JEE Main 2026 Jan 28 Shift 1, Chemistry Q7: Reactions of Various Orders

JEE Main2026Jan 28, Shift 1Chemistry

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 $\frac{t _{1/8}}{t _{1/10}} \times 10$ ?

( $lo g 2 = 0.3$ )

A
9
B
0.9
C
3
D
30

Solution

For a first-order reaction: $t = \frac{1}{k} ln \frac{A _{0}}{A _{t}}$ .

$t_{1/8} = \frac{1}{k} ln 8$ and $t_{1/10} = \frac{1}{k} ln 10$ .

$\frac{t _{1/8}}{t _{1/10}} = \frac{ln 8}{ln 10} = \frac{lo g 8}{lo g 10} = 3 lo g 2 = 0.9$ .

$∴ \frac{t _{1/8}}{t _{1/10}} \times 10 = 9$ . Option (1).

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JEE Main 2026 Jan 28 Shift 1, Chemistry Q7: Reactions of Various Orders

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