Chemical Kinetics and Nuclear ChemistryMedium
Question
If the rate constant of a reaction is $0.03{\text{ }s}^{- 1}$, how much time does it take for $7.2\text{ }mol{\text{ }L}^{- 1}$ concentration of the reactant to get reduced to $0.9\text{ }mol{\text{ }L}^{- 1}$ ?(Given: $log2 = 0.301$ )
Options
A.69.3 s
B.$\ 23.1\text{ }s$
C.210 s
D.$\ 21.0\text{ }s$
Solution
Sol. [Key Concept: First-order kinetics — time for the concentration to fall (t = 2.303/k·log)]
$k = 0.03{\text{ }s}^{- 1}$
$$\begin{matrix} t & \ = \frac{2.303}{k}log\frac{a}{a - x} \\ & \ = \frac{2.303}{0.03}log\frac{7.2}{0.9} \\ & \ = \frac{2.303}{0.03}log8 \\ & \ = \frac{2.303}{0.03} \times 3 \times log2 \\ & \ = \frac{2.303}{0.03} \times 3 \times 0.301 \\ & \ = 69.3\text{ }s \end{matrix}$$
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