A → product (First order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions: Run 1 ⇒ 100 mL of 10 M solution of reactant A Run 2 ⇒ 200 mL of 10 M solution of reactant A Run 3 ⇒ 100 mL of 10 M solution of reactant A + 100 mL of H₂O added. The correct variation of rate of reaction is

The correct variation of rate of reaction for a first order reaction

Q. A → product (First order reaction).

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

Run 1 ⇒ 100 mL of 10 M solution of reactant A

Run 2 ⇒ 200 mL of 10 M solution of reactant A

Run 3 ⇒ 100 mL of 10 M solution of reactant A + 100 mL of H₂O added.

The correct variation of rate of reaction is

A. Run 1 = Run 2 = Run 3

B. Run 3 < Run 1 < Run 2

C. Run 1 < Run 2 < Run 3

D. Run 3 < Run 1 = Run 2

Correct Answer: Run 3 < Run 1 = Run 2

Explanation

For a first order reaction, the rate of reaction is given by:

\[ \text{Rate} = k[A] \]

This shows that the rate depends only on the concentration of reactant A and not on the volume or total amount taken.

Run 1:
Concentration of A = 10 M
\[ \text{Rate}_1 = k \times 10 \]

Run 2:
Even though the volume is doubled to 200 mL, the concentration of A remains 10 M.
\[ \text{Rate}_2 = k \times 10 \]

Hence, \[ \text{Rate}_1 = \text{Rate}_2 \]

Run 3:
100 mL of 10 M solution is diluted by adding 100 mL of water.

Initial moles of A: \[ \text{Moles} = 10 \times 0.1 = 1 \]

Final volume = 0.2 L
New concentration: \[ [A] = \frac{1}{0.2} = 5 \text{ M} \]

Rate of reaction: \[ \text{Rate}_3 = k \times 5 \]

Thus, \[ \text{Rate}_3 < \text{Rate}_1 = \text{Rate}_2 \]

This concept is extremely important for JEE Main 2026, JEE Advanced and IIT JEE under Chemical Kinetics.

Explanation

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