CaO = O2 + Ca

CaO lime ⟶ O_2 oxygen + Ca calcium

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

⟶ +

lime ⟶ oxygen + calcium

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

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

Construct the equilibrium constant, K, expression for: CaO ⟶ O_2 + Ca 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 CaO ⟶ O_2 + 2 Ca 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 CaO | 2 | -2 O_2 | 1 | 1 Ca | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression CaO | 2 | -2 | ([CaO])^(-2) O_2 | 1 | 1 | [O2] Ca | 2 | 2 | ([Ca])^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 = ([CaO])^(-2) [O2] ([Ca])^2 = ([O2] ([Ca])^2)/([CaO])^2

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

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

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