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H2O + Th(NO3)4 = HNO3 + ThO2

Input interpretation

H_2O water + Th(NO_3)_4 thorium nitrate ⟶ HNO_3 nitric acid + ThO_2 thorium(IV) oxide
H_2O water + Th(NO_3)_4 thorium nitrate ⟶ HNO_3 nitric acid + ThO_2 thorium(IV) oxide

Balanced equation

Balance the chemical equation algebraically: H_2O + Th(NO_3)_4 ⟶ HNO_3 + ThO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Th(NO_3)_4 ⟶ c_3 HNO_3 + c_4 ThO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and Th: H: | 2 c_1 = c_3 O: | c_1 + 12 c_2 = 3 c_3 + 2 c_4 N: | 4 c_2 = c_3 Th: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 4 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + Th(NO_3)_4 ⟶ 4 HNO_3 + ThO_2
Balance the chemical equation algebraically: H_2O + Th(NO_3)_4 ⟶ HNO_3 + ThO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 Th(NO_3)_4 ⟶ c_3 HNO_3 + c_4 ThO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, N and Th: H: | 2 c_1 = c_3 O: | c_1 + 12 c_2 = 3 c_3 + 2 c_4 N: | 4 c_2 = c_3 Th: | c_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_2 = 1 and solve the system of equations for the remaining coefficients: c_1 = 2 c_2 = 1 c_3 = 4 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 H_2O + Th(NO_3)_4 ⟶ 4 HNO_3 + ThO_2

Structures

+ ⟶ +
+ ⟶ +

Names

water + thorium nitrate ⟶ nitric acid + thorium(IV) oxide
water + thorium nitrate ⟶ nitric acid + thorium(IV) oxide

Equilibrium constant

Construct the equilibrium constant, K, expression for: H_2O + Th(NO_3)_4 ⟶ HNO_3 + ThO_2 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_2O + Th(NO_3)_4 ⟶ 4 HNO_3 + ThO_2 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 | 2 | -2 Th(NO_3)_4 | 1 | -1 HNO_3 | 4 | 4 ThO_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) Th(NO_3)_4 | 1 | -1 | ([Th(NO3)4])^(-1) HNO_3 | 4 | 4 | ([HNO3])^4 ThO_2 | 1 | 1 | [ThO2] 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])^(-2) ([Th(NO3)4])^(-1) ([HNO3])^4 [ThO2] = (([HNO3])^4 [ThO2])/(([H2O])^2 [Th(NO3)4])
Construct the equilibrium constant, K, expression for: H_2O + Th(NO_3)_4 ⟶ HNO_3 + ThO_2 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_2O + Th(NO_3)_4 ⟶ 4 HNO_3 + ThO_2 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 | 2 | -2 Th(NO_3)_4 | 1 | -1 HNO_3 | 4 | 4 ThO_2 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression H_2O | 2 | -2 | ([H2O])^(-2) Th(NO_3)_4 | 1 | -1 | ([Th(NO3)4])^(-1) HNO_3 | 4 | 4 | ([HNO3])^4 ThO_2 | 1 | 1 | [ThO2] 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])^(-2) ([Th(NO3)4])^(-1) ([HNO3])^4 [ThO2] = (([HNO3])^4 [ThO2])/(([H2O])^2 [Th(NO3)4])

Rate of reaction

Construct the rate of reaction expression for: H_2O + Th(NO_3)_4 ⟶ HNO_3 + ThO_2 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_2O + Th(NO_3)_4 ⟶ 4 HNO_3 + ThO_2 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 | 2 | -2 Th(NO_3)_4 | 1 | -1 HNO_3 | 4 | 4 ThO_2 | 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) Th(NO_3)_4 | 1 | -1 | -(Δ[Th(NO3)4])/(Δt) HNO_3 | 4 | 4 | 1/4 (Δ[HNO3])/(Δt) ThO_2 | 1 | 1 | (Δ[ThO2])/(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[H2O])/(Δt) = -(Δ[Th(NO3)4])/(Δt) = 1/4 (Δ[HNO3])/(Δt) = (Δ[ThO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: H_2O + Th(NO_3)_4 ⟶ HNO_3 + ThO_2 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_2O + Th(NO_3)_4 ⟶ 4 HNO_3 + ThO_2 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 | 2 | -2 Th(NO_3)_4 | 1 | -1 HNO_3 | 4 | 4 ThO_2 | 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 H_2O | 2 | -2 | -1/2 (Δ[H2O])/(Δt) Th(NO_3)_4 | 1 | -1 | -(Δ[Th(NO3)4])/(Δt) HNO_3 | 4 | 4 | 1/4 (Δ[HNO3])/(Δt) ThO_2 | 1 | 1 | (Δ[ThO2])/(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[H2O])/(Δt) = -(Δ[Th(NO3)4])/(Δt) = 1/4 (Δ[HNO3])/(Δt) = (Δ[ThO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| water | thorium nitrate | nitric acid | thorium(IV) oxide formula | H_2O | Th(NO_3)_4 | HNO_3 | ThO_2 Hill formula | H_2O | N_4O_12Th | HNO_3 | O_2Th name | water | thorium nitrate | nitric acid | thorium(IV) oxide IUPAC name | water | thorium(+4) cation tetranitrate | nitric acid | oxygen(-2) anion; thorium(+4) cation
| water | thorium nitrate | nitric acid | thorium(IV) oxide formula | H_2O | Th(NO_3)_4 | HNO_3 | ThO_2 Hill formula | H_2O | N_4O_12Th | HNO_3 | O_2Th name | water | thorium nitrate | nitric acid | thorium(IV) oxide IUPAC name | water | thorium(+4) cation tetranitrate | nitric acid | oxygen(-2) anion; thorium(+4) cation

Substance properties

| water | thorium nitrate | nitric acid | thorium(IV) oxide molar mass | 18.015 g/mol | 480.05 g/mol | 63.012 g/mol | 264.036 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | 0 °C | 55 °C | -41.6 °C | 3390 °C boiling point | 99.9839 °C | 837 °C | 83 °C | 4400 °C density | 1 g/cm^3 | | 1.5129 g/cm^3 | 10 g/cm^3 solubility in water | | | miscible | insoluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 7.6×10^-4 Pa s (at 25 °C) | odor | odorless | | |
| water | thorium nitrate | nitric acid | thorium(IV) oxide molar mass | 18.015 g/mol | 480.05 g/mol | 63.012 g/mol | 264.036 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | solid (at STP) melting point | 0 °C | 55 °C | -41.6 °C | 3390 °C boiling point | 99.9839 °C | 837 °C | 83 °C | 4400 °C density | 1 g/cm^3 | | 1.5129 g/cm^3 | 10 g/cm^3 solubility in water | | | miscible | insoluble surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 7.6×10^-4 Pa s (at 25 °C) | odor | odorless | | |

Units

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