Cyclopropane (C3H6) reacts to form propene (C3H6) in the gas phase. The reaction is first order in cyclopropane and has a rate constant of 5.87 * 10^-4 s^-1 at 485 °C. If a 2.5-L reaction vessel initially contains 722 torr of cyclopropane at 485 °C, how long will it take for the partial pressure of cyclopropane to drop to below 1.00 * 10^2 torr?

Verified step by step guidance

1. Identify the reaction order and relevant equation: The reaction is first order, so we use the first-order rate equation: \( \ln \left( \frac{[A]_0}{[A]} \right) = kt \), where \([A]_0\) is the initial concentration (or pressure), \([A]\) is the concentration (or pressure) at time \(t\), \(k\) is the rate constant, and \(t\) is the time.
2. Convert pressures to the same units: The initial pressure \([A]_0\) is 722 torr, and the final pressure \([A]\) is 100 torr. Ensure both are in the same units for calculation.
3. Substitute the known values into the first-order rate equation: \( \ln \left( \frac{722}{100} \right) = (5.87 \times 10^{-4} \text{ s}^{-1}) \times t \).
4. Solve for \(t\): Rearrange the equation to solve for \(t\): \( t = \frac{\ln \left( \frac{722}{100} \right)}{5.87 \times 10^{-4}} \).
5. Calculate \(t\): Perform the calculation to find the time \(t\) it takes for the pressure to drop to below 100 torr.