Input interpretation
![H_2O (water) + NO_2 (nitrogen dioxide) ⟶ HNO_3 (nitric acid) + NO (nitric oxide)](../image_source/3af6e1cdf7c5680e6a636d21c0cd8238.png)
H_2O (water) + NO_2 (nitrogen dioxide) ⟶ HNO_3 (nitric acid) + NO (nitric oxide)
Balanced equation
![Balance the chemical equation algebraically: H_2O + NO_2 ⟶ HNO_3 + NO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NO_2 ⟶ c_3 HNO_3 + c_4 NO Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and N: H: | 2 c_1 = c_3 O: | c_1 + 2 c_2 = 3 c_3 + c_4 N: | c_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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + 3 NO_2 ⟶ 2 HNO_3 + NO](../image_source/938e26b36d58a2b589a5deb0c6db33f6.png)
Balance the chemical equation algebraically: H_2O + NO_2 ⟶ HNO_3 + NO Add stoichiometric coefficients, c_i, to the reactants and products: c_1 H_2O + c_2 NO_2 ⟶ c_3 HNO_3 + c_4 NO Set the number of atoms in the reactants equal to the number of atoms in the products for H, O and N: H: | 2 c_1 = c_3 O: | c_1 + 2 c_2 = 3 c_3 + c_4 N: | c_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_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | H_2O + 3 NO_2 ⟶ 2 HNO_3 + NO
Structures
![+ ⟶ +](../image_source/409d569e3383f3c48b3d47432b777769.png)
+ ⟶ +
Names
![water + nitrogen dioxide ⟶ nitric acid + nitric oxide](../image_source/21988c926685b2dce45c47d314278beb.png)
water + nitrogen dioxide ⟶ nitric acid + nitric oxide
Reaction thermodynamics
Gibbs free energy
![| water | nitrogen dioxide | nitric acid | nitric oxide molecular free energy | -237.1 kJ/mol | 51.3 kJ/mol | -80.7 kJ/mol | 87.6 kJ/mol total free energy | -237.1 kJ/mol | 153.9 kJ/mol | -161.4 kJ/mol | 87.6 kJ/mol | G_initial = -83.2 kJ/mol | | G_final = -73.8 kJ/mol | ΔG_rxn^0 | -73.8 kJ/mol - -83.2 kJ/mol = 9.4 kJ/mol (endergonic) | | |](../image_source/2c69a0b7b10461681f49406473efb33d.png)
| water | nitrogen dioxide | nitric acid | nitric oxide molecular free energy | -237.1 kJ/mol | 51.3 kJ/mol | -80.7 kJ/mol | 87.6 kJ/mol total free energy | -237.1 kJ/mol | 153.9 kJ/mol | -161.4 kJ/mol | 87.6 kJ/mol | G_initial = -83.2 kJ/mol | | G_final = -73.8 kJ/mol | ΔG_rxn^0 | -73.8 kJ/mol - -83.2 kJ/mol = 9.4 kJ/mol (endergonic) | | |
Entropy
![| water | nitrogen dioxide | nitric acid | nitric oxide molecular entropy | 69.91 J/(mol K) | 240 J/(mol K) | 156 J/(mol K) | 211 J/(mol K) total entropy | 69.91 J/(mol K) | 720 J/(mol K) | 312 J/(mol K) | 211 J/(mol K) | S_initial = 789.9 J/(mol K) | | S_final = 523 J/(mol K) | ΔS_rxn^0 | 523 J/(mol K) - 789.9 J/(mol K) = -266.9 J/(mol K) (exoentropic) | | |](../image_source/a062a22cd500884f193ccc6eceff99c0.png)
| water | nitrogen dioxide | nitric acid | nitric oxide molecular entropy | 69.91 J/(mol K) | 240 J/(mol K) | 156 J/(mol K) | 211 J/(mol K) total entropy | 69.91 J/(mol K) | 720 J/(mol K) | 312 J/(mol K) | 211 J/(mol K) | S_initial = 789.9 J/(mol K) | | S_final = 523 J/(mol K) | ΔS_rxn^0 | 523 J/(mol K) - 789.9 J/(mol K) = -266.9 J/(mol K) (exoentropic) | | |
Equilibrium constant
![Construct the equilibrium constant, K, expression for: H_2O + NO_2 ⟶ HNO_3 + NO 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: H_2O + 3 NO_2 ⟶ 2 HNO_3 + NO 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 | 1 | -1 NO_2 | 3 | -3 HNO_3 | 2 | 2 NO | 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 | 1 | -1 | ([H2O])^(-1) NO_2 | 3 | -3 | ([NO2])^(-3) HNO_3 | 2 | 2 | ([HNO3])^2 NO | 1 | 1 | [NO] 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])^(-1) ([NO2])^(-3) ([HNO3])^2 [NO] = (([HNO3])^2 [NO])/([H2O] ([NO2])^3)](../image_source/2ae5f2f1eb526ca6b0366a5a0ef68245.png)
Construct the equilibrium constant, K, expression for: H_2O + NO_2 ⟶ HNO_3 + NO 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: H_2O + 3 NO_2 ⟶ 2 HNO_3 + NO 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 | 1 | -1 NO_2 | 3 | -3 HNO_3 | 2 | 2 NO | 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 | 1 | -1 | ([H2O])^(-1) NO_2 | 3 | -3 | ([NO2])^(-3) HNO_3 | 2 | 2 | ([HNO3])^2 NO | 1 | 1 | [NO] 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])^(-1) ([NO2])^(-3) ([HNO3])^2 [NO] = (([HNO3])^2 [NO])/([H2O] ([NO2])^3)
Rate of reaction
![Construct the rate of reaction expression for: H_2O + NO_2 ⟶ HNO_3 + NO 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: H_2O + 3 NO_2 ⟶ 2 HNO_3 + NO 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 | 1 | -1 NO_2 | 3 | -3 HNO_3 | 2 | 2 NO | 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 | 1 | -1 | -(Δ[H2O])/(Δt) NO_2 | 3 | -3 | -1/3 (Δ[NO2])/(Δt) HNO_3 | 2 | 2 | 1/2 (Δ[HNO3])/(Δt) NO | 1 | 1 | (Δ[NO])/(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 = -(Δ[H2O])/(Δt) = -1/3 (Δ[NO2])/(Δt) = 1/2 (Δ[HNO3])/(Δt) = (Δ[NO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)](../image_source/9c09cf0edb92a665612177bc2f0a64ab.png)
Construct the rate of reaction expression for: H_2O + NO_2 ⟶ HNO_3 + NO 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: H_2O + 3 NO_2 ⟶ 2 HNO_3 + NO 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 | 1 | -1 NO_2 | 3 | -3 HNO_3 | 2 | 2 NO | 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 | 1 | -1 | -(Δ[H2O])/(Δt) NO_2 | 3 | -3 | -1/3 (Δ[NO2])/(Δt) HNO_3 | 2 | 2 | 1/2 (Δ[HNO3])/(Δt) NO | 1 | 1 | (Δ[NO])/(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 = -(Δ[H2O])/(Δt) = -1/3 (Δ[NO2])/(Δt) = 1/2 (Δ[HNO3])/(Δt) = (Δ[NO])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Chemical names and formulas
![| water | nitrogen dioxide | nitric acid | nitric oxide formula | H_2O | NO_2 | HNO_3 | NO name | water | nitrogen dioxide | nitric acid | nitric oxide IUPAC name | water | Nitrogen dioxide | nitric acid | nitric oxide](../image_source/9f5d0997ba4cae41621476cc72ef590d.png)
| water | nitrogen dioxide | nitric acid | nitric oxide formula | H_2O | NO_2 | HNO_3 | NO name | water | nitrogen dioxide | nitric acid | nitric oxide IUPAC name | water | Nitrogen dioxide | nitric acid | nitric oxide