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NH3 + H2O2 = H2O + N2

Input interpretation

NH_3 ammonia + H_2O_2 hydrogen peroxide ⟶ H_2O water + N_2 nitrogen
NH_3 ammonia + H_2O_2 hydrogen peroxide ⟶ H_2O water + N_2 nitrogen

Balanced equation

Balance the chemical equation algebraically: NH_3 + H_2O_2 ⟶ H_2O + N_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 H_2O_2 ⟶ 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: | 3 c_1 + 2 c_2 = 2 c_3 N: | c_1 = 2 c_4 O: | 2 c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | 2 NH_3 + 3 H_2O_2 ⟶ 6 H_2O + N_2
Balance the chemical equation algebraically: NH_3 + H_2O_2 ⟶ H_2O + N_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 H_2O_2 ⟶ 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: | 3 c_1 + 2 c_2 = 2 c_3 N: | c_1 = 2 c_4 O: | 2 c_2 = 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 3 c_3 = 6 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 NH_3 + 3 H_2O_2 ⟶ 6 H_2O + N_2

Structures

 + ⟶ +
+ ⟶ +

Names

ammonia + hydrogen peroxide ⟶ water + nitrogen
ammonia + hydrogen peroxide ⟶ water + nitrogen

Reaction thermodynamics

Gibbs free energy

 | ammonia | hydrogen peroxide | water | nitrogen molecular free energy | -16.4 kJ/mol | -120.4 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | -32.8 kJ/mol | -361.2 kJ/mol | -1423 kJ/mol | 0 kJ/mol  | G_initial = -394 kJ/mol | | G_final = -1423 kJ/mol |  ΔG_rxn^0 | -1423 kJ/mol - -394 kJ/mol = -1029 kJ/mol (exergonic) | | |
| ammonia | hydrogen peroxide | water | nitrogen molecular free energy | -16.4 kJ/mol | -120.4 kJ/mol | -237.1 kJ/mol | 0 kJ/mol total free energy | -32.8 kJ/mol | -361.2 kJ/mol | -1423 kJ/mol | 0 kJ/mol | G_initial = -394 kJ/mol | | G_final = -1423 kJ/mol | ΔG_rxn^0 | -1423 kJ/mol - -394 kJ/mol = -1029 kJ/mol (exergonic) | | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: NH_3 + H_2O_2 ⟶ 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: 2 NH_3 + 3 H_2O_2 ⟶ 6 H_2O + 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 NH_3 | 2 | -2 H_2O_2 | 3 | -3 H_2O | 6 | 6 N_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_3 | 2 | -2 | ([NH3])^(-2) H_2O_2 | 3 | -3 | ([H2O2])^(-3) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] 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 = ([NH3])^(-2) ([H2O2])^(-3) ([H2O])^6 [N2] = (([H2O])^6 [N2])/(([NH3])^2 ([H2O2])^3)
Construct the equilibrium constant, K, expression for: NH_3 + H_2O_2 ⟶ 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: 2 NH_3 + 3 H_2O_2 ⟶ 6 H_2O + 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 NH_3 | 2 | -2 H_2O_2 | 3 | -3 H_2O | 6 | 6 N_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_3 | 2 | -2 | ([NH3])^(-2) H_2O_2 | 3 | -3 | ([H2O2])^(-3) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] 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 = ([NH3])^(-2) ([H2O2])^(-3) ([H2O])^6 [N2] = (([H2O])^6 [N2])/(([NH3])^2 ([H2O2])^3)

Rate of reaction

Construct the rate of reaction expression for: NH_3 + H_2O_2 ⟶ 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: 2 NH_3 + 3 H_2O_2 ⟶ 6 H_2O + 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 NH_3 | 2 | -2 H_2O_2 | 3 | -3 H_2O | 6 | 6 N_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 NH_3 | 2 | -2 | -1/2 (Δ[NH3])/(Δt) H_2O_2 | 3 | -3 | -1/3 (Δ[H2O2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (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 = -1/2 (Δ[NH3])/(Δt) = -1/3 (Δ[H2O2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NH_3 + H_2O_2 ⟶ 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: 2 NH_3 + 3 H_2O_2 ⟶ 6 H_2O + 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 NH_3 | 2 | -2 H_2O_2 | 3 | -3 H_2O | 6 | 6 N_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 NH_3 | 2 | -2 | -1/2 (Δ[NH3])/(Δt) H_2O_2 | 3 | -3 | -1/3 (Δ[H2O2])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (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 = -1/2 (Δ[NH3])/(Δt) = -1/3 (Δ[H2O2])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | ammonia | hydrogen peroxide | water | nitrogen formula | NH_3 | H_2O_2 | H_2O | N_2 Hill formula | H_3N | H_2O_2 | H_2O | N_2 name | ammonia | hydrogen peroxide | water | nitrogen IUPAC name | ammonia | hydrogen peroxide | water | molecular nitrogen
| ammonia | hydrogen peroxide | water | nitrogen formula | NH_3 | H_2O_2 | H_2O | N_2 Hill formula | H_3N | H_2O_2 | H_2O | N_2 name | ammonia | hydrogen peroxide | water | nitrogen IUPAC name | ammonia | hydrogen peroxide | water | molecular nitrogen

Substance properties

 | ammonia | hydrogen peroxide | water | nitrogen molar mass | 17.031 g/mol | 34.014 g/mol | 18.015 g/mol | 28.014 g/mol phase | gas (at STP) | liquid (at STP) | liquid (at STP) | gas (at STP) melting point | -77.73 °C | -0.43 °C | 0 °C | -210 °C boiling point | -33.33 °C | 150.2 °C | 99.9839 °C | -195.79 °C density | 6.96×10^-4 g/cm^3 (at 25 °C) | 1.44 g/cm^3 | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) solubility in water | | miscible | | insoluble surface tension | 0.0234 N/m | 0.0804 N/m | 0.0728 N/m | 0.0066 N/m dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | 0.001249 Pa s (at 20 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) odor | | | odorless | odorless
| ammonia | hydrogen peroxide | water | nitrogen molar mass | 17.031 g/mol | 34.014 g/mol | 18.015 g/mol | 28.014 g/mol phase | gas (at STP) | liquid (at STP) | liquid (at STP) | gas (at STP) melting point | -77.73 °C | -0.43 °C | 0 °C | -210 °C boiling point | -33.33 °C | 150.2 °C | 99.9839 °C | -195.79 °C density | 6.96×10^-4 g/cm^3 (at 25 °C) | 1.44 g/cm^3 | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) solubility in water | | miscible | | insoluble surface tension | 0.0234 N/m | 0.0804 N/m | 0.0728 N/m | 0.0066 N/m dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | 0.001249 Pa s (at 20 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) odor | | | odorless | odorless

Units