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NH3 + Li = H2 + Li2NH

Input interpretation

NH_3 ammonia + Li lithium ⟶ H_2 hydrogen + Li2NH
NH_3 ammonia + Li lithium ⟶ H_2 hydrogen + Li2NH

Balanced equation

Balance the chemical equation algebraically: NH_3 + Li ⟶ H_2 + Li2NH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 Li ⟶ c_3 H_2 + c_4 Li2NH Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and Li: H: | 3 c_1 = 2 c_3 + c_4 N: | c_1 = c_4 Li: | c_2 = 2 c_4 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 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: |   | NH_3 + 2 Li ⟶ H_2 + Li2NH
Balance the chemical equation algebraically: NH_3 + Li ⟶ H_2 + Li2NH Add stoichiometric coefficients, c_i, to the reactants and products: c_1 NH_3 + c_2 Li ⟶ c_3 H_2 + c_4 Li2NH Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and Li: H: | 3 c_1 = 2 c_3 + c_4 N: | c_1 = c_4 Li: | c_2 = 2 c_4 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 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | NH_3 + 2 Li ⟶ H_2 + Li2NH

Structures

 + ⟶ + Li2NH
+ ⟶ + Li2NH

Names

ammonia + lithium ⟶ hydrogen + Li2NH
ammonia + lithium ⟶ hydrogen + Li2NH

Equilibrium constant

Construct the equilibrium constant, K, expression for: NH_3 + Li ⟶ H_2 + Li2NH 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_3 + 2 Li ⟶ H_2 + Li2NH 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 | 1 | -1 Li | 2 | -2 H_2 | 1 | 1 Li2NH | 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 | 1 | -1 | ([NH3])^(-1) Li | 2 | -2 | ([Li])^(-2) H_2 | 1 | 1 | [H2] Li2NH | 1 | 1 | [Li2NH] 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])^(-1) ([Li])^(-2) [H2] [Li2NH] = ([H2] [Li2NH])/([NH3] ([Li])^2)
Construct the equilibrium constant, K, expression for: NH_3 + Li ⟶ H_2 + Li2NH 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_3 + 2 Li ⟶ H_2 + Li2NH 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 | 1 | -1 Li | 2 | -2 H_2 | 1 | 1 Li2NH | 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 | 1 | -1 | ([NH3])^(-1) Li | 2 | -2 | ([Li])^(-2) H_2 | 1 | 1 | [H2] Li2NH | 1 | 1 | [Li2NH] 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])^(-1) ([Li])^(-2) [H2] [Li2NH] = ([H2] [Li2NH])/([NH3] ([Li])^2)

Rate of reaction

Construct the rate of reaction expression for: NH_3 + Li ⟶ H_2 + Li2NH 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_3 + 2 Li ⟶ H_2 + Li2NH 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 | 1 | -1 Li | 2 | -2 H_2 | 1 | 1 Li2NH | 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 | 1 | -1 | -(Δ[NH3])/(Δt) Li | 2 | -2 | -1/2 (Δ[Li])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) Li2NH | 1 | 1 | (Δ[Li2NH])/(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[NH3])/(Δt) = -1/2 (Δ[Li])/(Δt) = (Δ[H2])/(Δt) = (Δ[Li2NH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: NH_3 + Li ⟶ H_2 + Li2NH 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_3 + 2 Li ⟶ H_2 + Li2NH 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 | 1 | -1 Li | 2 | -2 H_2 | 1 | 1 Li2NH | 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 | 1 | -1 | -(Δ[NH3])/(Δt) Li | 2 | -2 | -1/2 (Δ[Li])/(Δt) H_2 | 1 | 1 | (Δ[H2])/(Δt) Li2NH | 1 | 1 | (Δ[Li2NH])/(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[NH3])/(Δt) = -1/2 (Δ[Li])/(Δt) = (Δ[H2])/(Δt) = (Δ[Li2NH])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

 | ammonia | lithium | hydrogen | Li2NH formula | NH_3 | Li | H_2 | Li2NH Hill formula | H_3N | Li | H_2 | HLi2N name | ammonia | lithium | hydrogen |  IUPAC name | ammonia | lithium | molecular hydrogen |
| ammonia | lithium | hydrogen | Li2NH formula | NH_3 | Li | H_2 | Li2NH Hill formula | H_3N | Li | H_2 | HLi2N name | ammonia | lithium | hydrogen | IUPAC name | ammonia | lithium | molecular hydrogen |

Substance properties

 | ammonia | lithium | hydrogen | Li2NH molar mass | 17.031 g/mol | 6.94 g/mol | 2.016 g/mol | 28.9 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) |  melting point | -77.73 °C | 180 °C | -259.2 °C |  boiling point | -33.33 °C | 1342 °C | -252.8 °C |  density | 6.96×10^-4 g/cm^3 (at 25 °C) | 0.534 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) |  solubility in water | | decomposes | |  surface tension | 0.0234 N/m | 0.3975 N/m | |  dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) |  odor | | | odorless |
| ammonia | lithium | hydrogen | Li2NH molar mass | 17.031 g/mol | 6.94 g/mol | 2.016 g/mol | 28.9 g/mol phase | gas (at STP) | solid (at STP) | gas (at STP) | melting point | -77.73 °C | 180 °C | -259.2 °C | boiling point | -33.33 °C | 1342 °C | -252.8 °C | density | 6.96×10^-4 g/cm^3 (at 25 °C) | 0.534 g/cm^3 | 8.99×10^-5 g/cm^3 (at 0 °C) | solubility in water | | decomposes | | surface tension | 0.0234 N/m | 0.3975 N/m | | dynamic viscosity | 1.009×10^-5 Pa s (at 25 °C) | | 8.9×10^-6 Pa s (at 25 °C) | odor | | | odorless |

Units