ZnO = O2 + Zn

ZnO zinc oxide ⟶ O_2 oxygen + Zn zinc

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

⟶ +

zinc oxide ⟶ oxygen + zinc

| zinc oxide | oxygen | zinc molecular enthalpy | -350.5 kJ/mol | 0 kJ/mol | 0 kJ/mol total enthalpy | -701 kJ/mol | 0 kJ/mol | 0 kJ/mol | H_initial = -701 kJ/mol | H_final = 0 kJ/mol | ΔH_rxn^0 | 0 kJ/mol - -701 kJ/mol = 701 kJ/mol (endothermic) | |

| zinc oxide | oxygen | zinc molecular entropy | 44 J/(mol K) | 205 J/(mol K) | 42 J/(mol K) total entropy | 88 J/(mol K) | 205 J/(mol K) | 84 J/(mol K) | S_initial = 88 J/(mol K) | S_final = 289 J/(mol K) | ΔS_rxn^0 | 289 J/(mol K) - 88 J/(mol K) = 201 J/(mol K) (endoentropic) | |

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

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

| zinc oxide | oxygen | zinc formula | ZnO | O_2 | Zn Hill formula | OZn | O_2 | Zn name | zinc oxide | oxygen | zinc IUPAC name | oxozinc | molecular oxygen | zinc

| zinc oxide | oxygen | zinc molar mass | 81.38 g/mol | 31.998 g/mol | 65.38 g/mol phase | solid (at STP) | gas (at STP) | solid (at STP) melting point | 1975 °C | -218 °C | 420 °C boiling point | 2360 °C | -183 °C | 907 °C density | 5.6 g/cm^3 | 0.001429 g/cm^3 (at 0 °C) | 7.14 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 | odorless | odorless