$$\matrix{ {[A]} & \to & {[B]} \cr {{\mathop{\rm Reactant}\nolimits} } & {} & {{\mathop{\rm Product}\nolimits} } \cr } $$
If formation of compound $[\mathrm{B}]$ follows the first order of kinetics and after 70 minutes the concentration of $[\mathrm{A}]$ was found to be half of its initial concentration. Then the rate constant of the reaction is $x \times 10^{-6} \mathrm{~s}^{-1}$. The value of $x$ is ______________. (Nearest Integer)
Answer (integer)
165
Solution
$\mathrm{K}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{70 \times 60}$<br/><br/>
$=\frac{6930}{7 \times 6} \times 10^{-6}$<br/><br/>
$=165 \times 10^{-6} \mathrm{~s}^{-1}$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.