = 17/3 - (2 c_1)/3 c_3 = 1 c_4 = c_1/3 + 5/3 c_5 = (2 c_1)/3 + 1/3 c_6 = 11/3 - (2 c_1)/3 c_7 = c_1/3 - 1/3 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 4 and solve for the remaining coefficients: c_1 = 4 c_2 = 3 c_3 = 1 c_4 = 3 c_5 = 3 c_6 = 1 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 KOH + 3 I_2 + CH_3COCH_3 ⟶ 3 H_2O + 3 KI + CHI_3 + CH_3COOK

+ + ⟶ + + +

potassium hydroxide + iodine + acetone ⟶ water + potassium iodide + iodoform + potassium acetate

Construct the equilibrium constant, K, expression for: KOH + I_2 + CH_3COCH_3 ⟶ H_2O + KI + CHI_3 + CH_3COOK 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: 4 KOH + 3 I_2 + CH_3COCH_3 ⟶ 3 H_2O + 3 KI + CHI_3 + CH_3COOK 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 | 4 | -4 I_2 | 3 | -3 CH_3COCH_3 | 1 | -1 H_2O | 3 | 3 KI | 3 | 3 CHI_3 | 1 | 1 CH_3COOK | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 4 | -4 | ([KOH])^(-4) I_2 | 3 | -3 | ([I2])^(-3) CH_3COCH_3 | 1 | -1 | ([CH3COCH3])^(-1) H_2O | 3 | 3 | ([H2O])^3 KI | 3 | 3 | ([KI])^3 CHI_3 | 1 | 1 | [CHI3] CH_3COOK | 1 | 1 | [CH3COOK] 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])^(-4) ([I2])^(-3) ([CH3COCH3])^(-1) ([H2O])^3 ([KI])^3 [CHI3] [CH3COOK] = (([H2O])^3 ([KI])^3 [CHI3] [CH3COOK])/(([KOH])^4 ([I2])^3 [CH3COCH3])

Construct the rate of reaction expression for: KOH + I_2 + CH_3COCH_3 ⟶ H_2O + KI + CHI_3 + CH_3COOK 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: 4 KOH + 3 I_2 + CH_3COCH_3 ⟶ 3 H_2O + 3 KI + CHI_3 + CH_3COOK 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 | 4 | -4 I_2 | 3 | -3 CH_3COCH_3 | 1 | -1 H_2O | 3 | 3 KI | 3 | 3 CHI_3 | 1 | 1 CH_3COOK | 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 | 4 | -4 | -1/4 (Δ[KOH])/(Δt) I_2 | 3 | -3 | -1/3 (Δ[I2])/(Δt) CH_3COCH_3 | 1 | -1 | -(Δ[CH3COCH3])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) KI | 3 | 3 | 1/3 (Δ[KI])/(Δt) CHI_3 | 1 | 1 | (Δ[CHI3])/(Δt) CH_3COOK | 1 | 1 | (Δ[CH3COOK])/(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 = -1/4 (Δ[KOH])/(Δt) = -1/3 (Δ[I2])/(Δt) = -(Δ[CH3COCH3])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/3 (Δ[KI])/(Δt) = (Δ[CHI3])/(Δt) = (Δ[CH3COOK])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | iodine | acetone | water | potassium iodide | iodoform | potassium acetate formula | KOH | I_2 | CH_3COCH_3 | H_2O | KI | CHI_3 | CH_3COOK Hill formula | HKO | I_2 | C_3H_6O | H_2O | IK | CHI_3 | C_2H_3KO_2 name | potassium hydroxide | iodine | acetone | water | potassium iodide | iodoform | potassium acetate IUPAC name | potassium hydroxide | molecular iodine | acetone | water | potassium iodide | iodoform | potassium acetate