NO + PH3 = N2 + H3PO4

NO nitric oxide + PH_3 phosphine ⟶ N_2 nitrogen + H_3PO_4 phosphoric acid

Balance the chemical equation algebraically: NO + PH_3 ⟶ N_2 + H_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NO + c_2 PH_3 ⟶ c_3 N_2 + c_4 H_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for N, O, H and P: N: | c_1 = 2 c_3 O: | c_1 = 4 c_4 H: | 3 c_2 = 3 c_4 P: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 NO + PH_3 ⟶ 2 N_2 + H_3PO_4

+ ⟶ +

nitric oxide + phosphine ⟶ nitrogen + phosphoric acid

| nitric oxide | phosphine | nitrogen | phosphoric acid molecular free energy | 87.6 kJ/mol | 13.5 kJ/mol | 0 kJ/mol | -1124 kJ/mol total free energy | 350.4 kJ/mol | 13.5 kJ/mol | 0 kJ/mol | -1124 kJ/mol | G_initial = 363.9 kJ/mol | | G_final = -1124 kJ/mol | ΔG_rxn^0 | -1124 kJ/mol - 363.9 kJ/mol = -1488 kJ/mol (exergonic) | | |

Construct the equilibrium constant, K, expression for: NO + PH_3 ⟶ N_2 + H_3PO_4 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: 4 NO + PH_3 ⟶ 2 N_2 + H_3PO_4 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 NO | 4 | -4 PH_3 | 1 | -1 N_2 | 2 | 2 H_3PO_4 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NO | 4 | -4 | ([NO])^(-4) PH_3 | 1 | -1 | ([PH3])^(-1) N_2 | 2 | 2 | ([N2])^2 H_3PO_4 | 1 | 1 | [H3PO4] 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 = ([NO])^(-4) ([PH3])^(-1) ([N2])^2 [H3PO4] = (([N2])^2 [H3PO4])/(([NO])^4 [PH3])

Construct the rate of reaction expression for: NO + PH_3 ⟶ N_2 + H_3PO_4 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: 4 NO + PH_3 ⟶ 2 N_2 + H_3PO_4 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 NO | 4 | -4 PH_3 | 1 | -1 N_2 | 2 | 2 H_3PO_4 | 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 NO | 4 | -4 | -1/4 (Δ[NO])/(Δt) PH_3 | 1 | -1 | -(Δ[PH3])/(Δt) N_2 | 2 | 2 | 1/2 (Δ[N2])/(Δt) H_3PO_4 | 1 | 1 | (Δ[H3PO4])/(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 = -1/4 (Δ[NO])/(Δt) = -(Δ[PH3])/(Δt) = 1/2 (Δ[N2])/(Δt) = (Δ[H3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric oxide | phosphine | nitrogen | phosphoric acid formula | NO | PH_3 | N_2 | H_3PO_4 Hill formula | NO | H_3P | N_2 | H_3O_4P name | nitric oxide | phosphine | nitrogen | phosphoric acid IUPAC name | nitric oxide | phosphine | molecular nitrogen | phosphoric acid

| nitric oxide | phosphine | nitrogen | phosphoric acid molar mass | 30.006 g/mol | 33.998 g/mol | 28.014 g/mol | 97.994 g/mol phase | gas (at STP) | gas (at STP) | gas (at STP) | liquid (at STP) melting point | -163.6 °C | -132.8 °C | -210 °C | 42.4 °C boiling point | -151.7 °C | -87.5 °C | -195.79 °C | 158 °C density | 0.001226 g/cm^3 (at 25 °C) | 0.00139 g/cm^3 (at 25 °C) | 0.001251 g/cm^3 (at 0 °C) | 1.685 g/cm^3 solubility in water | | slightly soluble | insoluble | very soluble surface tension | | | 0.0066 N/m | dynamic viscosity | 1.911×10^-5 Pa s (at 25 °C) | 1.1×10^-5 Pa s (at 0 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | | odorless | odorless