HCOOH + C6H5OH = H2O + C6H5COOH

HCOOH formic acid + C_6H_5OH phenol ⟶ H_2O water + C_6H_5COOH benzoic acid

Balance the chemical equation algebraically: HCOOH + C_6H_5OH ⟶ H_2O + C_6H_5COOH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCOOH + c_2 C_6H_5OH ⟶ c_3 H_2O + c_4 C_6H_5COOH Set the number of atoms in the reactants equal to the number of atoms in the products for C, H and O: C: | c_1 + 6 c_2 = 7 c_4 H: | 2 c_1 + 6 c_2 = 2 c_3 + 6 c_4 O: | 2 c_1 + c_2 = c_3 + 2 c_4 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 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | HCOOH + C_6H_5OH ⟶ H_2O + C_6H_5COOH

+ ⟶ +

formic acid + phenol ⟶ water + benzoic acid

| formic acid | phenol | water | benzoic acid molecular enthalpy | -425 kJ/mol | -165.1 kJ/mol | -285.8 kJ/mol | -385.2 kJ/mol total enthalpy | -425 kJ/mol | -165.1 kJ/mol | -285.8 kJ/mol | -385.2 kJ/mol | H_initial = -590.1 kJ/mol | | H_final = -671 kJ/mol | ΔH_rxn^0 | -671 kJ/mol - -590.1 kJ/mol = -80.93 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: HCOOH + C_6H_5OH ⟶ H_2O + C_6H_5COOH 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: HCOOH + C_6H_5OH ⟶ H_2O + C_6H_5COOH 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 HCOOH | 1 | -1 C_6H_5OH | 1 | -1 H_2O | 1 | 1 C_6H_5COOH | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCOOH | 1 | -1 | ([HCOOH])^(-1) C_6H_5OH | 1 | -1 | ([C6H5OH])^(-1) H_2O | 1 | 1 | [H2O] C_6H_5COOH | 1 | 1 | [C6H5COOH] 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 = ([HCOOH])^(-1) ([C6H5OH])^(-1) [H2O] [C6H5COOH] = ([H2O] [C6H5COOH])/([HCOOH] [C6H5OH])

Construct the rate of reaction expression for: HCOOH + C_6H_5OH ⟶ H_2O + C_6H_5COOH 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: HCOOH + C_6H_5OH ⟶ H_2O + C_6H_5COOH 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 HCOOH | 1 | -1 C_6H_5OH | 1 | -1 H_2O | 1 | 1 C_6H_5COOH | 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 HCOOH | 1 | -1 | -(Δ[HCOOH])/(Δt) C_6H_5OH | 1 | -1 | -(Δ[C6H5OH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) C_6H_5COOH | 1 | 1 | (Δ[C6H5COOH])/(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[HCOOH])/(Δt) = -(Δ[C6H5OH])/(Δt) = (Δ[H2O])/(Δt) = (Δ[C6H5COOH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| formic acid | phenol | water | benzoic acid formula | HCOOH | C_6H_5OH | H_2O | C_6H_5COOH Hill formula | CH_2O_2 | C_6H_6O | H_2O | C_7H_6O_2 name | formic acid | phenol | water | benzoic acid