At high temperatures, the decomposition of cyclobutane is a first-order reaction. Its activation energy is \(262 \mathrm{~kJ} / \mathrm{mol}\). At \(477^{\circ} \mathrm{C}\), its half-life is \(5.00 \mathrm{~min}\). What is its half-life (in seconds) at \(527^{\circ} \mathrm{C}\) ?

For a first-order reaction, we can use the half-life formula, \(t_{1/2} = \frac{0.693}{k}\). At \(477^\circ \mathrm{C}\), we are given the half-life as \(5.00 \mathrm{~min}\), which is equal to \(300 \mathrm{~s}\). So, we can rewrite the formula as \(300 \mathrm{~s} = \frac{0.693}{k}\), and we can solve for k. $$k = \frac{0.693}{300 \mathrm{~s}}$$ Now, calculate k: $$k \approx 2.31 \times 10^{-3} \mathrm{s}^{-1}$$

The Arrhenius equation is given by: $$k_2 = k_1 \cdot e^{\frac{-E_a}{R} \cdot (\frac{1}{T_2} - \frac{1}{T_1})}$$ where \(k_1\) is the initial rate constant at temperature \(T_1\), \(k_2\) is the rate constant at temperature \(T_2\), \(E_a\) is the activation energy, and \(R\) is the gas constant. In this problem, we have: - \(k_1 = 2.31 \times 10^{-3} \mathrm{s}^{-1}\) - \(E_a = 262 \, \mathrm{kJ/mol} = 262 \times 10^3 \, \mathrm{J/mol}\) - \(T_1 = 477^\circ \mathrm{C} = 750.15 \, \mathrm{K}\) - \(T_2 = 527^\circ \mathrm{C} = 800.15 \, \mathrm{K}\) - \(R = 8.314 \, \mathrm{J/(mol \, K)}\) Plug these values into the equation and solve for \(k_2\). $$k_2 = (2.31 \times 10^{-3} \mathrm{s}^{-1}) \cdot e^{\frac{-(262 \times 10^3 \mathrm{J/mol})}{(8.314 \mathrm{J/(mol \, K)})} \cdot (\frac{1}{800.15 \mathrm{K}} - \frac{1}{750.15 \mathrm{K}})}$$ After calculation, we obtain: $$k_2 \approx 1.624 \times 10^{-2} \mathrm{s}^{-1}$$

Now that we have found the rate constant at \(527^\circ \mathrm{C}\), we can calculate the half-life using the formula \(t_{1/2} = \frac{0.693}{k}\). $$t_{1/2} = \frac{0.693}{1.624 \times 10^{-2} \mathrm{s}^{-1}}$$ After calculation, we obtain: $$t_{1/2} \approx 42.67 \mathrm{s}$$ So the half-life of cyclobutane at \(527^\circ \mathrm{C}\) is approximately \(42.67 \mathrm{~seconds}\).