P2O3 = O2 + P

P_2O_3 phosphorus trioxide ⟶ O_2 oxygen + P red phosphorus

Balance the chemical equation algebraically: P_2O_3 ⟶ O_2 + P Add stoichiometric coefficients, c_i, to the reactants and products: c_1 P_2O_3 ⟶ c_2 O_2 + c_3 P Set the number of atoms in the reactants equal to the number of atoms in the products for O and P: O: | 3 c_1 = 2 c_2 P: | 2 c_1 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 P_2O_3 ⟶ 3 O_2 + 4 P

⟶ +

phosphorus trioxide ⟶ oxygen + red phosphorus

Construct the equilibrium constant, K, expression for: P_2O_3 ⟶ O_2 + P 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: 2 P_2O_3 ⟶ 3 O_2 + 4 P 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 P_2O_3 | 2 | -2 O_2 | 3 | 3 P | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression P_2O_3 | 2 | -2 | ([P2O3])^(-2) O_2 | 3 | 3 | ([O2])^3 P | 4 | 4 | ([P])^4 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 = ([P2O3])^(-2) ([O2])^3 ([P])^4 = (([O2])^3 ([P])^4)/([P2O3])^2

Construct the rate of reaction expression for: P_2O_3 ⟶ O_2 + P 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: 2 P_2O_3 ⟶ 3 O_2 + 4 P 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 P_2O_3 | 2 | -2 O_2 | 3 | 3 P | 4 | 4 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 P_2O_3 | 2 | -2 | -1/2 (Δ[P2O3])/(Δt) O_2 | 3 | 3 | 1/3 (Δ[O2])/(Δt) P | 4 | 4 | 1/4 (Δ[P])/(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 = -1/2 (Δ[P2O3])/(Δt) = 1/3 (Δ[O2])/(Δt) = 1/4 (Δ[P])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| phosphorus trioxide | oxygen | red phosphorus formula | P_2O_3 | O_2 | P Hill formula | O_3P_2 | O_2 | P name | phosphorus trioxide | oxygen | red phosphorus IUPAC name | | molecular oxygen | phosphorus

| phosphorus trioxide | oxygen | red phosphorus molar mass | 109.94 g/mol | 31.998 g/mol | 30.973761998 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 23.8 °C | -218 °C | 579.2 °C boiling point | 173.1 °C | -183 °C | density | 2.135 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 2.16 g/cm^3 solubility in water | decomposes | | insoluble surface tension | | 0.01347 N/m | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 7.6×10^-4 Pa s (at 20.2 °C) odor | | odorless |