HNO3 + Pb = H2O + Pb(NO3)2 + N2O

Input interpretation

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HNO_3 (nitric acid) + Pb (lead) ⟶ H_2O (water) + Pb(NO_3)_2 (lead(II) nitrate) + N_2O (nitrous oxide)

Balanced equation

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Balance the chemical equation algebraically: HNO_3 + Pb ⟶ H_2O + Pb(NO_3)_2 + N_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Pb ⟶ c_3 H_2O + c_4 Pb(NO_3)_2 + c_5 N_2O Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Pb: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 c_4 + c_5 Pb: | 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_5 = 1 and solve the system of equations for the remaining coefficients: c_1 = 10 c_2 = 4 c_3 = 5 c_4 = 4 c_5 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 10 HNO_3 + 4 Pb ⟶ 5 H_2O + 4 Pb(NO_3)_2 + N_2O

+ ⟶ + +

Names

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nitric acid + lead ⟶ water + lead(II) nitrate + nitrous oxide

Equilibrium constant

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Construct the equilibrium constant, K, expression for: HNO_3 + Pb ⟶ H_2O + Pb(NO_3)_2 + N_2O 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: 10 HNO_3 + 4 Pb ⟶ 5 H_2O + 4 Pb(NO_3)_2 + N_2O 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 HNO_3 | 10 | -10 Pb | 4 | -4 H_2O | 5 | 5 Pb(NO_3)_2 | 4 | 4 N_2O | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 10 | -10 | ([HNO3])^(-10) Pb | 4 | -4 | ([Pb])^(-4) H_2O | 5 | 5 | ([H2O])^5 Pb(NO_3)_2 | 4 | 4 | ([Pb(NO3)2])^4 N_2O | 1 | 1 | [N2O] 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 = ([HNO3])^(-10) ([Pb])^(-4) ([H2O])^5 ([Pb(NO3)2])^4 [N2O] = (([H2O])^5 ([Pb(NO3)2])^4 [N2O])/(([HNO3])^10 ([Pb])^4)

Rate of reaction

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Construct the rate of reaction expression for: HNO_3 + Pb ⟶ H_2O + Pb(NO_3)_2 + N_2O 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: 10 HNO_3 + 4 Pb ⟶ 5 H_2O + 4 Pb(NO_3)_2 + N_2O 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 HNO_3 | 10 | -10 Pb | 4 | -4 H_2O | 5 | 5 Pb(NO_3)_2 | 4 | 4 N_2O | 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 HNO_3 | 10 | -10 | -1/10 (Δ[HNO3])/(Δt) Pb | 4 | -4 | -1/4 (Δ[Pb])/(Δt) H_2O | 5 | 5 | 1/5 (Δ[H2O])/(Δt) Pb(NO_3)_2 | 4 | 4 | 1/4 (Δ[Pb(NO3)2])/(Δt) N_2O | 1 | 1 | (Δ[N2O])/(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 = -1/10 (Δ[HNO3])/(Δt) = -1/4 (Δ[Pb])/(Δt) = 1/5 (Δ[H2O])/(Δt) = 1/4 (Δ[Pb(NO3)2])/(Δt) = (Δ[N2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)