Chemical Kinetics and Nuclear ChemistryHard

Question

In a first-order reaction, the activity of reactant drops from 800 mol/dm3 to 50 mol/dm3 in 2 × 104 s. The rate constant of the reaction in s−1 is

Options

A.1.386 × 10−4
B.1.386 × 10−3
C.1.386 × 10−5
D.5.0 × 103

Solution

$K = \frac{1}{t}.\ln\frac{\left\lbrack A_{0} \right\rbrack}{\lbrack A\rbrack} = \frac{1}{2 \times 10^{4}\text{ s}}.\ln\frac{800}{50} = 1.386 \times 10^{- 4}s^{- 1}$

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