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NH4NO3 = H2O + N2O

Input interpretation

NH_4NO_3 (ammonium nitrate) ⟶ H_2O (water) + N_2O (nitrous oxide)
NH_4NO_3 (ammonium nitrate) ⟶ H_2O (water) + N_2O (nitrous oxide)

Balanced equation

Balance the chemical equation algebraically: NH_4NO_3 ⟶ H_2O + N_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_4NO_3 ⟶ c_2 H_2O + c_3 N_2O Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | 4 c_1 = 2 c_2 N: | 2 c_1 = 2 c_3 O: | 3 c_1 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | NH_4NO_3 ⟶ 2 H_2O + N_2O
Balance the chemical equation algebraically: NH_4NO_3 ⟶ H_2O + N_2O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_4NO_3 ⟶ c_2 H_2O + c_3 N_2O Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | 4 c_1 = 2 c_2 N: | 2 c_1 = 2 c_3 O: | 3 c_1 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 2 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | NH_4NO_3 ⟶ 2 H_2O + N_2O

Structures

 ⟶ +
⟶ +

Names

ammonium nitrate ⟶ water + nitrous oxide
ammonium nitrate ⟶ water + nitrous oxide

Reaction thermodynamics

Gibbs free energy

 | ammonium nitrate | water | nitrous oxide molecular free energy | -183.9 kJ/mol | -237.1 kJ/mol | 104 kJ/mol total free energy | -183.9 kJ/mol | -474.2 kJ/mol | 104 kJ/mol  | G_initial = -183.9 kJ/mol | G_final = -370.2 kJ/mol |  ΔG_rxn^0 | -370.2 kJ/mol - -183.9 kJ/mol = -186.3 kJ/mol (exergonic) | |
| ammonium nitrate | water | nitrous oxide molecular free energy | -183.9 kJ/mol | -237.1 kJ/mol | 104 kJ/mol total free energy | -183.9 kJ/mol | -474.2 kJ/mol | 104 kJ/mol | G_initial = -183.9 kJ/mol | G_final = -370.2 kJ/mol | ΔG_rxn^0 | -370.2 kJ/mol - -183.9 kJ/mol = -186.3 kJ/mol (exergonic) | |

Equilibrium constant

Construct the equilibrium constant, K, expression for: NH_4NO_3 ⟶ H_2O + N_2O 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: NH_4NO_3 ⟶ 2 H_2O + N_2O 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_4NO_3 | 1 | -1 H_2O | 2 | 2 N_2O | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_4NO_3 | 1 | -1 | ([NH4NO3])^(-1) H_2O | 2 | 2 | ([H2O])^2 N_2O | 1 | 1 | [N2O] 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 = ([NH4NO3])^(-1) ([H2O])^2 [N2O] = (([H2O])^2 [N2O])/([NH4NO3])
Construct the equilibrium constant, K, expression for: NH_4NO_3 ⟶ H_2O + N_2O 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: NH_4NO_3 ⟶ 2 H_2O + N_2O 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_4NO_3 | 1 | -1 H_2O | 2 | 2 N_2O | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression NH_4NO_3 | 1 | -1 | ([NH4NO3])^(-1) H_2O | 2 | 2 | ([H2O])^2 N_2O | 1 | 1 | [N2O] 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 = ([NH4NO3])^(-1) ([H2O])^2 [N2O] = (([H2O])^2 [N2O])/([NH4NO3])

Rate of reaction

Construct the rate of reaction expression for: NH_4NO_3 ⟶ H_2O + N_2O 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: NH_4NO_3 ⟶ 2 H_2O + N_2O 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_4NO_3 | 1 | -1 H_2O | 2 | 2 N_2O | 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_4NO_3 | 1 | -1 | -(Δ[NH4NO3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) N_2O | 1 | 1 | (Δ[N2O])/(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 = -(Δ[NH4NO3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[N2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NH_4NO_3 ⟶ H_2O + N_2O 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: NH_4NO_3 ⟶ 2 H_2O + N_2O 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_4NO_3 | 1 | -1 H_2O | 2 | 2 N_2O | 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_4NO_3 | 1 | -1 | -(Δ[NH4NO3])/(Δt) H_2O | 2 | 2 | 1/2 (Δ[H2O])/(Δt) N_2O | 1 | 1 | (Δ[N2O])/(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 = -(Δ[NH4NO3])/(Δt) = 1/2 (Δ[H2O])/(Δt) = (Δ[N2O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | ammonium nitrate | water | nitrous oxide formula | NH_4NO_3 | H_2O | N_2O Hill formula | H_4N_2O_3 | H_2O | N_2O name | ammonium nitrate | water | nitrous oxide
| ammonium nitrate | water | nitrous oxide formula | NH_4NO_3 | H_2O | N_2O Hill formula | H_4N_2O_3 | H_2O | N_2O name | ammonium nitrate | water | nitrous oxide

Substance properties

 | ammonium nitrate | water | nitrous oxide molar mass | 80.04 g/mol | 18.015 g/mol | 44.013 g/mol phase | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 169 °C | 0 °C | -91 °C boiling point | 210 °C | 99.9839 °C | -88 °C density | 1.73 g/cm^3 | 1 g/cm^3 | 0.001799 g/cm^3 (at 25 °C) surface tension | | 0.0728 N/m | 0.00175 N/m dynamic viscosity | | 8.9×10^-4 Pa s (at 25 °C) | 1.491×10^-5 Pa s (at 25 °C) odor | odorless | odorless |
| ammonium nitrate | water | nitrous oxide molar mass | 80.04 g/mol | 18.015 g/mol | 44.013 g/mol phase | solid (at STP) | liquid (at STP) | gas (at STP) melting point | 169 °C | 0 °C | -91 °C boiling point | 210 °C | 99.9839 °C | -88 °C density | 1.73 g/cm^3 | 1 g/cm^3 | 0.001799 g/cm^3 (at 25 °C) surface tension | | 0.0728 N/m | 0.00175 N/m dynamic viscosity | | 8.9×10^-4 Pa s (at 25 °C) | 1.491×10^-5 Pa s (at 25 °C) odor | odorless | odorless |

Units