HNO3 + Na = H2O + NaNO3 + N2O

nitric acid + sodium ⟶ water + sodium nitrate + nitrous oxide

Balance the chemical equation algebraically: + ⟶ + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 + c_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Na: H: | c_1 = 2 c_3 N: | c_1 = c_4 + 2 c_5 O: | 3 c_1 = c_3 + 3 c_4 + c_5 Na: | 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 8 c_3 = 5 c_4 = 8 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 + 8 ⟶ 5 + 8 +

+ ⟶ + +

nitric acid + sodium ⟶ water + sodium nitrate + nitrous oxide

| nitric acid | sodium | water | sodium nitrate | nitrous oxide molecular entropy | 156 J/(mol K) | 51 J/(mol K) | 69.91 J/(mol K) | 116 J/(mol K) | 220 J/(mol K) total entropy | 1560 J/(mol K) | 408 J/(mol K) | 349.6 J/(mol K) | 928 J/(mol K) | 220 J/(mol K) | S_initial = 1968 J/(mol K) | | S_final = 1498 J/(mol K) | | ΔS_rxn^0 | 1498 J/(mol K) - 1968 J/(mol K) = -470.5 J/(mol K) (exoentropic) | | | |

Construct the equilibrium constant, K, expression for: + ⟶ + + 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: 10 + 8 ⟶ 5 + 8 + 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 | 10 | -10 | 8 | -8 | 5 | 5 | 8 | 8 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression | 10 | -10 | ([HNO3])^(-10) | 8 | -8 | ([Na])^(-8) | 5 | 5 | ([H2O])^5 | 8 | 8 | ([NaNO3])^8 | 1 | 1 | [N2O] 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])^(-10) ([Na])^(-8) ([H2O])^5 ([NaNO3])^8 [N2O] = (([H2O])^5 ([NaNO3])^8 [N2O])/(([HNO3])^10 ([Na])^8)

Construct the rate of reaction expression for: + ⟶ + + 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: 10 + 8 ⟶ 5 + 8 + 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 | 10 | -10 | 8 | -8 | 5 | 5 | 8 | 8 | 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 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) | 8 | -8 | -1/8 (Δ[Na])/(Δt) | 5 | 5 | 1/5 (Δ[H2O])/(Δt) | 8 | 8 | 1/8 (Δ[NaNO3])/(Δt) | 1 | 1 | (Δ[N2O])/(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 = -1/10 (Δ[HNO3])/(Δt) = -1/8 (Δ[Na])/(Δt) = 1/5 (Δ[H2O])/(Δt) = 1/8 (Δ[NaNO3])/(Δt) = (Δ[N2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | sodium | water | sodium nitrate | nitrous oxide Hill formula | HNO_3 | Na | H_2O | NNaO_3 | N_2O name | nitric acid | sodium | water | sodium nitrate | nitrous oxide