Input interpretation
![HNO_3 (nitric acid) + FeCl_2 (iron(II) chloride) ⟶ H_2O (water) + Cl_2 (chlorine) + NO (nitric oxide) + Fe(NO_3)_3 (ferric nitrate)](../image_source/60c1be987b429aeff94ffbf5e4b4c38e.png)
HNO_3 (nitric acid) + FeCl_2 (iron(II) chloride) ⟶ H_2O (water) + Cl_2 (chlorine) + NO (nitric oxide) + Fe(NO_3)_3 (ferric nitrate)
Balanced equation
![Balance the chemical equation algebraically: HNO_3 + FeCl_2 ⟶ H_2O + Cl_2 + NO + Fe(NO_3)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeCl_2 ⟶ c_3 H_2O + c_4 Cl_2 + c_5 NO + c_6 Fe(NO_3)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Cl and Fe: H: | c_1 = 2 c_3 N: | c_1 = c_5 + 3 c_6 O: | 3 c_1 = c_3 + c_5 + 9 c_6 Cl: | 2 c_2 = 2 c_4 Fe: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 1 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + FeCl_2 ⟶ 2 H_2O + Cl_2 + NO + Fe(NO_3)_3](../image_source/62d24277b1dc789e37f5bc963f1d38b6.png)
Balance the chemical equation algebraically: HNO_3 + FeCl_2 ⟶ H_2O + Cl_2 + NO + Fe(NO_3)_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 FeCl_2 ⟶ c_3 H_2O + c_4 Cl_2 + c_5 NO + c_6 Fe(NO_3)_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O, Cl and Fe: H: | c_1 = 2 c_3 N: | c_1 = c_5 + 3 c_6 O: | 3 c_1 = c_3 + c_5 + 9 c_6 Cl: | 2 c_2 = 2 c_4 Fe: | c_2 = 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 4 c_2 = 1 c_3 = 2 c_4 = 1 c_5 = 1 c_6 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 4 HNO_3 + FeCl_2 ⟶ 2 H_2O + Cl_2 + NO + Fe(NO_3)_3
Structures
![+ ⟶ + + +](../image_source/a0e1aaada19d660311eb7c4e76ed74d6.png)
+ ⟶ + + +
Names
![nitric acid + iron(II) chloride ⟶ water + chlorine + nitric oxide + ferric nitrate](../image_source/1dadebd2a4e12a808f1b185372338472.png)
nitric acid + iron(II) chloride ⟶ water + chlorine + nitric oxide + ferric nitrate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: HNO_3 + FeCl_2 ⟶ H_2O + Cl_2 + NO + Fe(NO_3)_3 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 + FeCl_2 ⟶ 2 H_2O + Cl_2 + NO + Fe(NO_3)_3 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 FeCl_2 | 1 | -1 H_2O | 2 | 2 Cl_2 | 1 | 1 NO | 1 | 1 Fe(NO_3)_3 | 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) FeCl_2 | 1 | -1 | ([FeCl2])^(-1) H_2O | 2 | 2 | ([H2O])^2 Cl_2 | 1 | 1 | [Cl2] NO | 1 | 1 | [NO] Fe(NO_3)_3 | 1 | 1 | [Fe(NO3)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 = ([HNO3])^(-4) ([FeCl2])^(-1) ([H2O])^2 [Cl2] [NO] [Fe(NO3)3] = (([H2O])^2 [Cl2] [NO] [Fe(NO3)3])/(([HNO3])^4 [FeCl2])](../image_source/334ca5b83b4afb74a397d84751c28489.png)
Construct the equilibrium constant, K, expression for: HNO_3 + FeCl_2 ⟶ H_2O + Cl_2 + NO + Fe(NO_3)_3 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 + FeCl_2 ⟶ 2 H_2O + Cl_2 + NO + Fe(NO_3)_3 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 FeCl_2 | 1 | -1 H_2O | 2 | 2 Cl_2 | 1 | 1 NO | 1 | 1 Fe(NO_3)_3 | 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) FeCl_2 | 1 | -1 | ([FeCl2])^(-1) H_2O | 2 | 2 | ([H2O])^2 Cl_2 | 1 | 1 | [Cl2] NO | 1 | 1 | [NO] Fe(NO_3)_3 | 1 | 1 | [Fe(NO3)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 = ([HNO3])^(-4) ([FeCl2])^(-1) ([H2O])^2 [Cl2] [NO] [Fe(NO3)3] = (([H2O])^2 [Cl2] [NO] [Fe(NO3)3])/(([HNO3])^4 [FeCl2])
Rate of reaction
![Construct the rate of reaction expression for: HNO_3 + FeCl_2 ⟶ H_2O + Cl_2 + NO + Fe(NO_3)_3 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 + FeCl_2 ⟶ 2 H_2O + Cl_2 + NO + Fe(NO_3)_3 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 FeCl_2 | 1 | -1 H_2O | 2 | 2 Cl_2 | 1 | 1 NO | 1 | 1 Fe(NO_3)_3 | 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) FeCl_2 | 1 | -1 | -(Δ[FeCl2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Cl_2 | 1 | 1 | (Δ[Cl2])/(Δt) NO | 1 | 1 | (Δ[NO])/(Δt) Fe(NO_3)_3 | 1 | 1 | (Δ[Fe(NO3)3])/(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) = -(Δ[FeCl2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Cl2])/(Δt) = (Δ[NO])/(Δt) = (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/466df7a160bacdd269992aafd775e0ce.png)
Construct the rate of reaction expression for: HNO_3 + FeCl_2 ⟶ H_2O + Cl_2 + NO + Fe(NO_3)_3 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 + FeCl_2 ⟶ 2 H_2O + Cl_2 + NO + Fe(NO_3)_3 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 FeCl_2 | 1 | -1 H_2O | 2 | 2 Cl_2 | 1 | 1 NO | 1 | 1 Fe(NO_3)_3 | 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) FeCl_2 | 1 | -1 | -(Δ[FeCl2])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) Cl_2 | 1 | 1 | (Δ[Cl2])/(Δt) NO | 1 | 1 | (Δ[NO])/(Δt) Fe(NO_3)_3 | 1 | 1 | (Δ[Fe(NO3)3])/(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) = -(Δ[FeCl2])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[Cl2])/(Δt) = (Δ[NO])/(Δt) = (Δ[Fe(NO3)3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| nitric acid | iron(II) chloride | water | chlorine | nitric oxide | ferric nitrate formula | HNO_3 | FeCl_2 | H_2O | Cl_2 | NO | Fe(NO_3)_3 Hill formula | HNO_3 | Cl_2Fe | H_2O | Cl_2 | NO | FeN_3O_9 name | nitric acid | iron(II) chloride | water | chlorine | nitric oxide | ferric nitrate IUPAC name | nitric acid | dichloroiron | water | molecular chlorine | nitric oxide | iron(+3) cation trinitrate](../image_source/cf6791d10e938ed0a3442265d34f6f21.png)
| nitric acid | iron(II) chloride | water | chlorine | nitric oxide | ferric nitrate formula | HNO_3 | FeCl_2 | H_2O | Cl_2 | NO | Fe(NO_3)_3 Hill formula | HNO_3 | Cl_2Fe | H_2O | Cl_2 | NO | FeN_3O_9 name | nitric acid | iron(II) chloride | water | chlorine | nitric oxide | ferric nitrate IUPAC name | nitric acid | dichloroiron | water | molecular chlorine | nitric oxide | iron(+3) cation trinitrate