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For the first order reaction A -> Product, the concentration of A is measured at different times. A graph of ln[A] versus time t is plotted, which gives a straight line passing through ln[A] = -1 at t = 0 and ln[A] = -8 at t = 1000 seconds. Calculate the half-life of the reaction. (Given: ln 2 = 0.7)
- 10 sec
- 100 s
- 500 sec
- 1000 s
Correct answer: 100 s
Solution
From the ln[A] vs t graph, the slope = (-8 - (-1)) / (1000 - 0) = -7/1000 s⁻¹. So k = 7/1000 = 0.007 s⁻¹. Half-life t_(1/2) = ln2 / k = 0.7 / 0.007 = 100 s.
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