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In a first-order reaction, over a 10-minute interval the concentration of the reactant drops from 80% to 20% of its initial value. How long (from the start of the reaction) does it take for the concentration to fall to 50% of the initial value?

1. 20 min.
2. 49.15 min.
3. 5 min.
4. 15.28 min.

Correct answer: 5 min.

Solution

Using integrated rate law for first-order: ln(C1/C2) = k*(t2-t1). From 80% to 20% in 10 min, k = ln(4)/10. For 50% remaining: ln(100/50) = k*t1 => t1 = ln(2)/k = 10*ln(2)/ln(4) = 10*ln(2)/(2*ln(2)) = 5 min.

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