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Chemical Kinetics and Nuclear Chemistry question

2003 · Shift 0 · Q41
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Chemical Kinetics and Nuclear Chemistry question

2003 · Shift 0 · Q41

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The rate law for a reaction between the substances A and B is given by Rate = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as
  1. A
    (m + n)
  2. B
    (n - m)
  3. C
    2( n - m)
  4. D
    12(m+n){1 \over {{2^{(m + n)}}}}2(m+n)1​
View written solutionFree

Correct answer: CORRECT ANSWER SHOULD BE $2^{N-M}$ (NOT LISTED AMONG THE OPTIONS)

  1. Given rate law

    Rate=k[A]n[B]m\text{Rate} = k[A]^n[B]^mRate=k[A]n[B]m

  2. Initial rate

    Let the initial rate be

    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 \dfrac{[B]}{2}[B]→2[B]​

    So the new rate becomes

    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 the new rate

    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 of new rate to old rate

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

  6. Compare with options

    The required ratio is

    2n−m2^{n-m}2n−m

    But the given options are:

    • A: (m+n)(m+n)(m+n)
    • B: (n−m)(n-m)(n−m)
    • C: 2(n−m)2(n-m)2(n−m)
    • D: 12m+n\dfrac{1}{2^{m+n}}2m+n1​

    None of these matches 2n−m2^{n-m}2n−m.

  7. Conclusion

    The correct expression should be

    2n−m\boxed{2^{n-m}}2n−m​

    Hence, the stored answer C is not correct.

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