KOH + NO2 = H2O + KNO3 + KNO2

KOH (potassium hydroxide) + NO_2 (nitrogen dioxide) ⟶ H_2O (water) + KNO_3 (potassium nitrate) + KNO_2 (potassium nitrite)

Balance the chemical equation algebraically: KOH + NO_2 ⟶ H_2O + KNO_3 + KNO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 KOH + c_2 NO_2 ⟶ c_3 H_2O + c_4 KNO_3 + c_5 KNO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, K, O and N: H: | c_1 = 2 c_3 K: | c_1 = c_4 + c_5 O: | c_1 + 2 c_2 = c_3 + 3 c_4 + 2 c_5 N: | c_2 = c_4 + c_5 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_1 = 2 c_2 = 2 c_3 = 1 c_4 = 1 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 KOH + 2 NO_2 ⟶ H_2O + KNO_3 + KNO_2

+ ⟶ + +

potassium hydroxide + nitrogen dioxide ⟶ water + potassium nitrate + potassium nitrite

| potassium hydroxide | nitrogen dioxide | water | potassium nitrate | potassium nitrite molecular enthalpy | -424.6 kJ/mol | 33.2 kJ/mol | -285.8 kJ/mol | -494.6 kJ/mol | -369.8 kJ/mol total enthalpy | -849.2 kJ/mol | 66.4 kJ/mol | -285.8 kJ/mol | -494.6 kJ/mol | -369.8 kJ/mol | H_initial = -782.8 kJ/mol | | H_final = -1150 kJ/mol | | ΔH_rxn^0 | -1150 kJ/mol - -782.8 kJ/mol = -367.4 kJ/mol (exothermic) | | | |

| potassium hydroxide | nitrogen dioxide | water | potassium nitrate | potassium nitrite molecular free energy | -379.4 kJ/mol | 51.3 kJ/mol | -237.1 kJ/mol | -394.9 kJ/mol | -306.6 kJ/mol total free energy | -758.8 kJ/mol | 102.6 kJ/mol | -237.1 kJ/mol | -394.9 kJ/mol | -306.6 kJ/mol | G_initial = -656.2 kJ/mol | | G_final = -938.6 kJ/mol | | ΔG_rxn^0 | -938.6 kJ/mol - -656.2 kJ/mol = -282.4 kJ/mol (exergonic) | | | |

Construct the equilibrium constant, K, expression for: KOH + NO_2 ⟶ H_2O + KNO_3 + KNO_2 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 KOH + 2 NO_2 ⟶ H_2O + KNO_3 + KNO_2 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 | 2 | -2 NO_2 | 2 | -2 H_2O | 1 | 1 KNO_3 | 1 | 1 KNO_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression KOH | 2 | -2 | ([KOH])^(-2) NO_2 | 2 | -2 | ([NO2])^(-2) H_2O | 1 | 1 | [H2O] KNO_3 | 1 | 1 | [KNO3] KNO_2 | 1 | 1 | [KNO2] 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])^(-2) ([NO2])^(-2) [H2O] [KNO3] [KNO2] = ([H2O] [KNO3] [KNO2])/(([KOH])^2 ([NO2])^2)

Construct the rate of reaction expression for: KOH + NO_2 ⟶ H_2O + KNO_3 + KNO_2 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 KOH + 2 NO_2 ⟶ H_2O + KNO_3 + KNO_2 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 | 2 | -2 NO_2 | 2 | -2 H_2O | 1 | 1 KNO_3 | 1 | 1 KNO_2 | 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 | 2 | -2 | -1/2 (Δ[KOH])/(Δt) NO_2 | 2 | -2 | -1/2 (Δ[NO2])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KNO_3 | 1 | 1 | (Δ[KNO3])/(Δt) KNO_2 | 1 | 1 | (Δ[KNO2])/(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[KOH])/(Δt) = -1/2 (Δ[NO2])/(Δt) = (Δ[H2O])/(Δt) = (Δ[KNO3])/(Δt) = (Δ[KNO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| potassium hydroxide | nitrogen dioxide | water | potassium nitrate | potassium nitrite formula | KOH | NO_2 | H_2O | KNO_3 | KNO_2 Hill formula | HKO | NO_2 | H_2O | KNO_3 | KNO_2 name | potassium hydroxide | nitrogen dioxide | water | potassium nitrate | potassium nitrite IUPAC name | potassium hydroxide | Nitrogen dioxide | water | potassium nitrate | potassium nitrite

| potassium hydroxide | nitrogen dioxide | water | potassium nitrate | potassium nitrite molar mass | 56.105 g/mol | 46.005 g/mol | 18.015 g/mol | 101.1 g/mol | 85.103 g/mol phase | solid (at STP) | gas (at STP) | liquid (at STP) | solid (at STP) | solid (at STP) melting point | 406 °C | -11 °C | 0 °C | 334 °C | 350 °C boiling point | 1327 °C | 21 °C | 99.9839 °C | | density | 2.044 g/cm^3 | 0.00188 g/cm^3 (at 25 °C) | 1 g/cm^3 | | 1.915 g/cm^3 solubility in water | soluble | reacts | | soluble | surface tension | | | 0.0728 N/m | | dynamic viscosity | 0.001 Pa s (at 550 °C) | 4.02×10^-4 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | | odor | | | odorless | odorless |