HNO3 + Na = H2O + NH3 + NaNO3

HNO_3 nitric acid + Na sodium ⟶ H_2O water + NH_3 ammonia + NaNO_3 sodium nitrate

Balance the chemical equation algebraically: HNO_3 + Na ⟶ H_2O + NH_3 + NaNO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Na ⟶ c_3 H_2O + c_4 NH_3 + c_5 NaNO_3 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 + 3 c_4 N: | c_1 = c_4 + c_5 O: | 3 c_1 = c_3 + 3 c_5 Na: | c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 9 c_2 = 8 c_3 = 3 c_4 = 1 c_5 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 HNO_3 + 8 Na ⟶ 3 H_2O + NH_3 + 8 NaNO_3

+ ⟶ + +

nitric acid + sodium ⟶ water + ammonia + sodium nitrate

| nitric acid | sodium | water | ammonia | sodium nitrate molecular entropy | 156 J/(mol K) | 51 J/(mol K) | 69.91 J/(mol K) | 193 J/(mol K) | 116 J/(mol K) total entropy | 1404 J/(mol K) | 408 J/(mol K) | 209.7 J/(mol K) | 193 J/(mol K) | 928 J/(mol K) | S_initial = 1812 J/(mol K) | | S_final = 1331 J/(mol K) | | ΔS_rxn^0 | 1331 J/(mol K) - 1812 J/(mol K) = -481.3 J/(mol K) (exoentropic) | | | |

K_c = ([H2O]^3 [NH3] [NaNO3]^8)/([HNO3]^9 [Na]^8)

rate = -1/9 (Δ[HNO3])/(Δt) = -1/8 (Δ[Na])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[NH3])/(Δt) = 1/8 (Δ[NaNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | sodium | water | ammonia | sodium nitrate formula | HNO_3 | Na | H_2O | NH_3 | NaNO_3 Hill formula | HNO_3 | Na | H_2O | H_3N | NNaO_3 name | nitric acid | sodium | water | ammonia | sodium nitrate