A student monitored the reaction rate of a 0.600M sucrose solution, at 25 oC, and determined that the reaction is first order with respect to sucrose. If at 25.0 oC it takes 3.33-hours for the concentration of the 0.600M solution to drop to 0.300M. How many hours are required for the concentration of the sucrose solution to drop to 10.0 % of its initial concentration.

11.06 hours are required

Explanation:

Because the reaction is first order with respect of sucrose, we can use the formula:

ln [A]ₓ = - k*t + ln[A]₀

Where [A]₀ is the initial concentration of sucrose, k is a constant and [A]ₓ is the concentration remaining after a time t.

Putting the data given by the exercise we can calculate k:

ln (0.300) = - k * 3.33 + ln (0.600) - 0.6931 = - k * 3.33 k = 0.2082

Now with k, we can calculate t when [A]ₓ = 0.600 * 10/100 = 0.060 M

ln (0.060) = - 0.2082 * t + ln (0.600) - 2.302 = - 0.2082 * t t = 11.06 hours