HNO3 + KOH = H2O + KNO3

HNO_3 (nitric acid) + KOH (potassium hydroxide) ⟶ H_2O (water) + KNO_3 (potassium nitrate)

Balance the chemical equation algebraically: HNO_3 + KOH ⟶ H_2O + KNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 KOH ⟶ c_3 H_2O + c_4 KNO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and K: H: | c_1 + c_2 = 2 c_3 N: | c_1 = c_4 O: | 3 c_1 + c_2 = c_3 + 3 c_4 K: | 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: | | HNO_3 + KOH ⟶ H_2O + KNO_3

+ ⟶ +

nitric acid + potassium hydroxide ⟶ water + potassium nitrate

| nitric acid | potassium hydroxide | water | potassium nitrate molecular free energy | -80.7 kJ/mol | -379.4 kJ/mol | -237.1 kJ/mol | -394.9 kJ/mol total free energy | -80.7 kJ/mol | -379.4 kJ/mol | -237.1 kJ/mol | -394.9 kJ/mol | G_initial = -460.1 kJ/mol | | G_final = -632 kJ/mol | ΔG_rxn^0 | -632 kJ/mol - -460.1 kJ/mol = -171.9 kJ/mol (exergonic) | | |

Construct the equilibrium constant, K, expression for: HNO_3 + KOH ⟶ H_2O + KNO_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: HNO_3 + KOH ⟶ H_2O + KNO_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 HNO_3 | 1 | -1 KOH | 1 | -1 H_2O | 1 | 1 KNO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 1 | -1 | ([HNO3])^(-1) KOH | 1 | -1 | ([KOH])^(-1) H_2O | 1 | 1 | [H2O] KNO_3 | 1 | 1 | [KNO3] 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 = ([HNO3])^(-1) ([KOH])^(-1) [H2O] [KNO3] = ([H2O] [KNO3])/([HNO3] [KOH])

Construct the rate of reaction expression for: HNO_3 + KOH ⟶ H_2O + KNO_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: HNO_3 + KOH ⟶ H_2O + KNO_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 HNO_3 | 1 | -1 KOH | 1 | -1 H_2O | 1 | 1 KNO_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 HNO_3 | 1 | -1 | -(Δ[HNO3])/(Δt) KOH | 1 | -1 | -(Δ[KOH])/(Δt) H_2O | 1 | 1 | (Δ[H2O])/(Δt) KNO_3 | 1 | 1 | (Δ[KNO3])/(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[HNO3])/(Δt) = -(Δ[KOH])/(Δt) = (Δ[H2O])/(Δt) = (Δ[KNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | potassium hydroxide | water | potassium nitrate formula | HNO_3 | KOH | H_2O | KNO_3 Hill formula | HNO_3 | HKO | H_2O | KNO_3 name | nitric acid | potassium hydroxide | water | potassium nitrate