Input interpretation
![HNO_3 nitric acid + Sn white tin ⟶ H_2O water + N_2 nitrogen + Sn(NO3)2](../image_source/bbeac602d277a65303ec3df9a07295c9.png)
HNO_3 nitric acid + Sn white tin ⟶ H_2O water + N_2 nitrogen + Sn(NO3)2
Balanced equation
![Balance the chemical equation algebraically: HNO_3 + Sn ⟶ H_2O + N_2 + Sn(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sn ⟶ c_3 H_2O + c_4 N_2 + c_5 Sn(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Sn: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 c_5 Sn: | c_2 = c_5 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 5 c_3 = 6 c_4 = 1 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HNO_3 + 5 Sn ⟶ 6 H_2O + N_2 + 5 Sn(NO3)2](../image_source/1c0e4d9479349401612d17953bc0ce8b.png)
Balance the chemical equation algebraically: HNO_3 + Sn ⟶ H_2O + N_2 + Sn(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Sn ⟶ c_3 H_2O + c_4 N_2 + c_5 Sn(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Sn: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 c_5 Sn: | c_2 = c_5 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_4 = 1 and solve the system of equations for the remaining coefficients: c_1 = 12 c_2 = 5 c_3 = 6 c_4 = 1 c_5 = 5 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 12 HNO_3 + 5 Sn ⟶ 6 H_2O + N_2 + 5 Sn(NO3)2
Structures
![+ ⟶ + + Sn(NO3)2](../image_source/ba70fb91c11e650336282413b0bc0e9e.png)
+ ⟶ + + Sn(NO3)2
Names
![nitric acid + white tin ⟶ water + nitrogen + Sn(NO3)2](../image_source/358e59c3aa913fd9f4047241f8f18db1.png)
nitric acid + white tin ⟶ water + nitrogen + Sn(NO3)2
Equilibrium constant
![Construct the equilibrium constant, K, expression for: HNO_3 + Sn ⟶ H_2O + N_2 + Sn(NO3)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: 12 HNO_3 + 5 Sn ⟶ 6 H_2O + N_2 + 5 Sn(NO3)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 HNO_3 | 12 | -12 Sn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Sn(NO3)2 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 12 | -12 | ([HNO3])^(-12) Sn | 5 | -5 | ([Sn])^(-5) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] Sn(NO3)2 | 5 | 5 | ([Sn(NO3)2])^5 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])^(-12) ([Sn])^(-5) ([H2O])^6 [N2] ([Sn(NO3)2])^5 = (([H2O])^6 [N2] ([Sn(NO3)2])^5)/(([HNO3])^12 ([Sn])^5)](../image_source/298c186317defb8d52d09c4c74e4bbe6.png)
Construct the equilibrium constant, K, expression for: HNO_3 + Sn ⟶ H_2O + N_2 + Sn(NO3)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: 12 HNO_3 + 5 Sn ⟶ 6 H_2O + N_2 + 5 Sn(NO3)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 HNO_3 | 12 | -12 Sn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Sn(NO3)2 | 5 | 5 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 12 | -12 | ([HNO3])^(-12) Sn | 5 | -5 | ([Sn])^(-5) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] Sn(NO3)2 | 5 | 5 | ([Sn(NO3)2])^5 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])^(-12) ([Sn])^(-5) ([H2O])^6 [N2] ([Sn(NO3)2])^5 = (([H2O])^6 [N2] ([Sn(NO3)2])^5)/(([HNO3])^12 ([Sn])^5)
Rate of reaction
![Construct the rate of reaction expression for: HNO_3 + Sn ⟶ H_2O + N_2 + Sn(NO3)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: 12 HNO_3 + 5 Sn ⟶ 6 H_2O + N_2 + 5 Sn(NO3)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 HNO_3 | 12 | -12 Sn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Sn(NO3)2 | 5 | 5 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 | 12 | -12 | -1/12 (Δ[HNO3])/(Δt) Sn | 5 | -5 | -1/5 (Δ[Sn])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) Sn(NO3)2 | 5 | 5 | 1/5 (Δ[Sn(NO3)2])/(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 = -1/12 (Δ[HNO3])/(Δt) = -1/5 (Δ[Sn])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) = 1/5 (Δ[Sn(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/9070a9d7a7415b7388b65a7b9532724a.png)
Construct the rate of reaction expression for: HNO_3 + Sn ⟶ H_2O + N_2 + Sn(NO3)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: 12 HNO_3 + 5 Sn ⟶ 6 H_2O + N_2 + 5 Sn(NO3)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 HNO_3 | 12 | -12 Sn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Sn(NO3)2 | 5 | 5 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 | 12 | -12 | -1/12 (Δ[HNO3])/(Δt) Sn | 5 | -5 | -1/5 (Δ[Sn])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) Sn(NO3)2 | 5 | 5 | 1/5 (Δ[Sn(NO3)2])/(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 = -1/12 (Δ[HNO3])/(Δt) = -1/5 (Δ[Sn])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) = 1/5 (Δ[Sn(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| nitric acid | white tin | water | nitrogen | Sn(NO3)2 formula | HNO_3 | Sn | H_2O | N_2 | Sn(NO3)2 Hill formula | HNO_3 | Sn | H_2O | N_2 | N2O6Sn name | nitric acid | white tin | water | nitrogen | IUPAC name | nitric acid | tin | water | molecular nitrogen |](../image_source/a14b9620c136c6108fd2fe3cf761dd8e.png)
| nitric acid | white tin | water | nitrogen | Sn(NO3)2 formula | HNO_3 | Sn | H_2O | N_2 | Sn(NO3)2 Hill formula | HNO_3 | Sn | H_2O | N_2 | N2O6Sn name | nitric acid | white tin | water | nitrogen | IUPAC name | nitric acid | tin | water | molecular nitrogen |
Substance properties
![| nitric acid | white tin | water | nitrogen | Sn(NO3)2 molar mass | 63.012 g/mol | 118.71 g/mol | 18.015 g/mol | 28.014 g/mol | 242.72 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 231.9 °C | 0 °C | -210 °C | boiling point | 83 °C | 2602 °C | 99.9839 °C | -195.79 °C | density | 1.5129 g/cm^3 | 7.31 g/cm^3 | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) | solubility in water | miscible | insoluble | | insoluble | surface tension | | | 0.0728 N/m | 0.0066 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.001 Pa s (at 600 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | odorless |](../image_source/1d4dc53c341d902d5e63402b62f3855b.png)
| nitric acid | white tin | water | nitrogen | Sn(NO3)2 molar mass | 63.012 g/mol | 118.71 g/mol | 18.015 g/mol | 28.014 g/mol | 242.72 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 231.9 °C | 0 °C | -210 °C | boiling point | 83 °C | 2602 °C | 99.9839 °C | -195.79 °C | density | 1.5129 g/cm^3 | 7.31 g/cm^3 | 1 g/cm^3 | 0.001251 g/cm^3 (at 0 °C) | solubility in water | miscible | insoluble | | insoluble | surface tension | | | 0.0728 N/m | 0.0066 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 0.001 Pa s (at 600 °C) | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | odorless |
Units