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Chemistry · Chemical Kinetics

JEE Main 2025 — 4 April, Morning Shift — Question 7

Rate law for a reaction between AA and BB is given by r=k[ A]n[ B]m\mathrm{r}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}} If concentration of AA is doubled and concentration of BB is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction (r2r1)\left(\frac{r_{2}}{r_{1}}\right) is

  1. Option A:

    2(n−m)2^{(n-m)}

    Correct
  2. Option B:

    (m+n)(m+n)

  3. Option C:

    12m+n\frac{1}{2^{m+n}}

  4. Option D:

    (n−m)(n-m)

Answer: A

Step-by-step solution

r1=k[A]n[B]m\mathrm{r}_{1}=k[A]^{\mathrm{n}}[B]^{m}

r2=k(2[A])n(12[B])mr2=k[A]n[B]m⋅2n⋅(12)mr2r1=2n⋅(12)m=2(n−m)\begin{aligned} r_{2} & =k(2[A])^{n}\left(\frac{1}{2}[B]\right)^{m} r_{2} \\& =k[A]^{n}[B]^{m} \cdot 2^{n} \cdot\left(\frac{1}{2}\right)^{m} \frac{r_{2}}{r_{1}} \\& =2^{n} \cdot\left(\frac{1}{2}\right)^{m} \\& =2^{(n-m)} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws

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