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Chemical Kinetics and Nuclear Chemistry question

2021 · 17 Mar · Shift 2 · Q16
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Chemical Kinetics and Nuclear Chemistry question

2021 · 17 Mar · Shift 2 · Q16

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The reaction 2AAA + B2B_2B2​ →\to→ 2ABABAB is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ‾\underline{\hspace{2cm}}​. (Round off to the Nearest Integer).
Numerical answer
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Correct answer: 27

  1. Write the rate law for the elementary reaction

For the elementary reaction 2A+B2→2AB2A + B_2 \to 2AB2A+B2​→2AB the rate law is written directly from the stoichiometric coefficients: r=k[A]2[B2]r = k[A]^2[B_2]r=k[A]2[B2​]

  1. Effect of reducing volume

For a fixed amount of reactants, concentration is inversely proportional to volume: [Concentration]∝1V[\text{Concentration}] \propto \frac{1}{V}[Concentration]∝V1​

If the volume is reduced by a factor of 333, then each concentration becomes 333 times: [A]′=3[A],[B2]′=3[B2][A]' = 3[A], \qquad [B_2]' = 3[B_2][A]′=3[A],[B2​]′=3[B2​]

  1. Find the new rate

Using the rate law: r′=k[A]′2[B2]′r' = k[A]'^2[B_2]'r′=k[A]′2[B2​]′ Substitute the new concentrations: r′=k(3[A])2(3[B2])r' = k(3[A])^2(3[B_2])r′=k(3[A])2(3[B2​]) r′=k⋅9[A]2⋅3[B2]r' = k \cdot 9[A]^2 \cdot 3[B_2]r′=k⋅9[A]2⋅3[B2​] r′=27k[A]2[B2]r' = 27k[A]^2[B_2]r′=27k[A]2[B2​]

Thus, r′=27rr' = 27rr′=27r

  1. Rate increase factor

So, the rate increases by a factor of 27\boxed{27}27​

  1. Comparison with stored answer

Stored correct answer = 272727.

My derived answer matches the stored answer.

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