High School

A first-order reaction takes 69.3 minutes for 50% completion. What is the time needed for 80% of the reaction to be completed?

(Given: [tex]\log 5 = 0.6990[/tex], [tex]\log 8 = 0.9030[/tex], [tex]\log 2 = 0.3010[/tex])

A. 92.4 minutes
B. 99.9 minutes
C. 107.4 minutes
D. 114.9 minutes

Answer :

Final answer:

To find the time needed for 80% completion of a first-order reaction, use the equation ln(C0/C) = kt and substitute the given values to solve for t. the correct answer is a) 92.4 min.

Explanation:

To find the time needed for 80% completion of a first-order reaction, we can use the equation ln(C0/C) = kt, where C0 is the initial concentration, C is the concentration at a given time, k is the rate constant, and t is the time. In this case, since we're given the time for 50% completion, we can substitute the given values into the equation and solve for t when C/C0 = 0.8.

ln(0.5) = k(69.3)

ln(0.8) = k(t)

By rearranging the second equation and substituting the given value for k, we can solve for t:

t = ln(0.8)/k = 0.3010/1.0 × 10^7 = 0.0301 min

Therefore, the time needed for 80% completion of the reaction is approximately 0.0301 min.