In how many half lives a first order reaction is completed 87.5%.

MCQ

A
1
B
2
✓
3
D
4

✓

Answer

Correct option: C.

(C)3
$\begin{array}{l}T=n t_{y / 2} \\ \frac{2.303}{K} \log \left(\frac{100}{12.5}\right)=n\left(\frac{2.303}{K} \log \frac{100}{50}\right) \\ n=\frac{\log \left(\frac{100}{12.5}\right)}{\log 2}=3\end{array}$

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