Input interpretation
nitric acid + lead ⟶ water + nitric oxide + lead(II) nitrate
Balanced equation
Balance the chemical equation algebraically: + ⟶ + + Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 + c_5 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 = c_4 + 2 c_5 O: | 3 c_1 = c_3 + c_4 + 6 c_5 Pb: | c_2 = c_5 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 3/2 c_3 = 2 c_4 = 1 c_5 = 3/2 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 8 c_2 = 3 c_3 = 4 c_4 = 2 c_5 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 8 + 3 ⟶ 4 + 2 + 3
Structures
+ ⟶ + +
Names
nitric acid + lead ⟶ water + nitric oxide + lead(II) nitrate
Equilibrium constant
![K_c = ([H2O]^4 [NO]^2 [Pb(NO3)2]^3)/([HNO3]^8 [Pb]^3)](../image_source/102ade29f1fe0486c40b21dceac33091.png)
K_c = ([H2O]^4 [NO]^2 [Pb(NO3)2]^3)/([HNO3]^8 [Pb]^3)
Rate of reaction
![rate = -1/8 (Δ[HNO3])/(Δt) = -1/3 (Δ[Pb])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/3 (Δ[Pb(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/a9c4b2d951c7c010d2cf178786c9930a.png)
rate = -1/8 (Δ[HNO3])/(Δt) = -1/3 (Δ[Pb])/(Δt) = 1/4 (Δ[H2O])/(Δt) = 1/2 (Δ[NO])/(Δt) = 1/3 (Δ[Pb(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
| nitric acid | lead | water | nitric oxide | lead(II) nitrate Hill formula | HNO_3 | Pb | H_2O | NO | N_2O_6Pb name | nitric acid | lead | water | nitric oxide | lead(II) nitrate IUPAC name | nitric acid | lead | water | nitric oxide | plumbous dinitrate