8H2O = H2O

8 H_2O water ⟶ H_2O water

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

⟶

water ⟶ water

| water | water molecular enthalpy | -285.8 kJ/mol | -285.8 kJ/mol total enthalpy | -285.8 kJ/mol | -285.8 kJ/mol | H_initial = -285.8 kJ/mol | H_final = -285.8 kJ/mol ΔH_rxn^0 | -285.8 kJ/mol - -285.8 kJ/mol = 0 kJ/mol (equilibrium) |

| water | water molecular free energy | -237.1 kJ/mol | -237.1 kJ/mol total free energy | -237.1 kJ/mol | -237.1 kJ/mol | G_initial = -237.1 kJ/mol | G_final = -237.1 kJ/mol ΔG_rxn^0 | -237.1 kJ/mol - -237.1 kJ/mol = 0 kJ/mol (equilibrium) |

| water | water molecular entropy | 69.91 J/(mol K) | 69.91 J/(mol K) total entropy | 69.91 J/(mol K) | 69.91 J/(mol K) | S_initial = 69.91 J/(mol K) | S_final = 69.91 J/(mol K) ΔS_rxn^0 | 69.91 J/(mol K) - 69.91 J/(mol K) = 0 J/(mol K) (equilibrium) |

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

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

| water | water formula | H_2O | H_2O name | water | water

| water | water molar mass | 18.015 g/mol | 18.015 g/mol phase | liquid (at STP) | liquid (at STP) melting point | 0 °C | 0 °C boiling point | 99.9839 °C | 99.9839 °C density | 1 g/cm^3 | 1 g/cm^3 surface tension | 0.0728 N/m | 0.0728 N/m dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) odor | odorless | odorless