Input interpretation
![H_3PO_4 phosphoric acid + Li_2O lithium oxide ⟶ H_2O water + Li_3PO_4 lithium phosphate](../image_source/3c76f5d18bd72b752f95fccdf21df758.png)
H_3PO_4 phosphoric acid + Li_2O lithium oxide ⟶ H_2O water + Li_3PO_4 lithium phosphate
Balanced equation
![Balance the chemical equation algebraically: H_3PO_4 + Li_2O ⟶ H_2O + Li_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 Li_2O ⟶ c_3 H_2O + c_4 Li_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Li: H: | 3 c_1 = 2 c_3 O: | 4 c_1 + c_2 = c_3 + 4 c_4 P: | c_1 = c_4 Li: | 2 c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + 3 Li_2O ⟶ 3 H_2O + 2 Li_3PO_4](../image_source/8b3a273ed2d38a69f378ed76bbc39dae.png)
Balance the chemical equation algebraically: H_3PO_4 + Li_2O ⟶ H_2O + Li_3PO_4 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_3PO_4 + c_2 Li_2O ⟶ c_3 H_2O + c_4 Li_3PO_4 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, P and Li: H: | 3 c_1 = 2 c_3 O: | 4 c_1 + c_2 = c_3 + 4 c_4 P: | c_1 = c_4 Li: | 2 c_2 = 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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3/2 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 3 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_3PO_4 + 3 Li_2O ⟶ 3 H_2O + 2 Li_3PO_4
Structures
![+ ⟶ +](../image_source/6048e82f535fb6211580b45d3e5eafa4.png)
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Names
![phosphoric acid + lithium oxide ⟶ water + lithium phosphate](../image_source/39a05ffda5ed66c510467a07ecfccd43.png)
phosphoric acid + lithium oxide ⟶ water + lithium phosphate
Equilibrium constant
![Construct the equilibrium constant, K, expression for: H_3PO_4 + Li_2O ⟶ H_2O + Li_3PO_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: 2 H_3PO_4 + 3 Li_2O ⟶ 3 H_2O + 2 Li_3PO_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_3PO_4 | 2 | -2 Li_2O | 3 | -3 H_2O | 3 | 3 Li_3PO_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_3PO_4 | 2 | -2 | ([H3PO4])^(-2) Li_2O | 3 | -3 | ([Li2O])^(-3) H_2O | 3 | 3 | ([H2O])^3 Li_3PO_4 | 2 | 2 | ([Li3PO4])^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 = ([H3PO4])^(-2) ([Li2O])^(-3) ([H2O])^3 ([Li3PO4])^2 = (([H2O])^3 ([Li3PO4])^2)/(([H3PO4])^2 ([Li2O])^3)](../image_source/b989b572469da792f220377cd00cedfa.png)
Construct the equilibrium constant, K, expression for: H_3PO_4 + Li_2O ⟶ H_2O + Li_3PO_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: 2 H_3PO_4 + 3 Li_2O ⟶ 3 H_2O + 2 Li_3PO_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_3PO_4 | 2 | -2 Li_2O | 3 | -3 H_2O | 3 | 3 Li_3PO_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_3PO_4 | 2 | -2 | ([H3PO4])^(-2) Li_2O | 3 | -3 | ([Li2O])^(-3) H_2O | 3 | 3 | ([H2O])^3 Li_3PO_4 | 2 | 2 | ([Li3PO4])^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 = ([H3PO4])^(-2) ([Li2O])^(-3) ([H2O])^3 ([Li3PO4])^2 = (([H2O])^3 ([Li3PO4])^2)/(([H3PO4])^2 ([Li2O])^3)
Rate of reaction
![Construct the rate of reaction expression for: H_3PO_4 + Li_2O ⟶ H_2O + Li_3PO_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: 2 H_3PO_4 + 3 Li_2O ⟶ 3 H_2O + 2 Li_3PO_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_3PO_4 | 2 | -2 Li_2O | 3 | -3 H_2O | 3 | 3 Li_3PO_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_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) Li_2O | 3 | -3 | -1/3 (Δ[Li2O])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Li_3PO_4 | 2 | 2 | 1/2 (Δ[Li3PO4])/(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[H3PO4])/(Δt) = -1/3 (Δ[Li2O])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Li3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/58fa6c0ff44b35c7c7e7a733435ffdc7.png)
Construct the rate of reaction expression for: H_3PO_4 + Li_2O ⟶ H_2O + Li_3PO_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: 2 H_3PO_4 + 3 Li_2O ⟶ 3 H_2O + 2 Li_3PO_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_3PO_4 | 2 | -2 Li_2O | 3 | -3 H_2O | 3 | 3 Li_3PO_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_3PO_4 | 2 | -2 | -1/2 (Δ[H3PO4])/(Δt) Li_2O | 3 | -3 | -1/3 (Δ[Li2O])/(Δt) H_2O | 3 | 3 | 1/3 (Δ[H2O])/(Δt) Li_3PO_4 | 2 | 2 | 1/2 (Δ[Li3PO4])/(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[H3PO4])/(Δt) = -1/3 (Δ[Li2O])/(Δt) = 1/3 (Δ[H2O])/(Δt) = 1/2 (Δ[Li3PO4])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| phosphoric acid | lithium oxide | water | lithium phosphate formula | H_3PO_4 | Li_2O | H_2O | Li_3PO_4 Hill formula | H_3O_4P | Li_2O | H_2O | Li_3O_4P name | phosphoric acid | lithium oxide | water | lithium phosphate IUPAC name | phosphoric acid | dilithium oxygen(-2) anion | water | trilithium phosphate](../image_source/d5b76988be332e93a03dc120fde3412e.png)
| phosphoric acid | lithium oxide | water | lithium phosphate formula | H_3PO_4 | Li_2O | H_2O | Li_3PO_4 Hill formula | H_3O_4P | Li_2O | H_2O | Li_3O_4P name | phosphoric acid | lithium oxide | water | lithium phosphate IUPAC name | phosphoric acid | dilithium oxygen(-2) anion | water | trilithium phosphate
Substance properties
![| phosphoric acid | lithium oxide | water | lithium phosphate molar mass | 97.994 g/mol | 29.9 g/mol | 18.015 g/mol | 115.8 g/mol phase | liquid (at STP) | | liquid (at STP) | melting point | 42.4 °C | | 0 °C | boiling point | 158 °C | | 99.9839 °C | density | 1.685 g/cm^3 | 2.013 g/cm^3 | 1 g/cm^3 | solubility in water | very soluble | | | surface tension | | | 0.0728 N/m | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |](../image_source/4438c12afe7f9d69e0a5184759bb8d3b.png)
| phosphoric acid | lithium oxide | water | lithium phosphate molar mass | 97.994 g/mol | 29.9 g/mol | 18.015 g/mol | 115.8 g/mol phase | liquid (at STP) | | liquid (at STP) | melting point | 42.4 °C | | 0 °C | boiling point | 158 °C | | 99.9839 °C | density | 1.685 g/cm^3 | 2.013 g/cm^3 | 1 g/cm^3 | solubility in water | very soluble | | | surface tension | | | 0.0728 N/m | dynamic viscosity | | | 8.9×10^-4 Pa s (at 25 °C) | odor | odorless | | odorless |
Units