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H2O + NH3 + P2O5 = (NH4)2HPO4

Input interpretation

H_2O water + NH_3 ammonia + P2O5 ⟶ (NH_4)_2HPO_4 diammonium hydrogen phosphate
H_2O water + NH_3 ammonia + P2O5 ⟶ (NH_4)_2HPO_4 diammonium hydrogen phosphate

Balanced equation

Balance the chemical equation algebraically: H_2O + NH_3 + P2O5 ⟶ (NH_4)_2HPO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NH_3 + c_3 P2O5 ⟶ c_4 (NH_4)_2HPO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and P: H: | 2 c_1 + 3 c_2 = 9 c_4 O: | c_1 + 5 c_3 = 4 c_4 N: | c_2 = 2 c_4 P: | 2 c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 4 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + 4 NH_3 + P2O5 ⟶ 2 (NH_4)_2HPO_4
Balance the chemical equation algebraically: H_2O + NH_3 + P2O5 ⟶ (NH_4)_2HPO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NH_3 + c_3 P2O5 ⟶ c_4 (NH_4)_2HPO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and P: H: | 2 c_1 + 3 c_2 = 9 c_4 O: | c_1 + 5 c_3 = 4 c_4 N: | c_2 = 2 c_4 P: | 2 c_3 = 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_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 3 c_2 = 4 c_3 = 1 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 3 H_2O + 4 NH_3 + P2O5 ⟶ 2 (NH_4)_2HPO_4

Structures

+ + P2O5 ⟶
+ + P2O5 ⟶

Names

water + ammonia + P2O5 ⟶ diammonium hydrogen phosphate
water + ammonia + P2O5 ⟶ diammonium hydrogen phosphate

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + NH_3 + P2O5 ⟶ (NH_4)_2HPO_4 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 H_2O + 4 NH_3 + P2O5 ⟶ 2 (NH_4)_2HPO_4 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 H_2O | 3 | -3 NH_3 | 4 | -4 P2O5 | 1 | -1 (NH_4)_2HPO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) NH_3 | 4 | -4 | ([NH3])^(-4) P2O5 | 1 | -1 | ([P2O5])^(-1) (NH_4)_2HPO_4 | 2 | 2 | ([(NH4)2HPO4])^2 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 = ([H2O])^(-3) ([NH3])^(-4) ([P2O5])^(-1) ([(NH4)2HPO4])^2 = ([(NH4)2HPO4])^2/(([H2O])^3 ([NH3])^4 [P2O5])
Construct the equilibrium constant, K, expression for: H_2O + NH_3 + P2O5 ⟶ (NH_4)_2HPO_4 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 H_2O + 4 NH_3 + P2O5 ⟶ 2 (NH_4)_2HPO_4 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 H_2O | 3 | -3 NH_3 | 4 | -4 P2O5 | 1 | -1 (NH_4)_2HPO_4 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 3 | -3 | ([H2O])^(-3) NH_3 | 4 | -4 | ([NH3])^(-4) P2O5 | 1 | -1 | ([P2O5])^(-1) (NH_4)_2HPO_4 | 2 | 2 | ([(NH4)2HPO4])^2 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 = ([H2O])^(-3) ([NH3])^(-4) ([P2O5])^(-1) ([(NH4)2HPO4])^2 = ([(NH4)2HPO4])^2/(([H2O])^3 ([NH3])^4 [P2O5])

Rate of reaction

Construct the rate of reaction expression for: H_2O + NH_3 + P2O5 ⟶ (NH_4)_2HPO_4 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 H_2O + 4 NH_3 + P2O5 ⟶ 2 (NH_4)_2HPO_4 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 H_2O | 3 | -3 NH_3 | 4 | -4 P2O5 | 1 | -1 (NH_4)_2HPO_4 | 2 | 2 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 H_2O | 3 | -3 | -1/3 (Δ[H2O])/(Δt) NH_3 | 4 | -4 | -1/4 (Δ[NH3])/(Δt) P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) (NH_4)_2HPO_4 | 2 | 2 | 1/2 (Δ[(NH4)2HPO4])/(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[H2O])/(Δt) = -1/4 (Δ[NH3])/(Δt) = -(Δ[P2O5])/(Δt) = 1/2 (Δ[(NH4)2HPO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + NH_3 + P2O5 ⟶ (NH_4)_2HPO_4 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 H_2O + 4 NH_3 + P2O5 ⟶ 2 (NH_4)_2HPO_4 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 H_2O | 3 | -3 NH_3 | 4 | -4 P2O5 | 1 | -1 (NH_4)_2HPO_4 | 2 | 2 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 H_2O | 3 | -3 | -1/3 (Δ[H2O])/(Δt) NH_3 | 4 | -4 | -1/4 (Δ[NH3])/(Δt) P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) (NH_4)_2HPO_4 | 2 | 2 | 1/2 (Δ[(NH4)2HPO4])/(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[H2O])/(Δt) = -1/4 (Δ[NH3])/(Δt) = -(Δ[P2O5])/(Δt) = 1/2 (Δ[(NH4)2HPO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | ammonia | P2O5 | diammonium hydrogen phosphate formula | H_2O | NH_3 | P2O5 | (NH_4)_2HPO_4 Hill formula | H_2O | H_3N | O5P2 | H_9N_2O_4P name | water | ammonia | | diammonium hydrogen phosphate
| water | ammonia | P2O5 | diammonium hydrogen phosphate formula | H_2O | NH_3 | P2O5 | (NH_4)_2HPO_4 Hill formula | H_2O | H_3N | O5P2 | H_9N_2O_4P name | water | ammonia | | diammonium hydrogen phosphate

Substance properties

| water | ammonia | P2O5 | diammonium hydrogen phosphate molar mass | 18.015 g/mol | 17.031 g/mol | 141.94 g/mol | 132.06 g/mol phase | liquid (at STP) | gas (at STP) | | solid (at STP) melting point | 0 °C | -77.73 °C | | 155 °C boiling point | 99.9839 °C | -33.33 °C | | density | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | | 1.619 g/cm^3 surface tension | 0.0728 N/m | 0.0234 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | | odor | odorless | | | odorless
| water | ammonia | P2O5 | diammonium hydrogen phosphate molar mass | 18.015 g/mol | 17.031 g/mol | 141.94 g/mol | 132.06 g/mol phase | liquid (at STP) | gas (at STP) | | solid (at STP) melting point | 0 °C | -77.73 °C | | 155 °C boiling point | 99.9839 °C | -33.33 °C | | density | 1 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | | 1.619 g/cm^3 surface tension | 0.0728 N/m | 0.0234 N/m | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | | odor | odorless | | | odorless

Units

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