Q15 / 75 · JEE Main 2026 (22 January Shift 1) · Chemistry

$\mathrm{A} \rightarrow$ product (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 $\mathrm{A}+100 \mathrm{~mL}$ of $\mathrm{H}_{2} \mathrm{O}$ added.
The correct variation of rate of reaction is

  • A
    Run $1<$ Run $2<$ Run 3
  • B
    Run $3<$ Run $1<$ Run 2
  • Correct answer
    C
    Run $3<$ Run 1 = Run 2
  • D
    Run $1=$ Run $2=$ Run 3

Solution

For a first-order reaction, rate = $k[A]$, where rate is directly proportional to reactant concentration.
Run 1: 100 mL of 10 M solution
Concentration = 10 M
Rate: $r_1 = k \times 10$
Run 2: 200 mL of 10 M solution
Concentration remains 10 M (increasing volume doesn't change concentration)
Rate: $r_2 = k \times 10$
Therefore $r_1 = r_2$
Run 3: 100 mL of 10 M solution + 100 mL of H₂O
Total moles of A = 10 M × 0.1 L = 1 mol
Total volume = 0.2 L
New concentration = 1 mol / 0.2 L = 5 M
Rate: $r_3 = k \times 5$
Therefore $r_3 < r_1 = r_2$

Hence, the correct answer is option: 'C'.

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