O2 + P4 = P4O10

O_2 (oxygen) + P_4 (white phosphorus) ⟶ P_4O_10 (phosphorus pentoxide)

Balance the chemical equation algebraically: O_2 + P_4 ⟶ P_4O_10 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 P_4 ⟶ c_3 P_4O_10 Set the number of atoms in the reactants equal to the number of atoms in the products for O and P: O: | 2 c_1 = 10 c_3 P: | 4 c_2 = 4 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 O_2 + P_4 ⟶ P_4O_10

+ ⟶

oxygen + white phosphorus ⟶ phosphorus pentoxide

| oxygen | white phosphorus | phosphorus pentoxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -2984 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -2984 kJ/mol | H_initial = 0 kJ/mol | | H_final = -2984 kJ/mol ΔH_rxn^0 | -2984 kJ/mol - 0 kJ/mol = -2984 kJ/mol (exothermic) | |

| oxygen | white phosphorus | phosphorus pentoxide molecular entropy | 205 J/(mol K) | 41.1 J/(mol K) | 229 J/(mol K) total entropy | 1025 J/(mol K) | 41.1 J/(mol K) | 229 J/(mol K) | S_initial = 1066 J/(mol K) | | S_final = 229 J/(mol K) ΔS_rxn^0 | 229 J/(mol K) - 1066 J/(mol K) = -837.1 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: O_2 + P_4 ⟶ P_4O_10 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: 5 O_2 + P_4 ⟶ P_4O_10 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 O_2 | 5 | -5 P_4 | 1 | -1 P_4O_10 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression O_2 | 5 | -5 | ([O2])^(-5) P_4 | 1 | -1 | ([P4])^(-1) P_4O_10 | 1 | 1 | [P4O10] 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 = ([O2])^(-5) ([P4])^(-1) [P4O10] = ([P4O10])/(([O2])^5 [P4])

Construct the rate of reaction expression for: O_2 + P_4 ⟶ P_4O_10 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: 5 O_2 + P_4 ⟶ P_4O_10 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 O_2 | 5 | -5 P_4 | 1 | -1 P_4O_10 | 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 O_2 | 5 | -5 | -1/5 (Δ[O2])/(Δt) P_4 | 1 | -1 | -(Δ[P4])/(Δt) P_4O_10 | 1 | 1 | (Δ[P4O10])/(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 = -1/5 (Δ[O2])/(Δt) = -(Δ[P4])/(Δt) = (Δ[P4O10])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | white phosphorus | phosphorus pentoxide formula | O_2 | P_4 | P_4O_10 Hill formula | O_2 | P_4 | O_10P_4 name | oxygen | white phosphorus | phosphorus pentoxide IUPAC name | molecular oxygen | tetraphosphorus |

| oxygen | white phosphorus | phosphorus pentoxide molar mass | 31.998 g/mol | 123.89504799 g/mol | 283.89 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 44.15 °C | 340 °C boiling point | -183 °C | 280.5 °C | density | 0.001429 g/cm^3 (at 0 °C) | 1.823 g/cm^3 | 2.3 g/cm^3 solubility in water | | insoluble | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | 0.00169 Pa s (at 50 °C) | odor | odorless | odorless |