Pb(NO3)2 = O2 + NO2 + Pb

Pb(NO_3)_2 lead(II) nitrate ⟶ O_2 oxygen + NO_2 nitrogen dioxide + Pb lead

Balance the chemical equation algebraically: Pb(NO_3)_2 ⟶ O_2 + NO_2 + Pb Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Pb(NO_3)_2 ⟶ c_2 O_2 + c_3 NO_2 + c_4 Pb Set the number of atoms in the reactants equal to the number of atoms in the products for N, O and Pb: N: | 2 c_1 = c_3 O: | 6 c_1 = 2 c_2 + 2 c_3 Pb: | c_1 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | Pb(NO_3)_2 ⟶ O_2 + 2 NO_2 + Pb

⟶ + +

lead(II) nitrate ⟶ oxygen + nitrogen dioxide + lead

| lead(II) nitrate | oxygen | nitrogen dioxide | lead molecular enthalpy | -451.9 kJ/mol | 0 kJ/mol | 33.2 kJ/mol | 0 kJ/mol total enthalpy | -451.9 kJ/mol | 0 kJ/mol | 66.4 kJ/mol | 0 kJ/mol | H_initial = -451.9 kJ/mol | H_final = 66.4 kJ/mol | | ΔH_rxn^0 | 66.4 kJ/mol - -451.9 kJ/mol = 518.3 kJ/mol (endothermic) | | |

Construct the equilibrium constant, K, expression for: Pb(NO_3)_2 ⟶ O_2 + NO_2 + Pb 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: Pb(NO_3)_2 ⟶ O_2 + 2 NO_2 + Pb 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 Pb(NO_3)_2 | 1 | -1 O_2 | 1 | 1 NO_2 | 2 | 2 Pb | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Pb(NO_3)_2 | 1 | -1 | ([Pb(NO3)2])^(-1) O_2 | 1 | 1 | [O2] NO_2 | 2 | 2 | ([NO2])^2 Pb | 1 | 1 | [Pb] 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 = ([Pb(NO3)2])^(-1) [O2] ([NO2])^2 [Pb] = ([O2] ([NO2])^2 [Pb])/([Pb(NO3)2])

Construct the rate of reaction expression for: Pb(NO_3)_2 ⟶ O_2 + NO_2 + Pb 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: Pb(NO_3)_2 ⟶ O_2 + 2 NO_2 + Pb 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 Pb(NO_3)_2 | 1 | -1 O_2 | 1 | 1 NO_2 | 2 | 2 Pb | 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 Pb(NO_3)_2 | 1 | -1 | -(Δ[Pb(NO3)2])/(Δt) O_2 | 1 | 1 | (Δ[O2])/(Δt) NO_2 | 2 | 2 | 1/2 (Δ[NO2])/(Δt) Pb | 1 | 1 | (Δ[Pb])/(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[Pb(NO3)2])/(Δt) = (Δ[O2])/(Δt) = 1/2 (Δ[NO2])/(Δt) = (Δ[Pb])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| lead(II) nitrate | oxygen | nitrogen dioxide | lead formula | Pb(NO_3)_2 | O_2 | NO_2 | Pb Hill formula | N_2O_6Pb | O_2 | NO_2 | Pb name | lead(II) nitrate | oxygen | nitrogen dioxide | lead IUPAC name | plumbous dinitrate | molecular oxygen | Nitrogen dioxide | lead

| lead(II) nitrate | oxygen | nitrogen dioxide | lead molar mass | 331.2 g/mol | 31.998 g/mol | 46.005 g/mol | 207.2 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | solid (at STP) melting point | 470 °C | -218 °C | -11 °C | 327.4 °C boiling point | | -183 °C | 21 °C | 1740 °C density | | 0.001429 g/cm^3 (at 0 °C) | 0.00188 g/cm^3 (at 25 °C) | 11.34 g/cm^3 solubility in water | | | reacts | insoluble surface tension | | 0.01347 N/m | | dynamic viscosity | | 2.055×10^-5 Pa s (at 25 °C) | 4.02×10^-4 Pa s (at 25 °C) | 0.00183 Pa s (at 38 °C) odor | odorless | odorless | |