Chemical Kinetics and Nuclear ChemistryMedium
Question
The rate constant of a second-order reaction is 10−2 lit mole−1 s−1. The rate constant when expressed in ml molecule−1 min−1 is (NA = 6 × 1023)
Options
A.0.01
B.1.0 × 10−21
C.1.0 × 1021
D.3.6 × 1020
Solution
$K = 10^{- 2}\text{ L mo}\text{l}^{- 1}s^{- 1} = 10^{- 2} \times \left( 1000\text{ ml} \right) \times \frac{1}{6 \times 10^{23}\text{ molecule}} \times \frac{60}{\text{min}}$
$= 1.0 \times 10^{- 21}\text{ ml molecul}\text{e}^{- 1}\text{ mi}\text{n}^{- 1}$
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