JEE MAIN - Chemistry (2023 - 30th January Morning Shift - No. 15)
If compound A reacts with B following first order kinetics with rate constant $$2.011 \times 10^{-3} \mathrm{~s}^{-1}$$. The time taken by $$\mathrm{A}$$ (in seconds) to reduce from $$7 \mathrm{~g}$$ to $$2 \mathrm{~g}$$ will be ___________. (Nearest Integer)
$$[\log 5=0.698, \log 7=0.845, \log 2=0.301]$$
Answer
623
Explanation
$$t = {{2.303} \over k}\log {{{C_0}} \over {{C_t}}}$$
$$ = {{2.303} \over {2.011 \times {{10}^{ - 3}}}}\log {7 \over 2}$$
$$ = {{2.303 \times {{10}^3}} \over {2.011}}(.845 - .301)$$
$$ = 622.99$$
$$ \approx 623$$ sec.
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