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

Question

An organic compound undergoes first-order decomposition. The time taken for 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 $\frac{\left[t_{1 / 8}ight]}{\left[t_{1 / 10}ight]} 	imes 10$ ?\ $\left(ight.$ take $\left.\log _{10} 2=0.3ight)$\

Professor PC Thomas Classes · Answer

(9) : $k=\frac{2.303}{t} \log \frac{a_{0}}{a}$.\ $a=a_{0} / 8$ at $t_{1 / 8}$\ $t_{1 / 8}=\frac{2.303}{k} \log \frac{a_{0}}{a_{0} / 8}=\frac{2.303}{k} \log 8$\ When $t=t_{1 / 10}, a=a_{0} / 10$\ $t_{1 / 10}=\frac{2.303}{k} \log \frac{a_{0}}{a_{0} / 10}=\frac{2.303}{k} \log 10$ From eq. (i) and (ii),\ $\frac{\left[t_{1 / 8}ight]}{\left[t_{1 / 10}ight]} 	imes 10=\frac{2.303}{k} \log 8 	imes \frac{k}{2.303 	imes \log 10} 	imes 10$\ $=\frac{\log 8}{\log 10} 	imes 10=\frac{\log 2^{3}}{\log 10} 	imes 10=\frac{3 \log 2}{\log 10} 	imes 10$\ $=\frac{3 	imes 0.3 	imes 10}{1}=9$\