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If the rate constant of a reaction is $0.03\text{ s}^{-1}$, how much time does it take for $7.2\text{ mol L}^{-1}$ concentration of the reactant to get reduced to $0.9\text{ mol L}^{-1}$? (Given: $\log 2 = 0.301$)
A
69.3 s
B
23.1 s
C
210 s
D
21.0 s
Explanation
Since $[R]_0 / [R] = 7.2 / 0.9 = 8 = 2^3$, this represents 3 half-lives: $t = 3 \times t_{1/2} = 3 \times \frac{0.693}{0.03} = 3 \times 23.1 = 69.3\text{ s}$.
Detailed Solution
The unit of rate constant $k = 0.03\text{ s}^{-1}$ signifies a first order reaction. The first order integrated rate law is: $t = \frac{2.303}{k} \log\left(\frac{[R]_0}{[R]}\right)$. Given $[R]_0 = 7.2\text{ mol L}^{-1}$ and $[R] = 0.9\text{ mol L}^{-1}$: $\frac{[R]_0}{[R]} = \frac{7.2}{0.9} = 8 = 2^3$. Then $\log 8 = 3 \log 2 = 3(0.301) = 0.903$. Substituting: $t = \frac{2.303}{0.03} \times 0.903 = \frac{2.0796}{0.03} = 69.32\text{ s} \approx 69.3\text{ s}$.