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Ch.14 - Chemical Kinetics
Chapter 14, Problem 9

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([A]0[A])=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.
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.
Substitute the known values into the first-order rate equation: ln(722100)=(5.87×104 s1)×t.
Solve for t: Rearrange the equation to solve for t: t=ln(722100)5.87×104.
Calculate t: Perform the calculation to find the time t it takes for the pressure to drop to below 100 torr.