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 $\frac{\mathrm{t}_{1 / 8}}{\mathrm{t}_{1 / 10}} \times 10$ ? $(\log 2=0.3)$

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Chemical Kinetics

An organic compound undergoes first order decomposition. The time taken for decomposition to

(\frac{1}{8})^{th}

and

(\frac{1}{10})^{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

(lo g 2 = 0.3)

Answer: (A)

For a first-order reaction

t = \frac{1}{k} ln \frac{[ A ] _{0}}{[ A ]}

, so

t_{1/8} = \frac{1}{k} ln 8

and

t_{1/10} = \frac{1}{k} ln 10

. Their ratio is

\frac{t _{1/8}}{t _{1/10}} = \frac{ln 8}{ln 10} = lo g 8 = 3 lo g 2 = 3 (0.3) = 0.9

. Multiplying by 10 gives

\frac{t _{1/8}}{t _{1/10}} \times 10 = 9

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option