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QJEEM22Ch27ba: JEE Main 2022 Chemistry - Class 12 JEEM22Ch27ba Solution: [ A ] → [ B ] Reactant Product If formation of compound [ B ]…

[A]→[B]ReactantProduct

If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x×10−6 s−1. The value of x is ______________. (Nearest Integer)

Step-by-Step Verified Solution

K=0.693t1/2=0.69370×60\mathrm{K}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{70 \times 60}

=69307×6×10−6=\frac{6930}{7 \times 6} \times 10^{-6}

=165×10−6 s−1=165 \times 10^{-6} \mathrm{~s}^{-1}

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