Chemistry Q58 – JEE-MAIN 2026

Home Chemistry Chemical KineticsQuestion #258

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 ?

Clue 1 of 3: Understand Reaction Orders

Recall the integrated rate laws and definitions of half-life and completion time for zero-order and first-order reactions.

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● Video Walkthrough

Step 1: Determine relations for Zero-Order Reaction✦ Active

For a zero-order reaction, the integrated rate law is $[A]_{t} = [A]_{0} - k t$ . We need to find expressions for half-life ( $t_{1 / 2}$ ) and completion time ( $t_{100 %}$ ). For $t_{1 / 2}$ , $[A]_{t} = [A]_{0} / 2$ . For $t_{100 %}$ , $[A]_{t} = 0$ .

For t_{1 / 2} : [A]_{0} / 2 = [A]_{0} - k t_{1 / 2} ⟹ k t_{1 / 2} = [A]_{0} / 2 ⟹ t_{1 / 2} = \frac{[A]_{0}}{2 k}

For t_{100 %} : 0 = [A]_{0} - k t_{100 %} ⟹ k t_{100 %} = [A]_{0} ⟹ t_{100 %} = \frac{[A]_{0}}{k}

Comparing these, for zero-order reactions: t_{100 %} = 2 t_{1 / 2}

Step 2: Determine relations for First-Order Reaction○ Expand

For a first-order reaction, the integrated rate law is $\ln [A]_{t} = \ln [A]_{0} - k t$ . We find expressions for half-life ( $t_{1 / 2}$ ) and completion time ( $t_{100 %}$ ). For $t_{1 / 2}$ , $[A]_{t} = [A]_{0} / 2$ . For $t_{100 %}$ , $[A]_{t} = 0$ .

For t_{1 / 2} : \ln ([A]_{0} / 2) = \ln [A]_{0} - k t_{1 / 2} ⟹ k t_{1 / 2} = \ln [A]_{0} - \ln ([A]_{0} / 2) = \ln (\frac{[A]_{0}}{[A]_{0} / 2}) = \ln (2) ⟹ t_{1 / 2} = \frac{\ln (2)}{k}

For t_{100 %} : For [A]_{t} to be 0 in [A]_{t} = [A]_{0} e^{- k t}, time t must approach infinity.

Therefore, for first-order reactions: t_{100 %} = \infty

Step 3: Match with Options○ Expand

Based on the derivations:

For zero-order reaction: t_{100 %} = 2 t_{1 / 2}

For first-order reaction: t_{100 %} = \infty

Option 2 states " $t_{100 %} = 2 t_{1 / 2}$ and $t_{100 %} = (t_{1 / 2})^{\infty}$ ". Interpreting $(t_{1 / 2})^{\infty}$ as $\infty$ (since $t_{1 / 2}$ is a finite positive value), this option correctly represents the relations for zero and first order reactions respectively.

💡 Teacher's Secret Hint

Be careful with the interpretation of $(t_{1 / 2})^{\infty}$ as infinity for first-order reactions.

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Chemistry Question 58 – JEE-MAIN 2026

Clue 1 of 3: Understand Reaction Orders

🧠 Full Solution Path