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Chemical Kinetics and Nuclear Chemistry question

2025 · 4 Apr · Shift 1 · Q4
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Chemical Kinetics and Nuclear Chemistry question

2025 · 4 Apr · Shift 1 · Q4

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Rate law for a reaction between AAA and BBB is given by r=k[ A]n[ B]m\mathrm{r}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}r=k[ A]n[ B]m If concentration of AAA is doubled and concentration of BBB 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)(r1​r2​​) is
  1. A
    (n−m)(\mathrm{n}-\mathrm{m})(n−m)
  2. B
    2(n−m)2^{(\mathrm{n}-m)}2(n−m)
  3. C
    12m+n\frac{1}{2^{m+n}}2m+n1​
  4. D
    (m+n)(\mathrm{m}+\mathrm{n})(m+n)
View written solutionFree

Correct answer: B

  1. Given rate law

    r=k[A]n[B]mr = k[A]^n[B]^mr=k[A]n[B]m

  2. Initial rate

    Let the initial concentrations be [A][A][A] and [B][B][B]. Then,

    r1=k[A]n[B]mr_1 = k[A]^n[B]^mr1​=k[A]n[B]m

  3. New concentrations

    • Concentration of AAA is doubled: [A]→2[A][A] \to 2[A][A]→2[A]
    • Concentration of BBB is halved: [B]→[B]2[B] \to \frac{[B]}{2}[B]→2[B]​

    So the new rate is

    r2=k(2[A])n([B]2)mr_2 = k(2[A])^n\left(\frac{[B]}{2}\right)^mr2​=k(2[A])n(2[B]​)m

  4. Simplify r2r_2r2​

    r2=k 2n[A]n⋅[B]m2mr_2 = k\,2^n[A]^n \cdot \frac{[B]^m}{2^m}r2​=k2n[A]n⋅2m[B]m​

    r2=k[A]n[B]m⋅2n−mr_2 = k[A]^n[B]^m \cdot 2^{n-m}r2​=k[A]n[B]m⋅2n−m

  5. Find the ratio r2r1\frac{r_2}{r_1}r1​r2​​

    r2r1=k[A]n[B]m⋅2n−mk[A]n[B]m\frac{r_2}{r_1} = \frac{k[A]^n[B]^m \cdot 2^{n-m}}{k[A]^n[B]^m}r1​r2​​=k[A]n[B]mk[A]n[B]m⋅2n−m​

    r2r1=2n−m\frac{r_2}{r_1} = 2^{n-m}r1​r2​​=2n−m

  6. Match with options

    The correct option is:

    2(n−m)\boxed{2^{(n-m)}}2(n−m)​

    So, Option B is correct.

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