NaOH + I2 + C2H5OH = H2O + HI + CHI3 + HCOONa

NaOH sodium hydroxide + I_2 iodine + CH_3CH_2OH ethanol ⟶ H_2O water + HI hydrogen iodide + CHI_3 iodoform + HCOONa sodium formate

Balance the chemical equation algebraically: NaOH + I_2 + CH_3CH_2OH ⟶ H_2O + HI + CHI_3 + HCOONa Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 I_2 + c_3 CH_3CH_2OH ⟶ c_4 H_2O + c_5 HI + c_6 CHI_3 + c_7 HCOONa Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O, I and C: H: | c_1 + 6 c_3 = 2 c_4 + c_5 + c_6 + c_7 Na: | c_1 = c_7 O: | c_1 + c_3 = c_4 + 2 c_7 I: | 2 c_2 = c_5 + 3 c_6 C: | 2 c_3 = c_6 + 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_4 = 1 and solve the system of equations for the remaining coefficients: c_2 = 4 c_1 + 4 c_3 = c_1 + 1 c_4 = 1 c_5 = 5 c_1 + 2 c_6 = c_1 + 2 c_7 = c_1 The resulting system of equations is still underdetermined, so an additional coefficient must be set arbitrarily. Set c_1 = 2 and solve for the remaining coefficients: c_1 = 2 c_2 = 12 c_3 = 3 c_4 = 1 c_5 = 12 c_6 = 4 c_7 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NaOH + 12 I_2 + 3 CH_3CH_2OH ⟶ H_2O + 12 HI + 4 CHI_3 + 2 HCOONa

+ + ⟶ + + +

sodium hydroxide + iodine + ethanol ⟶ water + hydrogen iodide + iodoform + sodium formate

| sodium hydroxide | iodine | ethanol | water | hydrogen iodide | iodoform | sodium formate molecular enthalpy | -425.8 kJ/mol | 0 kJ/mol | -277.7 kJ/mol | -285.8 kJ/mol | 26.5 kJ/mol | -181.1 kJ/mol | -666.5 kJ/mol total enthalpy | -851.6 kJ/mol | 0 kJ/mol | -833.1 kJ/mol | -285.8 kJ/mol | 318 kJ/mol | -724.4 kJ/mol | -1333 kJ/mol | H_initial = -1685 kJ/mol | | | H_final = -2025 kJ/mol | | | ΔH_rxn^0 | -2025 kJ/mol - -1685 kJ/mol = -340.6 kJ/mol (exothermic) | | | | | |

Construct the equilibrium constant, K, expression for: NaOH + I_2 + CH_3CH_2OH ⟶ H_2O + HI + CHI_3 + HCOONa 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: 2 NaOH + 12 I_2 + 3 CH_3CH_2OH ⟶ H_2O + 12 HI + 4 CHI_3 + 2 HCOONa 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 NaOH | 2 | -2 I_2 | 12 | -12 CH_3CH_2OH | 3 | -3 H_2O | 1 | 1 HI | 12 | 12 CHI_3 | 4 | 4 HCOONa | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 2 | -2 | ([NaOH])^(-2) I_2 | 12 | -12 | ([I2])^(-12) CH_3CH_2OH | 3 | -3 | ([CH3CH2OH])^(-3) H_2O | 1 | 1 | [H2O] HI | 12 | 12 | ([HI])^12 CHI_3 | 4 | 4 | ([CHI3])^4 HCOONa | 2 | 2 | ([HCOONa])^2 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 = ([NaOH])^(-2) ([I2])^(-12) ([CH3CH2OH])^(-3) [H2O] ([HI])^12 ([CHI3])^4 ([HCOONa])^2 = ([H2O] ([HI])^12 ([CHI3])^4 ([HCOONa])^2)/(([NaOH])^2 ([I2])^12 ([CH3CH2OH])^3)

Construct the rate of reaction expression for: NaOH + I_2 + CH_3CH_2OH ⟶ H_2O + HI + CHI_3 + HCOONa 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: 2 NaOH + 12 I_2 + 3 CH_3CH_2OH ⟶ H_2O + 12 HI + 4 CHI_3 + 2 HCOONa 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 NaOH | 2 | -2 I_2 | 12 | -12 CH_3CH_2OH | 3 | -3 H_2O | 1 | 1 HI | 12 | 12 CHI_3 | 4 | 4 HCOONa | 2 | 2 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 NaOH | 2 | -2 | -1/2 (Δ[NaOH])/(Δt) I_2 | 12 | -12 | -1/12 (Δ[I2])/(Δt) CH_3CH_2OH | 3 | -3 | -1/3 (Δ[CH3CH2OH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) HI | 12 | 12 | 1/12 (Δ[HI])/(Δt) CHI_3 | 4 | 4 | 1/4 (Δ[CHI3])/(Δt) HCOONa | 2 | 2 | 1/2 (Δ[HCOONa])/(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 = -1/2 (Δ[NaOH])/(Δt) = -1/12 (Δ[I2])/(Δt) = -1/3 (Δ[CH3CH2OH])/(Δt) = (Δ[H2O])/(Δt) = 1/12 (Δ[HI])/(Δt) = 1/4 (Δ[CHI3])/(Δt) = 1/2 (Δ[HCOONa])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | iodine | ethanol | water | hydrogen iodide | iodoform | sodium formate formula | NaOH | I_2 | CH_3CH_2OH | H_2O | HI | CHI_3 | HCOONa Hill formula | HNaO | I_2 | C_2H_6O | H_2O | HI | CHI_3 | CHNaO_2 name | sodium hydroxide | iodine | ethanol | water | hydrogen iodide | iodoform | sodium formate IUPAC name | sodium hydroxide | molecular iodine | ethanol | water | hydrogen iodide | iodoform | sodium oxomethanolate