JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

$t_{100\%}$ is the time required for the $100\%$ completion of the reaction while $t_{1/2}$ is the time required for $50\%$ of the reaction to be completed. Which of the following option correctly represents the relation between $t_{100\%}$ and $t_{1/2}$ for zero and first order reactions respectively?

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$t_{100\%} = (t_{1/2})^2$ and $t_{100\%} = (t_{1/2})^{-\infty}$
B
$t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (t_{1/2})^\infty$
C
$t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (2t_{1/2})^2$
D
$t_{100\%} = (t_{1/2})^\infty$ and $t_{100\%} = 2t_{1/2}$

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Solution

For a zero-order reaction:

The integrated rate law is given by $t = \dfrac{[A]_0 - [A]_t}{k}$ .

For $50\%$ completion ( $t_{1/2}$ ), $[A]_t = \dfrac{[A]_0}{2}$ :

$t_{1/2} = \dfrac{[A]_0 - \dfrac{[A]_0}{2}}{k} = \dfrac{[A]_0}{2k}$

For $100\%$ completion ( $t_{100\%}$ ), $[A]_t = 0$ :

$t_{100\%} = \dfrac{[A]_0 - 0}{k} = \dfrac{[A]_0}{k}$

Comparing the two, we get:

$t_{100\%} = 2t_{1/2}$

For a first-order reaction:

The integrated rate law is given by $t = \dfrac{1}{k} \ln\left(\dfrac{[A]_0}{[A]_t}\right)$ .

For $100\%$ completion ( $t_{100\%}$ ), $[A]_t = 0$ :

$t_{100\%} = \dfrac{1}{k} \ln\left(\dfrac{[A]_0}{0}\right) = \infty$

A first-order reaction takes infinite time for $100\%$ completion. Among the given choices, this is symbolically represented as $\infty$ or $(t_{1/2})^\infty$ .

Thus, the correct relation is $t_{100\%} = 2t_{1/2}$ for zero-order and $t_{100\%} = \infty$ for first-order.

Answer: $t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (t_{1/2})^\infty$

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Options

$t_{100\%} = (t_{1/2})^2$ and $t_{100\%} = (t_{1/2})^{-\infty}$

$t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (t_{1/2})^\infty$

$t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (2t_{1/2})^2$

$t_{100\%} = (t_{1/2})^\infty$ and $t_{100\%} = 2t_{1/2}$

Solution

For a zero-order reaction:

The integrated rate law is given by $t = \dfrac{[A]_0 - [A]_t}{k}$ .

For $50\%$ completion ( $t_{1/2}$ ), $[A]_t = \dfrac{[A]_0}{2}$ :

$t_{1/2} = \dfrac{[A]_0 - \dfrac{[A]_0}{2}}{k} = \dfrac{[A]_0}{2k}$

For $100\%$ completion ( $t_{100\%}$ ), $[A]_t = 0$ :

$t_{100\%} = \dfrac{[A]_0 - 0}{k} = \dfrac{[A]_0}{k}$

Comparing the two, we get:

$t_{100\%} = 2t_{1/2}$

For a first-order reaction:

The integrated rate law is given by $t = \dfrac{1}{k} \ln\left(\dfrac{[A]_0}{[A]_t}\right)$ .

For $100\%$ completion ( $t_{100\%}$ ), $[A]_t = 0$ :

$t_{100\%} = \dfrac{1}{k} \ln\left(\dfrac{[A]_0}{0}\right) = \infty$

A first-order reaction takes infinite time for $100\%$ completion. Among the given choices, this is symbolically represented as $\infty$ or $(t_{1/2})^\infty$ .

Thus, the correct relation is $t_{100\%} = 2t_{1/2}$ for zero-order and $t_{100\%} = \infty$ for first-order.

Answer: $t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (t_{1/2})^\infty$

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry