O2 + V = V2O5

O_2 oxygen + V vanadium ⟶ V_2O_5 vanadium pentoxide

Balance the chemical equation algebraically: O_2 + V ⟶ V_2O_5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 V ⟶ c_3 V_2O_5 Set the number of atoms in the reactants equal to the number of atoms in the products for O and V: O: | 2 c_1 = 5 c_3 V: | c_2 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5/2 c_2 = 2 c_3 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 5 c_2 = 4 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 O_2 + 4 V ⟶ 2 V_2O_5

+ ⟶

oxygen + vanadium ⟶ vanadium pentoxide

| oxygen | vanadium | vanadium pentoxide molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -1551 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -3101 kJ/mol | H_initial = 0 kJ/mol | | H_final = -3101 kJ/mol ΔH_rxn^0 | -3101 kJ/mol - 0 kJ/mol = -3101 kJ/mol (exothermic) | |

Construct the equilibrium constant, K, expression for: O_2 + V ⟶ V_2O_5 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 + 4 V ⟶ 2 V_2O_5 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 V | 4 | -4 V_2O_5 | 2 | 2 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) V | 4 | -4 | ([V])^(-4) V_2O_5 | 2 | 2 | ([V2O5])^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 = ([O2])^(-5) ([V])^(-4) ([V2O5])^2 = ([V2O5])^2/(([O2])^5 ([V])^4)

Construct the rate of reaction expression for: O_2 + V ⟶ V_2O_5 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 + 4 V ⟶ 2 V_2O_5 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 V | 4 | -4 V_2O_5 | 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 O_2 | 5 | -5 | -1/5 (Δ[O2])/(Δt) V | 4 | -4 | -1/4 (Δ[V])/(Δt) V_2O_5 | 2 | 2 | 1/2 (Δ[V2O5])/(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) = -1/4 (Δ[V])/(Δt) = 1/2 (Δ[V2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| oxygen | vanadium | vanadium pentoxide formula | O_2 | V | V_2O_5 Hill formula | O_2 | V | O_5V_2 name | oxygen | vanadium | vanadium pentoxide IUPAC name | molecular oxygen | vanadium |

| oxygen | vanadium | vanadium pentoxide molar mass | 31.998 g/mol | 50.9415 g/mol | 181.88 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 1890 °C | 690 °C boiling point | -183 °C | 3380 °C | 1750 °C density | 0.001429 g/cm^3 (at 0 °C) | 6.11 g/cm^3 | 3.35 g/cm^3 solubility in water | | insoluble | surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | |