Consider the hypothetical reaction $\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow\( \)2 \mathrm{AB}(g),$ where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is $2.45 \times 10^{-4} \mathrm{L} / \mathrm{mol}\( \)\mathrm{s},\( and at \)508^{\circ} \mathrm{C}\( the rate constant is 0.891 \)\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$ . What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C} ?\)

The Arrhenius equation is given by: $$ k = Ae^{\frac{-E_a}{RT}} $$ where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant (8.314 J/mol K), and \(T\) is the absolute temperature in Kelvin.

To relate the rate constants and temperatures, we have to convert them to Kelvin using the following formula: $$ T(K) = T(^\circ C) + 273.15 $$ For the given temperatures: $$ T_1 = 302^\circ C + 273.15 = 575.15 K \\ T_2 = 508^\circ C + 273.15 = 781.15 K \\ T_3 = 375^\circ C + 273.15 = 648.15 K $$ where \(T_1\), \(T_2\), and \(T_3\) are the absolute temperatures at the given Celsius temperatures.

Using the rate constants provided at the two temperatures \(k_1\) at \(T_1\) and \(k_2\) at \(T_2\), we now have two equations: $$ k_1 = Ae^{\frac{-E_a}{R T_1}} \\ k_2 = Ae^{\frac{-E_a}{R T_2}} $$ where \(k_1=2.45 \times 10^{-4}\,\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) and \(k_2=0.891\, \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\).

To solve for the activation energy, divide the second equation by the first equation: $$ \frac{k_2}{k_1} = \frac{Ae^{\frac{-E_a}{R T_2}}}{Ae^{\frac{-E_a}{R T_1}}} $$ Solve for the activation energy, \(E_a\): $$ E_a = R\left(T_2-T_1\right) \frac{\ln \frac{k_1}{k_2}}{\ln \frac{k_1}{k_2} - 1} $$ Substitute the values and compute the activation energy, \(E_a\): $$ E_a = 8.314\, J \cdot mol^{-1}K^{-1} \cdot(781.15\,K-575.15\,K) \cdot \frac{\ln \frac{2.45 \times 10^{-4}}{0.891}}{\ln \frac{2.45 \times 10^{-4}}{0.891} - 1} \approx 149655\,J/mol $$ So, the activation energy is \(E_a \approx 149.7\,\mathrm{kJ}\cdot\mathrm{mol}^{-1}\).

Now, we need to find the pre-exponential factor, \(A\), using one of the given rate constants. We can use the equation for \(k_1\) and \(T_1\): $$ A = \frac{k_1}{e^{\frac{-E_a}{R T_1}}} $$ Substitute the values and compute \(A\): $$ A = \frac{2.45 \times 10^{-4}\,\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}}{e^{\frac{-149655\,J\cdot mol^{-1}}{8.314\, J \cdot mol^{-1}K^{-1}\cdot 575.15\, K}}} \approx 4.88 \times 10^{-6}\, \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} $$ Now that we have the pre-exponential factor, we can calculate the rate constant at \(T_3\): $$ k_3 = Ae^{\frac{-E_a}{RT_3}} $$ Substitute the values and calculate \(k_3\): $$ k_3 = 4.88 \times 10^{-6}\, \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} \cdot e^{\frac{-149655\, J/mol}{8.314\, J / mol\cdot K\cdot 648.15\,K}} \approx 0.0075\, \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} $$ Therefore, the value of the rate constant for this reaction at \(375^\circ \mathrm{C}\) is approximately \(0.0075\, \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\).