H2 + GeO2 = H2O + Ge

H_2 hydrogen + GeO_2 germanium dioxide ⟶ H_2O water + Ge germanium

Balance the chemical equation algebraically: H_2 + GeO_2 ⟶ H_2O + Ge Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2 + c_2 GeO_2 ⟶ c_3 H_2O + c_4 Ge Set the number of atoms in the reactants equal to the number of atoms in the products for H, Ge and O: H: | 2 c_1 = 2 c_3 Ge: | c_2 = c_4 O: | 2 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 = 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_2 + GeO_2 ⟶ 2 H_2O + Ge

+ ⟶ +

hydrogen + germanium dioxide ⟶ water + germanium

| hydrogen | germanium dioxide | water | germanium molecular enthalpy | 0 kJ/mol | -580 kJ/mol | -285.8 kJ/mol | 0 kJ/mol total enthalpy | 0 kJ/mol | -580 kJ/mol | -571.7 kJ/mol | 0 kJ/mol | H_initial = -580 kJ/mol | | H_final = -571.7 kJ/mol | ΔH_rxn^0 | -571.7 kJ/mol - -580 kJ/mol = 8.34 kJ/mol (endothermic) | | |

Construct the equilibrium constant, K, expression for: H_2 + GeO_2 ⟶ H_2O + Ge 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_2 + GeO_2 ⟶ 2 H_2O + Ge 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_2 | 2 | -2 GeO_2 | 1 | -1 H_2O | 2 | 2 Ge | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2 | 2 | -2 | ([H2])^(-2) GeO_2 | 1 | -1 | ([GeO2])^(-1) H_2O | 2 | 2 | ([H2O])^2 Ge | 1 | 1 | [Ge] 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 = ([H2])^(-2) ([GeO2])^(-1) ([H2O])^2 [Ge] = (([H2O])^2 [Ge])/(([H2])^2 [GeO2])

Construct the rate of reaction expression for: H_2 + GeO_2 ⟶ H_2O + Ge 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_2 + GeO_2 ⟶ 2 H_2O + Ge 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_2 | 2 | -2 GeO_2 | 1 | -1 H_2O | 2 | 2 Ge | 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_2 | 2 | -2 | -1/2 (Δ[H2])/(Δt) GeO_2 | 1 | -1 | -(Δ[GeO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Ge | 1 | 1 | (Δ[Ge])/(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[H2])/(Δt) = -(Δ[GeO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Ge])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| hydrogen | germanium dioxide | water | germanium formula | H_2 | GeO_2 | H_2O | Ge name | hydrogen | germanium dioxide | water | germanium IUPAC name | molecular hydrogen | | water | germanium

| hydrogen | germanium dioxide | water | germanium molar mass | 2.016 g/mol | 104.63 g/mol | 18.015 g/mol | 72.63 g/mol phase | gas (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | -259.2 °C | 400 °C | 0 °C | 937 °C boiling point | -252.8 °C | | 99.9839 °C | 2830 °C density | 8.99×10^-5 g/cm^3 (at 0 °C) | 6.239 g/cm^3 | 1 g/cm^3 | 5.35 g/cm^3 surface tension | | | 0.0728 N/m | dynamic viscosity | 8.9×10^-6 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |