Chemical Kinetics - Result Question 56
####56. An organic compound undergoes first order decomposition. The time taken for its decomposition to $1 / 8$ and $1 / 10$ of its initial concentration are $t _{1 / 8}$ and $t _{1 / 10}$ respectively. What is the value of $\frac{\left[t _{1 / 8}\right]}{\left[t _{1 / 10}\right]} \times 10 ?\left(\log _{10} 2=0.3\right)$
(2012)
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Answer:
Correct Answer: 56. $\left(3.26 \times 10^{-3} mol L^{-1} min^{-1}\right)$
Solution:
- For a first order process $k t=\ln \frac{[A] _0}{[A]}$
where, $[A] _0=$ initial concentration.
$[A]=$ concentration of reactant remaining at time " $f$ “.
$\Rightarrow \quad k t _{1 / 8}=\ln \frac{[A] _0}{[A] _0 / 8}=\ln 8$
and
$$ k t _{1 / 10}=\ln \frac{[A] _0}{[A] _0 / 10}=\ln 10 $$
Therefore, $\quad \frac{t _{1 / 8}}{t _{1 / 10}}=\frac{\ln 8}{\ln 10}=\log 8=3 \log 2=3 \times 0.3=0.9$
$$ \Rightarrow \quad \frac{t _{1 / 8}}{t _{1 / 10}} \times 10=0.9 \times 10=9 $$
