JEE Main 2025 Jan 28 Shift 1, Chemistry Q7: Reactions of Various Orders

JEE Main2025Jan 28, Shift 1Chemistry

For a given reaction $R \to P$ , $t_{1/2}$ is related to $[A]_{0}$ as given in table :

$[A]_{0} / mol L^{- 1}$	$t_{1/2} / min$
0.100	200
0.025	100

Given : $lo g 2 = 0.30$

Which of the following is true ?

A. The order of the reaction is $\frac{1}{2}$ .

B. If $[A]_{0}$ is 1 M, then $t_{1/2}$ is $20010$ min.

C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.

D. $t_{1/2}$ is 800 min for $[A]_{0} = 1.6$ M.

Choose the correct answer from the options given below :

A
A and C only
B
A and B only
C
A, B and D only
D
C and D only

Solution

For an $n$ -th order reaction, $t_{1/2} \propto [A]_{0}^{1 - n}$ . Using the two given data points:

$\frac{200}{100} = (\frac{0.100}{0.025})^{n - 1} \Rightarrow 2 = 4^{n - 1}$

Hence $n - 1 = \frac{1}{2}$ , giving $n = \frac{1}{2}$ . Statement A is correct.

With order $1/2$ , $t_{1/2} \propto [A]_{0}$ . Using the anchor (0.1 M, 200 min):

At $[A]_{0} = 1$ M: $t_{1/2} = 200 \times \frac{1}{0.1} = 20010$ min. Statement B is correct.

The order of a reaction is a kinetic constant for a given mechanism; merely changing the initial concentration does not change $n$ . Statement C is false.

At $[A]_{0} = 1.6$ M: $t_{1/2} = 200 \times \frac{1.6}{0.1} = 200 \times 4 = 800$ min. Statement D is correct.

$∴$ A, B and D are true.

Concept behind this question

Reactions of Various OrdersNotes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 28 Shift 1, Chemistry Q7: Reactions of Various Orders

Concept behind this question

More previous year questions on Reactions of Various Orders

Practice more Chemistry