NH3 + P4S10 = (NH4)2S + P3N5

NH_3 ammonia + P_2S_5 phosphorus pentasulfide ⟶ (NH_4)_2S diammonium sulfide + P3N5

Balance the chemical equation algebraically: NH_3 + P_2S_5 ⟶ (NH_4)_2S + P3N5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 P_2S_5 ⟶ c_3 (NH_4)_2S + c_4 P3N5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, P and S: H: | 3 c_1 = 8 c_3 N: | c_1 = 2 c_3 + 5 c_4 P: | 2 c_2 = 3 c_4 S: | 5 c_2 = c_3 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 = 20 c_2 = 3/2 c_3 = 15/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 40 c_2 = 3 c_3 = 15 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 40 NH_3 + 3 P_2S_5 ⟶ 15 (NH_4)_2S + 2 P3N5

+ ⟶ + P3N5

ammonia + phosphorus pentasulfide ⟶ diammonium sulfide + P3N5

Construct the equilibrium constant, K, expression for: NH_3 + P_2S_5 ⟶ (NH_4)_2S + P3N5 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: 40 NH_3 + 3 P_2S_5 ⟶ 15 (NH_4)_2S + 2 P3N5 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 NH_3 | 40 | -40 P_2S_5 | 3 | -3 (NH_4)_2S | 15 | 15 P3N5 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_3 | 40 | -40 | ([NH3])^(-40) P_2S_5 | 3 | -3 | ([P2S5])^(-3) (NH_4)_2S | 15 | 15 | ([(NH4)2S])^15 P3N5 | 2 | 2 | ([P3N5])^2 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 = ([NH3])^(-40) ([P2S5])^(-3) ([(NH4)2S])^15 ([P3N5])^2 = (([(NH4)2S])^15 ([P3N5])^2)/(([NH3])^40 ([P2S5])^3)

Construct the rate of reaction expression for: NH_3 + P_2S_5 ⟶ (NH_4)_2S + P3N5 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: 40 NH_3 + 3 P_2S_5 ⟶ 15 (NH_4)_2S + 2 P3N5 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 NH_3 | 40 | -40 P_2S_5 | 3 | -3 (NH_4)_2S | 15 | 15 P3N5 | 2 | 2 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 NH_3 | 40 | -40 | -1/40 (Δ[NH3])/(Δt) P_2S_5 | 3 | -3 | -1/3 (Δ[P2S5])/(Δt) (NH_4)_2S | 15 | 15 | 1/15 (Δ[(NH4)2S])/(Δt) P3N5 | 2 | 2 | 1/2 (Δ[P3N5])/(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 = -1/40 (Δ[NH3])/(Δt) = -1/3 (Δ[P2S5])/(Δt) = 1/15 (Δ[(NH4)2S])/(Δt) = 1/2 (Δ[P3N5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| ammonia | phosphorus pentasulfide | diammonium sulfide | P3N5 formula | NH_3 | P_2S_5 | (NH_4)_2S | P3N5 Hill formula | H_3N | P_2S_5 | H_8N_2S | N5P3 name | ammonia | phosphorus pentasulfide | diammonium sulfide |

| ammonia | phosphorus pentasulfide | diammonium sulfide | P3N5 molar mass | 17.031 g/mol | 222.3 g/mol | 68.14 g/mol | 162.96 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | melting point | -77.73 °C | 282 °C | -18 °C | boiling point | -33.33 °C | | | density | 6.96×10^-4 g/cm^3 (at 25 °C) | 2.09 g/cm^3 | 0.997 g/cm^3 | solubility in water | | | very soluble | surface tension | 0.0234 N/m | | | dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | | |