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

Answer 1

Answer: (B)

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

Answer 2

$k = \frac{2.303}{t}\log\frac{a}{a - x}$

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