NaOH + HClO3 = H2O + NaClO3

NaOH sodium hydroxide + HClO3 ⟶ H_2O water + NaClO_3 sodium chlorate

Balance the chemical equation algebraically: NaOH + HClO3 ⟶ H_2O + NaClO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NaOH + c_2 HClO3 ⟶ c_3 H_2O + c_4 NaClO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, Na, O and Cl: H: | c_1 + c_2 = 2 c_3 Na: | c_1 = c_4 O: | c_1 + 3 c_2 = c_3 + 3 c_4 Cl: | c_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: | | NaOH + HClO3 ⟶ H_2O + NaClO_3

+ HClO3 ⟶ +

sodium hydroxide + HClO3 ⟶ water + sodium chlorate

Construct the equilibrium constant, K, expression for: NaOH + HClO3 ⟶ H_2O + NaClO_3 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: NaOH + HClO3 ⟶ H_2O + NaClO_3 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 | 1 | -1 HClO3 | 1 | -1 H_2O | 1 | 1 NaClO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NaOH | 1 | -1 | ([NaOH])^(-1) HClO3 | 1 | -1 | ([HClO3])^(-1) H_2O | 1 | 1 | [H2O] NaClO_3 | 1 | 1 | [NaClO3] 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])^(-1) ([HClO3])^(-1) [H2O] [NaClO3] = ([H2O] [NaClO3])/([NaOH] [HClO3])

Construct the rate of reaction expression for: NaOH + HClO3 ⟶ H_2O + NaClO_3 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: NaOH + HClO3 ⟶ H_2O + NaClO_3 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 | 1 | -1 HClO3 | 1 | -1 H_2O | 1 | 1 NaClO_3 | 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 NaOH | 1 | -1 | -(Δ[NaOH])/(Δt) HClO3 | 1 | -1 | -(Δ[HClO3])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) NaClO_3 | 1 | 1 | (Δ[NaClO3])/(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[NaOH])/(Δt) = -(Δ[HClO3])/(Δt) = (Δ[H2O])/(Δt) = (Δ[NaClO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| sodium hydroxide | HClO3 | water | sodium chlorate formula | NaOH | HClO3 | H_2O | NaClO_3 Hill formula | HNaO | HClO3 | H_2O | ClNaO_3 name | sodium hydroxide | | water | sodium chlorate

| sodium hydroxide | HClO3 | water | sodium chlorate molar mass | 39.997 g/mol | 84.45 g/mol | 18.015 g/mol | 106.4 g/mol phase | solid (at STP) | | liquid (at STP) | liquid (at STP) melting point | 323 °C | | 0 °C | boiling point | 1390 °C | | 99.9839 °C | 106 °C density | 2.13 g/cm^3 | | 1 g/cm^3 | 1.3 g/cm^3 solubility in water | soluble | | | very soluble surface tension | 0.07435 N/m | | 0.0728 N/m | dynamic viscosity | 0.004 Pa s (at 350 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 0.00542 Pa s (at 286 °C) odor | | | odorless | odorless