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 (integer)
623
Solution
<p>$t = {{2.303} \over k}\log {{{C_0}} \over {{C_t}}}$</p>
<p>$= {{2.303} \over {2.011 \times {{10}^{ - 3}}}}\log {7 \over 2}$</p>
<p>$= {{2.303 \times {{10}^3}} \over {2.011}}(.845 - .301)$</p>
<p>$= 622.99$</p>
<p>$\approx 623$ sec.</p>
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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