Discussion Forum

1)

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 

$\frac{[t_{1/8}]}{[t_{1/10}]}\times10?  (\log_{10} 2=0.3)$

Answer:

Option D

Explanation:

The first order process  $kt=ln \frac{[A]_{0}}{[A]}$   where ,$[A]_{0}$=initial concentration

[A]=concentration of reactant remaining at time 't'

$\Rightarrow$    $kt_{1/8}=ln \frac{[A]_{0}}{[A]_{0}/8}=ln 8$..........(i)

and  $kt _{1/10}=ln \frac{[A]_{0}}{[A]_{0}/10}=ln 10$

Therefore,

$\frac{t_{1/8}}{t_{1/10}} =\frac{ln 8}{ln 10}=\log 8=3 \log 2$

                         =$3 \times 0.3=0.9$

$\Rightarrow$  $\frac{t_{1/8}}{t_{1/10}}  \times 10=0.9 \times 10=9$