An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(

Question

An organic compound undergoes first order decomposition. The time taken for decomposition to $\left( \frac{1}{8} \right)^{\text{th~}}$ and $\left( \frac{1}{10} \right)^{\text{th~}}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively.

What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$ ?

$$(log2 = 0.3) $$(1) 9

Accepted Answer

$t = \frac{1}{k}ln\frac{A_{0}}{A_{t}}$$$\begin{matrix} & t_{1/8} = \frac{1}{k}ln\frac{A_{0}}{A_{0}/8} = \frac{1}{k}ln8 \ & t_{1/10} = \frac{1}{k}ln\frac{A_{0}}{A_{0}/10} = \frac{1}{k}ln10 \ & \frac{t_{1/8}}{t_{1/10}} = \frac{ln8}{ln10} = \frac{log8}{log10} \ & \frac{t_{1/8}}{t_{1/10}} = log8 = 3log2 = 0.9 \ & \frac{t_{1/8}}{t_{1/10}} 	imes 10 = 9 \end{matrix}$$

Question

Options

Solution

More Chemical Kinetics and Nuclear Chemistry Questions