O2 + Ca = CaO

O_2 (oxygen) + Ca (calcium) ⟶ CaO (lime)

Balance the chemical equation algebraically: O_2 + Ca ⟶ CaO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 O_2 + c_2 Ca ⟶ c_3 CaO Set the number of atoms in the reactants equal to the number of atoms in the products for O and Ca: O: | 2 c_1 = c_3 Ca: | 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | O_2 + 2 Ca ⟶ 2 CaO

+ ⟶

oxygen + calcium ⟶ lime

| oxygen | calcium | lime molecular enthalpy | 0 kJ/mol | 0 kJ/mol | -634.9 kJ/mol total enthalpy | 0 kJ/mol | 0 kJ/mol | -1270 kJ/mol | H_initial = 0 kJ/mol | | H_final = -1270 kJ/mol ΔH_rxn^0 | -1270 kJ/mol - 0 kJ/mol = -1270 kJ/mol (exothermic) | |

| oxygen | calcium | lime molecular entropy | 205 J/(mol K) | 41 J/(mol K) | 40 J/(mol K) total entropy | 205 J/(mol K) | 82 J/(mol K) | 80 J/(mol K) | S_initial = 287 J/(mol K) | | S_final = 80 J/(mol K) ΔS_rxn^0 | 80 J/(mol K) - 287 J/(mol K) = -207 J/(mol K) (exoentropic) | |

Construct the equilibrium constant, K, expression for: O_2 + Ca ⟶ CaO 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: O_2 + 2 Ca ⟶ 2 CaO 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 | 1 | -1 Ca | 2 | -2 CaO | 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 | 1 | -1 | ([O2])^(-1) Ca | 2 | -2 | ([Ca])^(-2) CaO | 2 | 2 | ([CaO])^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])^(-1) ([Ca])^(-2) ([CaO])^2 = ([CaO])^2/([O2] ([Ca])^2)

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

| oxygen | calcium | lime formula | O_2 | Ca | CaO name | oxygen | calcium | lime IUPAC name | molecular oxygen | calcium |

| oxygen | calcium | lime molar mass | 31.998 g/mol | 40.078 g/mol | 56.077 g/mol phase | gas (at STP) | solid (at STP) | solid (at STP) melting point | -218 °C | 850 °C | 2580 °C boiling point | -183 °C | 1484 °C | 2850 °C density | 0.001429 g/cm^3 (at 0 °C) | 1.54 g/cm^3 | 3.3 g/cm^3 solubility in water | | decomposes | reacts surface tension | 0.01347 N/m | | dynamic viscosity | 2.055×10^-5 Pa s (at 25 °C) | | odor | odorless | |