The half-life of a first-order reaction is 693 s — JEE Chemistry
The half-life of a first-order reaction is 693 s. Calculate the rate constant and the time for 90% completion.
1 Answer
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· 1mo ago
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$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{693} = 1.0 \times 10^{-3} s^{-1}$$
For 90% completion, 10% remains ($[A]/[A]_0 = 0.1$):
$$t = \frac{2.303}{k}\log\frac{1}{0.1} = \frac{2.303}{10^{-3}} \times 1 = 2303 s$$
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Discussion (2)
N
Does this approach generalise to the JEE Advanced version of this question?
RV
Why do we take the positive value only in the last step?