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Chemical Kinetics and Nuclear Chemistry question

2025 · 28 Jan · Shift 2 · Q17
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Chemical Kinetics and Nuclear Chemistry question

2025 · 28 Jan · Shift 2 · Q17

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Consider an elementary reaction A( g)+B( g)→C( g)+D( g)\mathrm{A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{~g})+\mathrm{D}(\mathrm{~g})A( g)+B( g)→C( g)+D( g) If the volume of reaction mixture is suddenly reduced to 13\frac{1}{3}31​ of its initial volume, the reaction rate will become 'x′x^{\prime}x′ times of the original reaction rate. The value of xxx is :
  1. A
    3
  2. B
    9
  3. C
    13\frac{1}{3}31​
  4. D
    19\frac{1}{9}91​
View written solutionFree

Correct answer: B

  1. For an elementary reaction A(g)+B(g)→C(g)+D(g)\mathrm{A(g) + B(g) \to C(g) + D(g)}A(g)+B(g)→C(g)+D(g) the rate law is directly written from the stoichiometric coefficients: r=k[A][B]r = k[A][B]r=k[A][B]

  2. Suppose initially the concentrations are [A][A][A] and [B][B][B]. Then the initial rate is r0=k[A][B]r_0 = k[A][B]r0​=k[A][B]

  3. The volume is suddenly reduced to 13\dfrac{1}{3}31​ of the initial volume.

    Since concentration is inversely proportional to volume, [A]′=3[A],[B]′=3[B][A]' = 3[A], \qquad [B]' = 3[B][A]′=3[A],[B]′=3[B]

  4. New rate: r′=k[A]′[B]′=k(3[A])(3[B])r' = k[A]'[B]' = k(3[A])(3[B])r′=k[A]′[B]′=k(3[A])(3[B]) r′=9k[A][B]=9r0r' = 9k[A][B] = 9r_0r′=9k[A][B]=9r0​

  5. Therefore, the reaction rate becomes 999 times the original rate.

Hence, x=9x = 9x=9

So the correct option is B.

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