Input interpretation
![HNO_3 nitric acid + Fe iron ⟶ H_2O water + NH_3 ammonia + FeNO3](../image_source/b81997b2fc5cbdc97c1c994882dedea4.png)
HNO_3 nitric acid + Fe iron ⟶ H_2O water + NH_3 ammonia + FeNO3
Balanced equation
![Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NH_3 + FeNO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 NH_3 + c_5 FeNO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 + 3 c_4 N: | c_1 = c_4 + c_5 O: | 3 c_1 = c_3 + 3 c_5 Fe: | 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 = 9 c_2 = 8 c_3 = 3 c_4 = 1 c_5 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 HNO_3 + 8 Fe ⟶ 3 H_2O + NH_3 + 8 FeNO3](../image_source/d81a2d9eb2268452a0c10631207c8549.png)
Balance the chemical equation algebraically: HNO_3 + Fe ⟶ H_2O + NH_3 + FeNO3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Fe ⟶ c_3 H_2O + c_4 NH_3 + c_5 FeNO3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Fe: H: | c_1 = 2 c_3 + 3 c_4 N: | c_1 = c_4 + c_5 O: | 3 c_1 = c_3 + 3 c_5 Fe: | 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 = 9 c_2 = 8 c_3 = 3 c_4 = 1 c_5 = 8 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 9 HNO_3 + 8 Fe ⟶ 3 H_2O + NH_3 + 8 FeNO3
Structures
![+ ⟶ + + FeNO3](../image_source/2d03aa0fb05d62e05763388d74613fb8.png)
+ ⟶ + + FeNO3
Names
![nitric acid + iron ⟶ water + ammonia + FeNO3](../image_source/f0e96b77c1298ab1a8e84c1db8203fd4.png)
nitric acid + iron ⟶ water + ammonia + FeNO3
Equilibrium constant
![Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + FeNO3 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: 9 HNO_3 + 8 Fe ⟶ 3 H_2O + NH_3 + 8 FeNO3 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 | 9 | -9 Fe | 8 | -8 H_2O | 3 | 3 NH_3 | 1 | 1 FeNO3 | 8 | 8 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 9 | -9 | ([HNO3])^(-9) Fe | 8 | -8 | ([Fe])^(-8) H_2O | 3 | 3 | ([H2O])^3 NH_3 | 1 | 1 | [NH3] FeNO3 | 8 | 8 | ([FeNO3])^8 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])^(-9) ([Fe])^(-8) ([H2O])^3 [NH3] ([FeNO3])^8 = (([H2O])^3 [NH3] ([FeNO3])^8)/(([HNO3])^9 ([Fe])^8)](../image_source/fbc0914a973b7ec748092f5ea420e210.png)
Construct the equilibrium constant, K, expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + FeNO3 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: 9 HNO_3 + 8 Fe ⟶ 3 H_2O + NH_3 + 8 FeNO3 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 | 9 | -9 Fe | 8 | -8 H_2O | 3 | 3 NH_3 | 1 | 1 FeNO3 | 8 | 8 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 9 | -9 | ([HNO3])^(-9) Fe | 8 | -8 | ([Fe])^(-8) H_2O | 3 | 3 | ([H2O])^3 NH_3 | 1 | 1 | [NH3] FeNO3 | 8 | 8 | ([FeNO3])^8 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])^(-9) ([Fe])^(-8) ([H2O])^3 [NH3] ([FeNO3])^8 = (([H2O])^3 [NH3] ([FeNO3])^8)/(([HNO3])^9 ([Fe])^8)
Rate of reaction
![Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + FeNO3 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: 9 HNO_3 + 8 Fe ⟶ 3 H_2O + NH_3 + 8 FeNO3 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 | 9 | -9 Fe | 8 | -8 H_2O | 3 | 3 NH_3 | 1 | 1 FeNO3 | 8 | 8 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 | 9 | -9 | -1/9 (Δ[HNO3])/(Δt) Fe | 8 | -8 | -1/8 (Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NH_3 | 1 | 1 | (Δ[NH3])/(Δt) FeNO3 | 8 | 8 | 1/8 (Δ[FeNO3])/(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 = -1/9 (Δ[HNO3])/(Δt) = -1/8 (Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[NH3])/(Δt) = 1/8 (Δ[FeNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/10831347cec50edb51a1a0f8e6049eb0.png)
Construct the rate of reaction expression for: HNO_3 + Fe ⟶ H_2O + NH_3 + FeNO3 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: 9 HNO_3 + 8 Fe ⟶ 3 H_2O + NH_3 + 8 FeNO3 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 | 9 | -9 Fe | 8 | -8 H_2O | 3 | 3 NH_3 | 1 | 1 FeNO3 | 8 | 8 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 | 9 | -9 | -1/9 (Δ[HNO3])/(Δt) Fe | 8 | -8 | -1/8 (Δ[Fe])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) NH_3 | 1 | 1 | (Δ[NH3])/(Δt) FeNO3 | 8 | 8 | 1/8 (Δ[FeNO3])/(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 = -1/9 (Δ[HNO3])/(Δt) = -1/8 (Δ[Fe])/(Δt) = 1/3 (Δ[H2O])/(Δt) = (Δ[NH3])/(Δt) = 1/8 (Δ[FeNO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| nitric acid | iron | water | ammonia | FeNO3 formula | HNO_3 | Fe | H_2O | NH_3 | FeNO3 Hill formula | HNO_3 | Fe | H_2O | H_3N | FeNO3 name | nitric acid | iron | water | ammonia |](../image_source/6c836f188375218580a4e9b3f09fab27.png)
| nitric acid | iron | water | ammonia | FeNO3 formula | HNO_3 | Fe | H_2O | NH_3 | FeNO3 Hill formula | HNO_3 | Fe | H_2O | H_3N | FeNO3 name | nitric acid | iron | water | ammonia |
Substance properties
![| nitric acid | iron | water | ammonia | FeNO3 molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 17.031 g/mol | 117.85 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1535 °C | 0 °C | -77.73 °C | boiling point | 83 °C | 2750 °C | 99.9839 °C | -33.33 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | 0.0234 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | odor | | | odorless | |](../image_source/b6c59ae8211524c77e724e5c02618f0a.png)
| nitric acid | iron | water | ammonia | FeNO3 molar mass | 63.012 g/mol | 55.845 g/mol | 18.015 g/mol | 17.031 g/mol | 117.85 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 1535 °C | 0 °C | -77.73 °C | boiling point | 83 °C | 2750 °C | 99.9839 °C | -33.33 °C | density | 1.5129 g/cm^3 | 7.874 g/cm^3 | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | solubility in water | miscible | insoluble | | | surface tension | | | 0.0728 N/m | 0.0234 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | odor | | | odorless | |
Units