Question

The decomposition of a substance follows first order kinetics. If its concentration is reduced to $1/8$ of its initial value in $12$ minutes, the rate constant of the decomposition system is ?

• A. $\left(\frac{2.303}{12}\log \frac{1}{8}\right)min{.}^{-1}$
• B. $\left(\frac{2.303}{12}\log 8\right)min{.}^{-1}$
• C. $\left(\frac{0.693}{12}\right)min{.}^{-1}$
• D. $\left(\frac{1}{12}\log 8\right)min{.}^{-1}$

Answer 1

The correct option is B $\left(\frac{2.303}{12}\log 8\right)min{.}^{-1}$

$k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$ (for first order)

$k=\frac{2.303}{12}\log \left(\frac{1}{1/8}\right)=\left(\frac{2.303}{12}\log 8\right)min{.}^{-1}$.