H2O + Zr = H2 + ZrO2

H_2O (water) + Zr (zirconium) ⟶ H_2 (hydrogen) + ZrO_2 (zirconium(IV) oxide)

Balance the chemical equation algebraically: H_2O + Zr ⟶ H_2 + ZrO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Zr ⟶ c_3 H_2 + c_4 ZrO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and Zr: H: | 2 c_1 = 2 c_3 O: | c_1 = 2 c_4 Zr: | c_2 = c_4 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 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + Zr ⟶ 2 H_2 + ZrO_2

+ ⟶ +

water + zirconium ⟶ hydrogen + zirconium(IV) oxide

| water | zirconium | hydrogen | zirconium(IV) oxide molecular enthalpy | -285.8 kJ/mol | 0 kJ/mol | 0 kJ/mol | -1101 kJ/mol total enthalpy | -571.7 kJ/mol | 0 kJ/mol | 0 kJ/mol | -1101 kJ/mol | H_initial = -571.7 kJ/mol | | H_final = -1101 kJ/mol | ΔH_rxn^0 | -1101 kJ/mol - -571.7 kJ/mol = -528.9 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: H_2O + Zr ⟶ H_2 + ZrO_2 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 H_2O + Zr ⟶ 2 H_2 + ZrO_2 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 H_2O | 2 | -2 Zr | 1 | -1 H_2 | 2 | 2 ZrO_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) Zr | 1 | -1 | ([Zr])^(-1) H_2 | 2 | 2 | ([H2])^2 ZrO_2 | 1 | 1 | [ZrO2] 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 = ([H2O])^(-2) ([Zr])^(-1) ([H2])^2 [ZrO2] = (([H2])^2 [ZrO2])/(([H2O])^2 [Zr])

Construct the rate of reaction expression for: H_2O + Zr ⟶ H_2 + ZrO_2 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 H_2O + Zr ⟶ 2 H_2 + ZrO_2 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 H_2O | 2 | -2 Zr | 1 | -1 H_2 | 2 | 2 ZrO_2 | 1 | 1 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) Zr | 1 | -1 | -(Δ[Zr])/(Δt) H_2 | 2 | 2 | 1/2 (Δ[H2])/(Δt) ZrO_2 | 1 | 1 | (Δ[ZrO2])/(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[H2O])/(Δt) = -(Δ[Zr])/(Δt) = 1/2 (Δ[H2])/(Δt) = (Δ[ZrO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | zirconium | hydrogen | zirconium(IV) oxide formula | H_2O | Zr | H_2 | ZrO_2 Hill formula | H_2O | Zr | H_2 | O_2Zr name | water | zirconium | hydrogen | zirconium(IV) oxide IUPAC name | water | zirconium | molecular hydrogen | dioxozirconium

| water | zirconium | hydrogen | zirconium(IV) oxide molar mass | 18.015 g/mol | 91.224 g/mol | 2.016 g/mol | 123.22 g/mol phase | liquid (at STP) | solid (at STP) | gas (at STP) | solid (at STP) melting point | 0 °C | 1852 °C | -259.2 °C | 2700 °C boiling point | 99.9839 °C | 4377 °C | -252.8 °C | 5000 °C density | 1 g/cm^3 | 6.511 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | 5.89 g/cm^3 solubility in water | | insoluble | | surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | | odorless |