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HNO3 + KBr + PbO2 = H2O + Br2 + KNO3 + Pb(NO3)2

Input interpretation

HNO_3 nitric acid + KBr potassium bromide + PbO_2 lead dioxide ⟶ H_2O water + Br_2 bromine + KNO_3 potassium nitrate + Pb(NO_3)_2 lead(II) nitrate
HNO_3 nitric acid + KBr potassium bromide + PbO_2 lead dioxide ⟶ H_2O water + Br_2 bromine + KNO_3 potassium nitrate + Pb(NO_3)_2 lead(II) nitrate

Balanced equation

Balance the chemical equation algebraically: HNO_3 + KBr + PbO_2 ⟶ H_2O + Br_2 + KNO_3 + Pb(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 KBr + c_3 PbO_2 ⟶ c_4 H_2O + c_5 Br_2 + c_6 KNO_3 + c_7 Pb(NO_3)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Br, K and Pb: H: | c_1 = 2 c_4 N: | c_1 = c_6 + 2 c_7 O: | 3 c_1 + 2 c_3 = c_4 + 3 c_6 + 6 c_7 Br: | c_2 = 2 c_5 K: | c_2 = c_6 Pb: | c_3 = c_7 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 1 c_6 = 2 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + 2 KBr + PbO_2 ⟶ 2 H_2O + Br_2 + 2 KNO_3 + Pb(NO_3)_2
Balance the chemical equation algebraically: HNO_3 + KBr + PbO_2 ⟶ H_2O + Br_2 + KNO_3 + Pb(NO_3)_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 KBr + c_3 PbO_2 ⟶ c_4 H_2O + c_5 Br_2 + c_6 KNO_3 + c_7 Pb(NO_3)_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Br, K and Pb: H: | c_1 = 2 c_4 N: | c_1 = c_6 + 2 c_7 O: | 3 c_1 + 2 c_3 = c_4 + 3 c_6 + 6 c_7 Br: | c_2 = 2 c_5 K: | c_2 = c_6 Pb: | c_3 = c_7 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 2 c_3 = 1 c_4 = 2 c_5 = 1 c_6 = 2 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + 2 KBr + PbO_2 ⟶ 2 H_2O + Br_2 + 2 KNO_3 + Pb(NO_3)_2

Structures

+ + ⟶ + + +
+ + ⟶ + + +

Names

nitric acid + potassium bromide + lead dioxide ⟶ water + bromine + potassium nitrate + lead(II) nitrate
nitric acid + potassium bromide + lead dioxide ⟶ water + bromine + potassium nitrate + lead(II) nitrate

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + KBr + PbO_2 ⟶ H_2O + Br_2 + KNO_3 + Pb(NO_3)_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: 4 HNO_3 + 2 KBr + PbO_2 ⟶ 2 H_2O + Br_2 + 2 KNO_3 + Pb(NO_3)_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 HNO_3 | 4 | -4 KBr | 2 | -2 PbO_2 | 1 | -1 H_2O | 2 | 2 Br_2 | 1 | 1 KNO_3 | 2 | 2 Pb(NO_3)_2 | 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 | 4 | -4 | ([HNO3])^(-4) KBr | 2 | -2 | ([KBr])^(-2) PbO_2 | 1 | -1 | ([PbO2])^(-1) H_2O | 2 | 2 | ([H2O])^2 Br_2 | 1 | 1 | [Br2] KNO_3 | 2 | 2 | ([KNO3])^2 Pb(NO_3)_2 | 1 | 1 | [Pb(NO3)2] 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])^(-4) ([KBr])^(-2) ([PbO2])^(-1) ([H2O])^2 [Br2] ([KNO3])^2 [Pb(NO3)2] = (([H2O])^2 [Br2] ([KNO3])^2 [Pb(NO3)2])/(([HNO3])^4 ([KBr])^2 [PbO2])
Construct the equilibrium constant, K, expression for: HNO_3 + KBr + PbO_2 ⟶ H_2O + Br_2 + KNO_3 + Pb(NO_3)_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: 4 HNO_3 + 2 KBr + PbO_2 ⟶ 2 H_2O + Br_2 + 2 KNO_3 + Pb(NO_3)_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 HNO_3 | 4 | -4 KBr | 2 | -2 PbO_2 | 1 | -1 H_2O | 2 | 2 Br_2 | 1 | 1 KNO_3 | 2 | 2 Pb(NO_3)_2 | 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 | 4 | -4 | ([HNO3])^(-4) KBr | 2 | -2 | ([KBr])^(-2) PbO_2 | 1 | -1 | ([PbO2])^(-1) H_2O | 2 | 2 | ([H2O])^2 Br_2 | 1 | 1 | [Br2] KNO_3 | 2 | 2 | ([KNO3])^2 Pb(NO_3)_2 | 1 | 1 | [Pb(NO3)2] 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])^(-4) ([KBr])^(-2) ([PbO2])^(-1) ([H2O])^2 [Br2] ([KNO3])^2 [Pb(NO3)2] = (([H2O])^2 [Br2] ([KNO3])^2 [Pb(NO3)2])/(([HNO3])^4 ([KBr])^2 [PbO2])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + KBr + PbO_2 ⟶ H_2O + Br_2 + KNO_3 + Pb(NO_3)_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: 4 HNO_3 + 2 KBr + PbO_2 ⟶ 2 H_2O + Br_2 + 2 KNO_3 + Pb(NO_3)_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 HNO_3 | 4 | -4 KBr | 2 | -2 PbO_2 | 1 | -1 H_2O | 2 | 2 Br_2 | 1 | 1 KNO_3 | 2 | 2 Pb(NO_3)_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 HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) KBr | 2 | -2 | -1/2 (Δ[KBr])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(Δt) Pb(NO_3)_2 | 1 | 1 | (Δ[Pb(NO3)2])/(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 = -1/4 (Δ[HNO3])/(Δt) = -1/2 (Δ[KBr])/(Δt) = -(Δ[PbO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Br2])/(Δt) = 1/2 (Δ[KNO3])/(Δt) = (Δ[Pb(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + KBr + PbO_2 ⟶ H_2O + Br_2 + KNO_3 + Pb(NO_3)_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: 4 HNO_3 + 2 KBr + PbO_2 ⟶ 2 H_2O + Br_2 + 2 KNO_3 + Pb(NO_3)_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 HNO_3 | 4 | -4 KBr | 2 | -2 PbO_2 | 1 | -1 H_2O | 2 | 2 Br_2 | 1 | 1 KNO_3 | 2 | 2 Pb(NO_3)_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 HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) KBr | 2 | -2 | -1/2 (Δ[KBr])/(Δt) PbO_2 | 1 | -1 | -(Δ[PbO2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Br_2 | 1 | 1 | (Δ[Br2])/(Δt) KNO_3 | 2 | 2 | 1/2 (Δ[KNO3])/(Δt) Pb(NO_3)_2 | 1 | 1 | (Δ[Pb(NO3)2])/(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 = -1/4 (Δ[HNO3])/(Δt) = -1/2 (Δ[KBr])/(Δt) = -(Δ[PbO2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Br2])/(Δt) = 1/2 (Δ[KNO3])/(Δt) = (Δ[Pb(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | potassium bromide | lead dioxide | water | bromine | potassium nitrate | lead(II) nitrate formula | HNO_3 | KBr | PbO_2 | H_2O | Br_2 | KNO_3 | Pb(NO_3)_2 Hill formula | HNO_3 | BrK | O_2Pb | H_2O | Br_2 | KNO_3 | N_2O_6Pb name | nitric acid | potassium bromide | lead dioxide | water | bromine | potassium nitrate | lead(II) nitrate IUPAC name | nitric acid | potassium bromide | | water | molecular bromine | potassium nitrate | plumbous dinitrate
| nitric acid | potassium bromide | lead dioxide | water | bromine | potassium nitrate | lead(II) nitrate formula | HNO_3 | KBr | PbO_2 | H_2O | Br_2 | KNO_3 | Pb(NO_3)_2 Hill formula | HNO_3 | BrK | O_2Pb | H_2O | Br_2 | KNO_3 | N_2O_6Pb name | nitric acid | potassium bromide | lead dioxide | water | bromine | potassium nitrate | lead(II) nitrate IUPAC name | nitric acid | potassium bromide | | water | molecular bromine | potassium nitrate | plumbous dinitrate

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