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. The value of $\frac{\left[t_{1/8}\right]}{\left[t_{1/10}\right]}\times 10$ is ______ (take $lo{g}_{10}2=0.3$)

Answer 1

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

For decomposition to 1/8 of its initial concentration
$t_{1/8}=\frac{2.303}{k}log\left(\frac{1}{1/8}\right)=\frac{2.08}{k}$

For decomposition to 1/10 of its initial concentration

$t_{1/10}=\frac{2.303}{k}log\left(\frac{1}{1/10}\right)=\frac{2.303}{k}$

Hence, $\frac{\left[t_{1/8}\right]}{\left[t_{1/10}\right]}\times 10=\frac{\frac{2.08}{k}}{\frac{2.303}{k}}\times 10=9$