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HNO3 + Zn = H2O + N2 + Zn(NO3)

Input interpretation

HNO_3 nitric acid + Zn zinc ⟶ H_2O water + N_2 nitrogen + Zn(NO3)2
HNO_3 nitric acid + Zn zinc ⟶ H_2O water + N_2 nitrogen + Zn(NO3)2

Balanced equation

Balance the chemical equation algebraically: HNO_3 + Zn ⟶ H_2O + N_2 + Zn(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Zn ⟶ c_3 H_2O + c_4 N_2 + c_5 Zn(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Zn: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 c_5 Zn: | 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 Zn ⟶ 6 H_2O + N_2 + 5 Zn(NO3)2
Balance the chemical equation algebraically: HNO_3 + Zn ⟶ H_2O + N_2 + Zn(NO3)2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 Zn ⟶ c_3 H_2O + c_4 N_2 + c_5 Zn(NO3)2 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N, O and Zn: H: | c_1 = 2 c_3 N: | c_1 = 2 c_4 + 2 c_5 O: | 3 c_1 = c_3 + 6 c_5 Zn: | 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 Zn ⟶ 6 H_2O + N_2 + 5 Zn(NO3)2

Structures

+ ⟶ + + Zn(NO3)2
+ ⟶ + + Zn(NO3)2

Names

nitric acid + zinc ⟶ water + nitrogen + Zn(NO3)2
nitric acid + zinc ⟶ water + nitrogen + Zn(NO3)2

Equilibrium constant

Construct the equilibrium constant, K, expression for: HNO_3 + Zn ⟶ H_2O + N_2 + Zn(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 Zn ⟶ 6 H_2O + N_2 + 5 Zn(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 Zn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Zn(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) Zn | 5 | -5 | ([Zn])^(-5) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] Zn(NO3)2 | 5 | 5 | ([Zn(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) ([Zn])^(-5) ([H2O])^6 [N2] ([Zn(NO3)2])^5 = (([H2O])^6 [N2] ([Zn(NO3)2])^5)/(([HNO3])^12 ([Zn])^5)
Construct the equilibrium constant, K, expression for: HNO_3 + Zn ⟶ H_2O + N_2 + Zn(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 Zn ⟶ 6 H_2O + N_2 + 5 Zn(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 Zn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Zn(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) Zn | 5 | -5 | ([Zn])^(-5) H_2O | 6 | 6 | ([H2O])^6 N_2 | 1 | 1 | [N2] Zn(NO3)2 | 5 | 5 | ([Zn(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) ([Zn])^(-5) ([H2O])^6 [N2] ([Zn(NO3)2])^5 = (([H2O])^6 [N2] ([Zn(NO3)2])^5)/(([HNO3])^12 ([Zn])^5)

Rate of reaction

Construct the rate of reaction expression for: HNO_3 + Zn ⟶ H_2O + N_2 + Zn(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 Zn ⟶ 6 H_2O + N_2 + 5 Zn(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 Zn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Zn(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) Zn | 5 | -5 | -1/5 (Δ[Zn])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) Zn(NO3)2 | 5 | 5 | 1/5 (Δ[Zn(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[Zn])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) = 1/5 (Δ[Zn(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HNO_3 + Zn ⟶ H_2O + N_2 + Zn(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 Zn ⟶ 6 H_2O + N_2 + 5 Zn(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 Zn | 5 | -5 H_2O | 6 | 6 N_2 | 1 | 1 Zn(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) Zn | 5 | -5 | -1/5 (Δ[Zn])/(Δt) H_2O | 6 | 6 | 1/6 (Δ[H2O])/(Δt) N_2 | 1 | 1 | (Δ[N2])/(Δt) Zn(NO3)2 | 5 | 5 | 1/5 (Δ[Zn(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[Zn])/(Δt) = 1/6 (Δ[H2O])/(Δt) = (Δ[N2])/(Δt) = 1/5 (Δ[Zn(NO3)2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| nitric acid | zinc | water | nitrogen | Zn(NO3)2 formula | HNO_3 | Zn | H_2O | N_2 | Zn(NO3)2 Hill formula | HNO_3 | Zn | H_2O | N_2 | N2O6Zn name | nitric acid | zinc | water | nitrogen | IUPAC name | nitric acid | zinc | water | molecular nitrogen |
| nitric acid | zinc | water | nitrogen | Zn(NO3)2 formula | HNO_3 | Zn | H_2O | N_2 | Zn(NO3)2 Hill formula | HNO_3 | Zn | H_2O | N_2 | N2O6Zn name | nitric acid | zinc | water | nitrogen | IUPAC name | nitric acid | zinc | water | molecular nitrogen |

Substance properties

| nitric acid | zinc | water | nitrogen | Zn(NO3)2 molar mass | 63.012 g/mol | 65.38 g/mol | 18.015 g/mol | 28.014 g/mol | 189.4 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 420 °C | 0 °C | -210 °C | boiling point | 83 °C | 907 °C | 99.9839 °C | -195.79 °C | density | 1.5129 g/cm^3 | 7.14 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) | | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | odorless |
| nitric acid | zinc | water | nitrogen | Zn(NO3)2 molar mass | 63.012 g/mol | 65.38 g/mol | 18.015 g/mol | 28.014 g/mol | 189.4 g/mol phase | liquid (at STP) | solid (at STP) | liquid (at STP) | gas (at STP) | melting point | -41.6 °C | 420 °C | 0 °C | -210 °C | boiling point | 83 °C | 907 °C | 99.9839 °C | -195.79 °C | density | 1.5129 g/cm^3 | 7.14 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) | | 8.9×10^-4 Pa s (at 25 °C) | 1.78×10^-5 Pa s (at 25 °C) | odor | | odorless | odorless | odorless |

Units

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