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HNO3 + NH3 = H2O + N2

Input interpretation

HNO_3 nitric acid + NH_3 ammonia ⟶ H_2O water + N_2 nitrogen
HNO_3 nitric acid + NH_3 ammonia ⟶ H_2O water + N_2 nitrogen

Balanced equation

Balance the chemical equation algebraically: HNO_3 + NH_3 ⟶ H_2O + N_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 NH_3 ⟶ c_3 H_2O + c_4 N_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | c_1 + 3 c_2 = 2 c_3 N: | c_1 + c_2 = 2 c_4 O: | 3 c_1 = c_3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 5/3 c_3 = 3 c_4 = 4/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 5 c_3 = 9 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 3 HNO_3 + 5 NH_3 ⟶ 9 H_2O + 4 N_2
Balance the chemical equation algebraically: HNO_3 + NH_3 ⟶ H_2O + N_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 NH_3 ⟶ c_3 H_2O + c_4 N_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | c_1 + 3 c_2 = 2 c_3 N: | c_1 + c_2 = 2 c_4 O: | 3 c_1 = c_3 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 5/3 c_3 = 3 c_4 = 4/3 Multiply by the least common denominator, 3, to eliminate fractional coefficients: c_1 = 3 c_2 = 5 c_3 = 9 c_4 = 4 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 HNO_3 + 5 NH_3 ⟶ 9 H_2O + 4 N_2

Structures

 + ⟶ +
+ ⟶ +

Names

nitric acid + ammonia ⟶ water + nitrogen
nitric acid + ammonia ⟶ water + nitrogen

Reaction thermodynamics

Gibbs free energy

 | nitric acid | ammonia | water | nitrogen molecular free energy | -80.7 kJ/mol | -16.4 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | -242.1 kJ/mol | -82 kJ/mol | -2134 kJ/mol | 0 kJ/mol  | G_initial = -324.1 kJ/mol | | G_final = -2134 kJ/mol |  ΔG_rxn^0 | -2134 kJ/mol - -324.1 kJ/mol = -1810 kJ/mol (exergonic) | | |
| nitric acid | ammonia | water | nitrogen molecular free energy | -80.7 kJ/mol | -16.4 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | -242.1 kJ/mol | -82 kJ/mol | -2134 kJ/mol | 0 kJ/mol | G_initial = -324.1 kJ/mol | | G_final = -2134 kJ/mol | ΔG_rxn^0 | -2134 kJ/mol - -324.1 kJ/mol = -1810 kJ/mol (exergonic) | | |

Entropy

 | nitric acid | ammonia | water | nitrogen molecular entropy | 156 J/(mol K) | 193 J/(mol K) | 69.91 J/(mol K) | 192 J/(mol K) total entropy | 468 J/(mol K) | 965 J/(mol K) | 629.2 J/(mol K) | 768 J/(mol K)  | S_initial = 1433 J/(mol K) | | S_final = 1397 J/(mol K) |  ΔS_rxn^0 | 1397 J/(mol K) - 1433 J/(mol K) = -35.81 J/(mol K) (exoentropic) | | |
| nitric acid | ammonia | water | nitrogen molecular entropy | 156 J/(mol K) | 193 J/(mol K) | 69.91 J/(mol K) | 192 J/(mol K) total entropy | 468 J/(mol K) | 965 J/(mol K) | 629.2 J/(mol K) | 768 J/(mol K) | S_initial = 1433 J/(mol K) | | S_final = 1397 J/(mol K) | ΔS_rxn^0 | 1397 J/(mol K) - 1433 J/(mol K) = -35.81 J/(mol K) (exoentropic) | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + NH_3 ⟶ H_2O + N_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: 3 HNO_3 + 5 NH_3 ⟶ 9 H_2O + 4 N_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 | 3 | -3 NH_3 | 5 | -5 H_2O | 9 | 9 N_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 3 | -3 | ([HNO3])^(-3) NH_3 | 5 | -5 | ([NH3])^(-5) H_2O | 9 | 9 | ([H2O])^9 N_2 | 4 | 4 | ([N2])^4 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])^(-3) ([NH3])^(-5) ([H2O])^9 ([N2])^4 = (([H2O])^9 ([N2])^4)/(([HNO3])^3 ([NH3])^5)
Construct the equilibrium constant, K, expression for: HNO_3 + NH_3 ⟶ H_2O + N_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: 3 HNO_3 + 5 NH_3 ⟶ 9 H_2O + 4 N_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 | 3 | -3 NH_3 | 5 | -5 H_2O | 9 | 9 N_2 | 4 | 4 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 3 | -3 | ([HNO3])^(-3) NH_3 | 5 | -5 | ([NH3])^(-5) H_2O | 9 | 9 | ([H2O])^9 N_2 | 4 | 4 | ([N2])^4 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])^(-3) ([NH3])^(-5) ([H2O])^9 ([N2])^4 = (([H2O])^9 ([N2])^4)/(([HNO3])^3 ([NH3])^5)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + NH_3 ⟶ H_2O + N_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: 3 HNO_3 + 5 NH_3 ⟶ 9 H_2O + 4 N_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 | 3 | -3 NH_3 | 5 | -5 H_2O | 9 | 9 N_2 | 4 | 4 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 | 3 | -3 | -1/3 (Δ[HNO3])/(Δt) NH_3 | 5 | -5 | -1/5 (Δ[NH3])/(Δt) H_2O | 9 | 9 | 1/9 (Δ[H2O])/(Δt) N_2 | 4 | 4 | 1/4 (Δ[N2])/(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 = -1/3 (Δ[HNO3])/(Δt) = -1/5 (Δ[NH3])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/4 (Δ[N2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + NH_3 ⟶ H_2O + N_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: 3 HNO_3 + 5 NH_3 ⟶ 9 H_2O + 4 N_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 | 3 | -3 NH_3 | 5 | -5 H_2O | 9 | 9 N_2 | 4 | 4 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 | 3 | -3 | -1/3 (Δ[HNO3])/(Δt) NH_3 | 5 | -5 | -1/5 (Δ[NH3])/(Δt) H_2O | 9 | 9 | 1/9 (Δ[H2O])/(Δt) N_2 | 4 | 4 | 1/4 (Δ[N2])/(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 = -1/3 (Δ[HNO3])/(Δt) = -1/5 (Δ[NH3])/(Δt) = 1/9 (Δ[H2O])/(Δt) = 1/4 (Δ[N2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | nitric acid | ammonia | water | nitrogen formula | HNO_3 | NH_3 | H_2O | N_2 Hill formula | HNO_3 | H_3N | H_2O | N_2 name | nitric acid | ammonia | water | nitrogen IUPAC name | nitric acid | ammonia | water | molecular nitrogen
| nitric acid | ammonia | water | nitrogen formula | HNO_3 | NH_3 | H_2O | N_2 Hill formula | HNO_3 | H_3N | H_2O | N_2 name | nitric acid | ammonia | water | nitrogen IUPAC name | nitric acid | ammonia | water | molecular nitrogen

Substance properties

 | nitric acid | ammonia | water | nitrogen molar mass | 63.012 g/mol | 17.031 g/mol | 18.015 g/mol | 28.014 g/mol phase | liquid (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) melting point | -41.6 °C | -77.73 °C | 0 °C | -210 °C boiling point | 83 °C | -33.33 °C | 99.9839 °C | -195.79 °C density | 1.5129 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) solubility in water | miscible | | | insoluble surface tension | | 0.0234 N/m | 0.0728 N/m | 0.0066 N/m dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) odor | | | odorless | odorless
| nitric acid | ammonia | water | nitrogen molar mass | 63.012 g/mol | 17.031 g/mol | 18.015 g/mol | 28.014 g/mol phase | liquid (at STP) | gas (at STP) | liquid (at STP) | gas (at STP) melting point | -41.6 °C | -77.73 °C | 0 °C | -210 °C boiling point | 83 °C | -33.33 °C | 99.9839 °C | -195.79 °C density | 1.5129 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) solubility in water | miscible | | | insoluble surface tension | | 0.0234 N/m | 0.0728 N/m | 0.0066 N/m dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) odor | | | odorless | odorless

Units