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HNO3 + P2O5 = H3PO4 + N2O5

Input interpretation

HNO_3 nitric acid + P2O5 ⟶ H_3PO_4 phosphoric acid + N_2O_5 dinitrogen pentoxide
HNO_3 nitric acid + P2O5 ⟶ H_3PO_4 phosphoric acid + N_2O_5 dinitrogen pentoxide

Balanced equation

Balance the chemical equation algebraically: HNO_3 + P2O5 ⟶ H_3PO_4 + N_2O_5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 P2O5 ⟶ c_3 H_3PO_4 + c_4 N_2O_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and P: H: | c_1 = 3 c_3 N: | c_1 = 2 c_4 O: | 3 c_1 + 5 c_2 = 4 c_3 + 5 c_4 P: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HNO_3 + P2O5 ⟶ 2 H_3PO_4 + 3 N_2O_5
Balance the chemical equation algebraically: HNO_3 + P2O5 ⟶ H_3PO_4 + N_2O_5 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 P2O5 ⟶ c_3 H_3PO_4 + c_4 N_2O_5 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and P: H: | c_1 = 3 c_3 N: | c_1 = 2 c_4 O: | 3 c_1 + 5 c_2 = 4 c_3 + 5 c_4 P: | 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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 6 c_2 = 1 c_3 = 2 c_4 = 3 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 6 HNO_3 + P2O5 ⟶ 2 H_3PO_4 + 3 N_2O_5

Structures

+ P2O5 ⟶ +
+ P2O5 ⟶ +

Names

nitric acid + P2O5 ⟶ phosphoric acid + dinitrogen pentoxide
nitric acid + P2O5 ⟶ phosphoric acid + dinitrogen pentoxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + P2O5 ⟶ H_3PO_4 + N_2O_5 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: 6 HNO_3 + P2O5 ⟶ 2 H_3PO_4 + 3 N_2O_5 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 | 6 | -6 P2O5 | 1 | -1 H_3PO_4 | 2 | 2 N_2O_5 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 6 | -6 | ([HNO3])^(-6) P2O5 | 1 | -1 | ([P2O5])^(-1) H_3PO_4 | 2 | 2 | ([H3PO4])^2 N_2O_5 | 3 | 3 | ([N2O5])^3 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])^(-6) ([P2O5])^(-1) ([H3PO4])^2 ([N2O5])^3 = (([H3PO4])^2 ([N2O5])^3)/(([HNO3])^6 [P2O5])
Construct the equilibrium constant, K, expression for: HNO_3 + P2O5 ⟶ H_3PO_4 + N_2O_5 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: 6 HNO_3 + P2O5 ⟶ 2 H_3PO_4 + 3 N_2O_5 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 | 6 | -6 P2O5 | 1 | -1 H_3PO_4 | 2 | 2 N_2O_5 | 3 | 3 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 6 | -6 | ([HNO3])^(-6) P2O5 | 1 | -1 | ([P2O5])^(-1) H_3PO_4 | 2 | 2 | ([H3PO4])^2 N_2O_5 | 3 | 3 | ([N2O5])^3 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])^(-6) ([P2O5])^(-1) ([H3PO4])^2 ([N2O5])^3 = (([H3PO4])^2 ([N2O5])^3)/(([HNO3])^6 [P2O5])

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + P2O5 ⟶ H_3PO_4 + N_2O_5 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: 6 HNO_3 + P2O5 ⟶ 2 H_3PO_4 + 3 N_2O_5 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 | 6 | -6 P2O5 | 1 | -1 H_3PO_4 | 2 | 2 N_2O_5 | 3 | 3 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 | 6 | -6 | -1/6 (Δ[HNO3])/(Δt) P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) H_3PO_4 | 2 | 2 | 1/2 (Δ[H3PO4])/(Δt) N_2O_5 | 3 | 3 | 1/3 (Δ[N2O5])/(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 = -1/6 (Δ[HNO3])/(Δt) = -(Δ[P2O5])/(Δt) = 1/2 (Δ[H3PO4])/(Δt) = 1/3 (Δ[N2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + P2O5 ⟶ H_3PO_4 + N_2O_5 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: 6 HNO_3 + P2O5 ⟶ 2 H_3PO_4 + 3 N_2O_5 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 | 6 | -6 P2O5 | 1 | -1 H_3PO_4 | 2 | 2 N_2O_5 | 3 | 3 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 | 6 | -6 | -1/6 (Δ[HNO3])/(Δt) P2O5 | 1 | -1 | -(Δ[P2O5])/(Δt) H_3PO_4 | 2 | 2 | 1/2 (Δ[H3PO4])/(Δt) N_2O_5 | 3 | 3 | 1/3 (Δ[N2O5])/(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 = -1/6 (Δ[HNO3])/(Δt) = -(Δ[P2O5])/(Δt) = 1/2 (Δ[H3PO4])/(Δt) = 1/3 (Δ[N2O5])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | P2O5 | phosphoric acid | dinitrogen pentoxide formula | HNO_3 | P2O5 | H_3PO_4 | N_2O_5 Hill formula | HNO_3 | O5P2 | H_3O_4P | N_2O_5 name | nitric acid | | phosphoric acid | dinitrogen pentoxide IUPAC name | nitric acid | | phosphoric acid | nitro nitrate
| nitric acid | P2O5 | phosphoric acid | dinitrogen pentoxide formula | HNO_3 | P2O5 | H_3PO_4 | N_2O_5 Hill formula | HNO_3 | O5P2 | H_3O_4P | N_2O_5 name | nitric acid | | phosphoric acid | dinitrogen pentoxide IUPAC name | nitric acid | | phosphoric acid | nitro nitrate

Substance properties

| nitric acid | P2O5 | phosphoric acid | dinitrogen pentoxide molar mass | 63.012 g/mol | 141.94 g/mol | 97.994 g/mol | 108.01 g/mol phase | liquid (at STP) | | liquid (at STP) | solid (at STP) melting point | -41.6 °C | | 42.4 °C | 30 °C boiling point | 83 °C | | 158 °C | 47 °C density | 1.5129 g/cm^3 | | 1.685 g/cm^3 | 2.05 g/cm^3 solubility in water | miscible | | very soluble | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | odor | | | odorless |
| nitric acid | P2O5 | phosphoric acid | dinitrogen pentoxide molar mass | 63.012 g/mol | 141.94 g/mol | 97.994 g/mol | 108.01 g/mol phase | liquid (at STP) | | liquid (at STP) | solid (at STP) melting point | -41.6 °C | | 42.4 °C | 30 °C boiling point | 83 °C | | 158 °C | 47 °C density | 1.5129 g/cm^3 | | 1.685 g/cm^3 | 2.05 g/cm^3 solubility in water | miscible | | very soluble | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | | | odor | | | odorless |

Units

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