Ca + Nb2O5 = CaO + Nb

Ca calcium + Nb_2O_5 niobium pentoxide ⟶ CaO lime + Nb niobium

Balance the chemical equation algebraically: Ca + Nb_2O_5 ⟶ CaO + Nb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Ca + c_2 Nb_2O_5 ⟶ c_3 CaO + c_4 Nb Set the number of atoms in the reactants equal to the number of atoms in the products for Ca, Nb and O: Ca: | c_1 = c_3 Nb: | 2 c_2 = c_4 O: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 5 c_2 = 1 c_3 = 5 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 5 Ca + Nb_2O_5 ⟶ 5 CaO + 2 Nb

+ ⟶ +

calcium + niobium pentoxide ⟶ lime + niobium

| calcium | niobium pentoxide | lime | niobium molecular enthalpy | 0 kJ/mol | -1900 kJ/mol | -634.9 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | -1900 kJ/mol | -3175 kJ/mol | 0 kJ/mol | H_initial = -1900 kJ/mol | | H_final = -3175 kJ/mol | ΔH_rxn^0 | -3175 kJ/mol - -1900 kJ/mol = -1275 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: Ca + Nb_2O_5 ⟶ CaO + Nb 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 Ca + Nb_2O_5 ⟶ 5 CaO + 2 Nb 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 Ca | 5 | -5 Nb_2O_5 | 1 | -1 CaO | 5 | 5 Nb | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Ca | 5 | -5 | ([Ca])^(-5) Nb_2O_5 | 1 | -1 | ([Nb2O5])^(-1) CaO | 5 | 5 | ([CaO])^5 Nb | 2 | 2 | ([Nb])^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 = ([Ca])^(-5) ([Nb2O5])^(-1) ([CaO])^5 ([Nb])^2 = (([CaO])^5 ([Nb])^2)/(([Ca])^5 [Nb2O5])

Construct the rate of reaction expression for: Ca + Nb_2O_5 ⟶ CaO + Nb 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 Ca + Nb_2O_5 ⟶ 5 CaO + 2 Nb 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 Ca | 5 | -5 Nb_2O_5 | 1 | -1 CaO | 5 | 5 Nb | 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 Ca | 5 | -5 | -1/5 (Δ[Ca])/(Δt) Nb_2O_5 | 1 | -1 | -(Δ[Nb2O5])/(Δt) CaO | 5 | 5 | 1/5 (Δ[CaO])/(Δt) Nb | 2 | 2 | 1/2 (Δ[Nb])/(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[Ca])/(Δt) = -(Δ[Nb2O5])/(Δt) = 1/5 (Δ[CaO])/(Δt) = 1/2 (Δ[Nb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| calcium | niobium pentoxide | lime | niobium formula | Ca | Nb_2O_5 | CaO | Nb name | calcium | niobium pentoxide | lime | niobium

| calcium | niobium pentoxide | lime | niobium molar mass | 40.078 g/mol | 265.808 g/mol | 56.077 g/mol | 92.90637 g/mol phase | solid (at STP) | solid (at STP) | solid (at STP) | solid (at STP) melting point | 850 °C | 1496 °C | 2580 °C | 2468 °C boiling point | 1484 °C | | 2850 °C | 4742 °C density | 1.54 g/cm^3 | 4.47 g/cm^3 | 3.3 g/cm^3 | 8.57 g/cm^3 solubility in water | decomposes | | reacts | insoluble