Question

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 $\frac{\left[t_1/8\right]}{\left[t_1/10\right]}\times 10?(lo{g}_{10}2=0.3)$___

Answer 1

For a first order process $kt=In\frac{{\left[A\right]}_0}{\left[A\right]}$
Where ${\left[A\right]}_0=$ intial concentration.
[A] = concentration ofreactant remaining at time "t".
$\Rightarrow k{t}_{1/8}=In\frac{{\left[A\right]}_0}{{\left[A\right]}_0/8}=In8andk{t}_{1/10}=In\frac{{\left[A\right]}_0}{{\left[A\right]}_0/10}=In10Therefore,\frac{t_{1/8}}{t_1/10}=\frac{In8}{In10}=log8=3log2=3\times 0.3=0.9\Rightarrow \frac{t_{1/8}}{t_{1/10}}\times 10=0.9\times 10=9.$