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 $\left[\frac{t_{1 / 8}}{t_{1 / 10}} ight] imes 10$ ? $\left(\log _{10} 2=0…

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 $\left[\frac{t_{1 / 8}}{t_{1 / 10}}ight] 	imes 10$ ? $\left(\log _{10} 2=0.3ight)$

Aicharya · Accepted Answer

$\begin{array}{l} t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 	imes 3 \log 2}{k} \ t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \ {\left[\frac{t_{1 / 8}}{t_{1 / 10}}ight] 	imes 10=\frac{\left(\frac{2.303 	imes 3 \log 2}{k}ight)}{\left(\frac{2.303}{k}ight)} 	imes 10=9} \end{array}$

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1 /$ 8 and…

Solution