The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at \(273 \mathrm{~K}\) is \(2.3 \times 10^{-12} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\), and the activation energy is \(111 \mathrm{~kJ} / \mathrm{mol}\). How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of \(2.5\) atm to \(1.5\) atm at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

We will first convert the given initial and final partial pressures of NO₂(g) to initial and final concentrations. Using the ideal gas law: \(C = \frac{P}{RT}\) where C is the concentration, P is the partial pressure, R is the ideal gas constant (0.08206 L⋅atm/mol⋅K), and T is the temperature. Initial concentration: \(C_{1} = \frac{2.5\;\text{atm}}{(0.08206\;\text{L}\cdot\text{atm/mol} \cdot \text{K})(500\;\text{K})} = 0.06091\;\text{mol/L}\) Final concentration: \(C_{2} = \frac{1.5\;\text{atm}}{(0.08206\;\text{L}\cdot\text{atm/mol} \cdot \text{K})(500\;\text{K})} = 0.03654\;\text{mol/L}\)

We will use the Arrhenius equation to find the rate constant (k) at 500 K: \(k = k_0 e^{\frac{-E_a}{RT}}\) where k₀ is the rate constant at 273 K (given as \(2.3 \times 10^{-12}\; \text{L/mol}\cdot\text{s}\)), Eₐ is the activation energy (given as \(111 \times 10^3\; \text{J/mol}\)), R is the gas constant in J/mol⋅K (8.314 J/mol⋅K), and T is the temperature at which the reaction takes place (500 K). \(k = (2.3 \times 10^{-12}) e^{\frac{-(111 \times 10^3)}{(8.314)(500)}} = 9.33 \times 10^{-10}\; \text{L/mol}\cdot\text{s}\)

Given a bimolecular elementary reaction: \(2\;\text{NO}_2 (g) \longrightarrow 2\;\text{NO} (g) + \text{O}_2 (g)\) The rate expression for the decomposition of NO₂(g) is: \(-\frac{d[\text{NO}_2]}{dt} = k[\text{NO}_2]^2\)

Now, we will integrate the rate expression to find the time (t) taken for the concentration of NO₂(g) to decrease from C₁ to C₂: \(\int_{C_1}^{C_2} \frac{d[\text{NO}_2]}{[\text{NO}_2]^2} = -\int_{0}^{t} k \; dt\) Solving the integral, we get: \(\left[-\frac{1}{[\text{NO}_2]}\right]_{C_1}^{C_2} = -k(t - 0)\) Substitute the values of C₁, C₂, and k: \(\left[-\frac{1}{0.03654} + \frac{1}{0.06091}\right] = -(9.33 \times 10^{-10})(t)\) Solve for t: \(t = \frac{-1}{9.33 \times 10^{-10}}\left[-\frac{1}{0.03654} + \frac{1}{0.06091}\right]\) \(t = 1.04 \times 10^7\; \text{s}\) So, it will take approximately \(1.04 \times 10^7\) seconds for the concentration of NO₂(g) to decrease from an initial partial pressure of 2.5 atm to 1.5 atm at 500 K, assuming ideal gas behavior.