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A first order reaction takes 40 min for 30% decomposition. What is its half-life? (log(10/7) = 0.155)
A92.4 min
B77.7 min
C120.0 min
D60.0 min
Answer & Solution
Correct answer: B. 77.7 min
1. After 30% decomposition, $[R] = 0.70[R]_0$, so $\dfrac{[R]_0}{[R]} = \dfrac{1}{0.7} = \dfrac{10}{7}$.
2. $k = \dfrac{2.303}{40}\log\dfrac{10}{7} = \dfrac{2.303}{40}\times0.155 = 8.92\times10^{-3}\ \text{min}^{-1}$.
3. $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{8.92\times10^{-3}}$.
4. $= 77.7\ \text{min}$.
5. Distractor A uses a wrong log (e.g. treats it as 50%); distractor C miscomputes k.
_Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.25_
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