N2O4 = NO2

N_2O_4 dinitrogen tetroxide ⟶ NO_2 nitrogen dioxide

Balance the chemical equation algebraically: N_2O_4 ⟶ NO_2 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 N_2O_4 ⟶ c_2 NO_2 Set the number of atoms in the reactants equal to the number of atoms in the products for N and O: N: | 2 c_1 = c_2 O: | 4 c_1 = 2 c_2 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 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | N_2O_4 ⟶ 2 NO_2

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dinitrogen tetroxide ⟶ nitrogen dioxide

| dinitrogen tetroxide | nitrogen dioxide molecular enthalpy | 11.1 kJ/mol | 33.2 kJ/mol total enthalpy | 11.1 kJ/mol | 66.4 kJ/mol | H_initial = 11.1 kJ/mol | H_final = 66.4 kJ/mol ΔH_rxn^0 | 66.4 kJ/mol - 11.1 kJ/mol = 55.3 kJ/mol (endothermic) |

| dinitrogen tetroxide | nitrogen dioxide molecular free energy | 99.8 kJ/mol | 51.3 kJ/mol total free energy | 99.8 kJ/mol | 102.6 kJ/mol | G_initial = 99.8 kJ/mol | G_final = 102.6 kJ/mol ΔG_rxn^0 | 102.6 kJ/mol - 99.8 kJ/mol = 2.8 kJ/mol (endergonic) |

| dinitrogen tetroxide | nitrogen dioxide molecular entropy | 304.3 J/(mol K) | 240 J/(mol K) total entropy | 304.3 J/(mol K) | 480 J/(mol K) | S_initial = 304.3 J/(mol K) | S_final = 480 J/(mol K) ΔS_rxn^0 | 480 J/(mol K) - 304.3 J/(mol K) = 175.7 J/(mol K) (endoentropic) |

Construct the equilibrium constant, K, expression for: N_2O_4 ⟶ NO_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: N_2O_4 ⟶ 2 NO_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 N_2O_4 | 1 | -1 NO_2 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression N_2O_4 | 1 | -1 | ([N2O4])^(-1) NO_2 | 2 | 2 | ([NO2])^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 = ([N2O4])^(-1) ([NO2])^2 = ([NO2])^2/([N2O4])

Construct the rate of reaction expression for: N_2O_4 ⟶ NO_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: N_2O_4 ⟶ 2 NO_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 N_2O_4 | 1 | -1 NO_2 | 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 N_2O_4 | 1 | -1 | -(Δ[N2O4])/(Δt) NO_2 | 2 | 2 | 1/2 (Δ[NO2])/(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[N2O4])/(Δt) = 1/2 (Δ[NO2])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| dinitrogen tetroxide | nitrogen dioxide formula | N_2O_4 | NO_2 name | dinitrogen tetroxide | nitrogen dioxide IUPAC name | | Nitrogen dioxide

| dinitrogen tetroxide | nitrogen dioxide molar mass | 92.01 g/mol | 46.005 g/mol phase | gas (at STP) | gas (at STP) melting point | -15 °C | -11 °C boiling point | 21.2 °C | 21 °C density | 1.45 g/cm^3 (at 20 °C) | 0.00188 g/cm^3 (at 25 °C) solubility in water | reacts | reacts surface tension | 0.0275 N/m | dynamic viscosity | 4.18×10^-4 Pa s (at 20 °C) | 4.02×10^-4 Pa s (at 25 °C)