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HCl + HNO3 + Pt = H2O + NO + PtCl4

Input interpretation

HCl hydrogen chloride + HNO_3 nitric acid + Pt platinum ⟶ H_2O water + NO nitric oxide + PtCl_4 platinum(IV) chloride
HCl hydrogen chloride + HNO_3 nitric acid + Pt platinum ⟶ H_2O water + NO nitric oxide + PtCl_4 platinum(IV) chloride

Balanced equation

Balance the chemical equation algebraically: HCl + HNO_3 + Pt ⟶ H_2O + NO + PtCl_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 HNO_3 + c_3 Pt ⟶ c_4 H_2O + c_5 NO + c_6 PtCl_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, N, O and Pt: Cl: | c_1 = 4 c_6 H: | c_1 + c_2 = 2 c_4 N: | c_2 = c_5 O: | 3 c_2 = c_4 + c_5 Pt: | c_3 = c_6 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 = 4/3 c_3 = 1 c_4 = 8/3 c_5 = 4/3 c_6 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 12 c_2 = 4 c_3 = 3 c_4 = 8 c_5 = 4 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 12 HCl + 4 HNO_3 + 3 Pt ⟶ 8 H_2O + 4 NO + 3 PtCl_4
Balance the chemical equation algebraically: HCl + HNO_3 + Pt ⟶ H_2O + NO + PtCl_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 HNO_3 + c_3 Pt ⟶ c_4 H_2O + c_5 NO + c_6 PtCl_4 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, N, O and Pt: Cl: | c_1 = 4 c_6 H: | c_1 + c_2 = 2 c_4 N: | c_2 = c_5 O: | 3 c_2 = c_4 + c_5 Pt: | c_3 = c_6 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 = 4/3 c_3 = 1 c_4 = 8/3 c_5 = 4/3 c_6 = 1 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 12 c_2 = 4 c_3 = 3 c_4 = 8 c_5 = 4 c_6 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HCl + 4 HNO_3 + 3 Pt ⟶ 8 H_2O + 4 NO + 3 PtCl_4

Structures

 + + ⟶ + +
+ + ⟶ + +

Names

hydrogen chloride + nitric acid + platinum ⟶ water + nitric oxide + platinum(IV) chloride
hydrogen chloride + nitric acid + platinum ⟶ water + nitric oxide + platinum(IV) chloride

Equilibrium constant

Construct the equilibrium constant, K, expression for: HCl + HNO_3 + Pt ⟶ H_2O + NO + PtCl_4 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: 12 HCl + 4 HNO_3 + 3 Pt ⟶ 8 H_2O + 4 NO + 3 PtCl_4 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 HCl | 12 | -12 HNO_3 | 4 | -4 Pt | 3 | -3 H_2O | 8 | 8 NO | 4 | 4 PtCl_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 12 | -12 | ([HCl])^(-12) HNO_3 | 4 | -4 | ([HNO3])^(-4) Pt | 3 | -3 | ([Pt])^(-3) H_2O | 8 | 8 | ([H2O])^8 NO | 4 | 4 | ([NO])^4 PtCl_4 | 3 | 3 | ([PtCl4])^3 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 = ([HCl])^(-12) ([HNO3])^(-4) ([Pt])^(-3) ([H2O])^8 ([NO])^4 ([PtCl4])^3 = (([H2O])^8 ([NO])^4 ([PtCl4])^3)/(([HCl])^12 ([HNO3])^4 ([Pt])^3)
Construct the equilibrium constant, K, expression for: HCl + HNO_3 + Pt ⟶ H_2O + NO + PtCl_4 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: 12 HCl + 4 HNO_3 + 3 Pt ⟶ 8 H_2O + 4 NO + 3 PtCl_4 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 HCl | 12 | -12 HNO_3 | 4 | -4 Pt | 3 | -3 H_2O | 8 | 8 NO | 4 | 4 PtCl_4 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 12 | -12 | ([HCl])^(-12) HNO_3 | 4 | -4 | ([HNO3])^(-4) Pt | 3 | -3 | ([Pt])^(-3) H_2O | 8 | 8 | ([H2O])^8 NO | 4 | 4 | ([NO])^4 PtCl_4 | 3 | 3 | ([PtCl4])^3 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 = ([HCl])^(-12) ([HNO3])^(-4) ([Pt])^(-3) ([H2O])^8 ([NO])^4 ([PtCl4])^3 = (([H2O])^8 ([NO])^4 ([PtCl4])^3)/(([HCl])^12 ([HNO3])^4 ([Pt])^3)

Rate of reaction

Construct the rate of reaction expression for: HCl + HNO_3 + Pt ⟶ H_2O + NO + PtCl_4 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: 12 HCl + 4 HNO_3 + 3 Pt ⟶ 8 H_2O + 4 NO + 3 PtCl_4 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 HCl | 12 | -12 HNO_3 | 4 | -4 Pt | 3 | -3 H_2O | 8 | 8 NO | 4 | 4 PtCl_4 | 3 | 3 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 HCl | 12 | -12 | -1/12 (Δ[HCl])/(Δt) HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Pt | 3 | -3 | -1/3 (Δ[Pt])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(Δt) PtCl_4 | 3 | 3 | 1/3 (Δ[PtCl4])/(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 = -1/12 (Δ[HCl])/(Δt) = -1/4 (Δ[HNO3])/(Δt) = -1/3 (Δ[Pt])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/4 (Δ[NO])/(Δt) = 1/3 (Δ[PtCl4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HCl + HNO_3 + Pt ⟶ H_2O + NO + PtCl_4 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: 12 HCl + 4 HNO_3 + 3 Pt ⟶ 8 H_2O + 4 NO + 3 PtCl_4 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 HCl | 12 | -12 HNO_3 | 4 | -4 Pt | 3 | -3 H_2O | 8 | 8 NO | 4 | 4 PtCl_4 | 3 | 3 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 HCl | 12 | -12 | -1/12 (Δ[HCl])/(Δt) HNO_3 | 4 | -4 | -1/4 (Δ[HNO3])/(Δt) Pt | 3 | -3 | -1/3 (Δ[Pt])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) NO | 4 | 4 | 1/4 (Δ[NO])/(Δt) PtCl_4 | 3 | 3 | 1/3 (Δ[PtCl4])/(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 = -1/12 (Δ[HCl])/(Δt) = -1/4 (Δ[HNO3])/(Δt) = -1/3 (Δ[Pt])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/4 (Δ[NO])/(Δt) = 1/3 (Δ[PtCl4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | hydrogen chloride | nitric acid | platinum | water | nitric oxide | platinum(IV) chloride formula | HCl | HNO_3 | Pt | H_2O | NO | PtCl_4 Hill formula | ClH | HNO_3 | Pt | H_2O | NO | Cl_4Pt name | hydrogen chloride | nitric acid | platinum | water | nitric oxide | platinum(IV) chloride IUPAC name | hydrogen chloride | nitric acid | platinum | water | nitric oxide | tetrachloroplatinum
| hydrogen chloride | nitric acid | platinum | water | nitric oxide | platinum(IV) chloride formula | HCl | HNO_3 | Pt | H_2O | NO | PtCl_4 Hill formula | ClH | HNO_3 | Pt | H_2O | NO | Cl_4Pt name | hydrogen chloride | nitric acid | platinum | water | nitric oxide | platinum(IV) chloride IUPAC name | hydrogen chloride | nitric acid | platinum | water | nitric oxide | tetrachloroplatinum