For each reaction in List-I, the rate…

For each reaction in List-I, the rate law (or mechanism) is given. Match with the effect in List-II on the half-life t1/2 when the initial…

Single correct option

For each reaction in List-I, the rate law (or mechanism) is given. Match with the effect in List-II on the half-life

t_{1/2}

when the initial concentration of the reactant is doubled (temperature constant).

| List-I | List-II |
|---|---|
| P.

\mathrm{2A\to}

products, rate

=k[\mathrm{A}]^2

| 1.

t_{1/2}

becomes twice |
| Q.

\mathrm{3A\to}

products;

\mathrm{2A\rightleftharpoons A_2}

(fast, constant

K

), then

\mathrm{A_2+A\to}

products (slow) | 2.

t_{1/2}

becomes

\sqrt2

times |
| R.

\mathrm{A\to}

products, rate

=k[\mathrm{A}]^{1/2}

| 3.

t_{1/2}

is unchanged |
| S.

\mathrm{A\to}

products on a saturated catalytic surface, rate

=k

| 4.

t_{1/2}

becomes half |
| | 5.

t_{1/2}

becomes one-fourth |

Show answer and solution

Answer: (A) P→4, Q→5, R→2, S→1

For rate

=k[\mathrm{A}]^n

(

n\ne1

t_{1/2}\propto c_0^{\,1-n}

. P:

n=2

t_{1/2}\propto c_0^{-1}

, so doubling

c_0

halves

t_{1/2}

: 4. Q: the fast pre-equilibrium gives

[\mathrm{A_2}]=K[\mathrm{A}]^2

, so rate

=k_2K[\mathrm{A}]^2[\mathrm{A}]=k'[\mathrm{A}]^3

n=3

t_{1/2}\propto c_0^{-2}

, so

t_{1/2}

becomes

1/4

: 5. R:

n=1/2

t_{1/2}\propto c_0^{1/2}

, factor

\sqrt2

: 2. S: zero order,

t_{1/2}=c_0/2k

, doubled: 1. Option 3 (unchanged) applies only to first order, which is not in the list. The unique correct code is P→4, Q→5, R→2, S→1.

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake