QJEEAdv12Ch661f: JEE Advanced 2012 Chemistry - Class 12 JEEAdv12Ch661f Solution: An organic compound undergoes first-order decomposition. The…

The expression for first-order reaction is

$t=\frac{2.303}{k} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}$

When the compound is decomposed to $1 / 8$ th of its initial value then the time taken is

$t_{1 / 8}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 8)}=\left(\frac{2.303}{k}\right) \log 8$ .........(1)

When the compound is decomposed to $1 / 10$ th of its initial value then the time taken is

$t_{1 / 10}=\left(\frac{2.303}{k}\right) \log \frac{1}{(1 / 10)}=\left(\frac{2.303}{k}\right) \log 10$ ........(2)

Dividing Eq. (1) by Eq. (2), we get

$\frac{t_{1 / 8}}{t_{1 / 10}}=\frac{\log 8}{\log 10}=\log \left(2^3\right)=3 \times 0.3=0.9$

So, the value of

$\frac{\left[t_{1 / 8}\right]}{\left[t_{1 / 10}\right]} \times 10=9$