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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.

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Answer:

11.06 hours are required

Explanation:

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

ln [A]â‚“ = -k*t + ln[A]â‚€

Where [A]â‚€ is the initial concentration of sucrose, k is a constant and [A]â‚“ 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]â‚“ = 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