A → products (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run $1 \Rightarrow 100 \mathrm{~mL}$ of 10 M solution of reactant A Run $2 \Rightarrow 200 \mathrm{~mL}$ of 10 M solution of reactant A Run $3 \Rightarrow 100 \mathrm{~mL}$ of 10 M solution of reactant $\mathrm{A}+$ 100 mL of $\mathrm{H}_{2} \mathrm{O}$ added. The correct variation of rate of reaction is

(B) Run $3<$ Run $1=$ Run 2

(A) Run $1=$ Run $2=$ Run 3

(C) Run $3<$ Run $1<$ Run 2

(D) Run $1<\operatorname{Run} 2<\operatorname{Run} 3$

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Chemical Kinetics

A → products (First order reaction).

Three sets of experiment were performed for a reaction under similar experimental conditions.
Run

1 \Rightarrow 100 mL

of 10 M solution of reactant A
Run

2 \Rightarrow 200 mL

of 10 M solution of reactant A
Run

3 \Rightarrow 100 mL

of 10 M solution of reactant

A +

100 mL of

H_{2} O

added.
The correct variation of rate of reaction is

(A)Run $1 =$ Run $2 =$ Run 3
(B)Run $3 <$ Run $1 =$ Run 2✓ Correct
(C)Run $3 <$ Run $1 <$ Run 2
(D)Run $1 < Run 2 < Run 3$

Answer: (B)

Solution

For a first-order reaction the rate is

k [A]

, set by the concentration and not by the total volume. In Run 1 the concentration is 10 M, and in Run 2 it is still 10 M, since taking a larger volume of the same solution does not change its concentration, so these two runs have equal rates. In Run 3 the 100 mL of 10 M solution is diluted with an equal volume of water to 200 mL, halving the concentration to 5 M and lowering the rate. The rates therefore follow Run 3

<

Run 1

=

Run 2.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Chemical Kinetics

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