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Chemical Kinetics and Nuclear Chemistry question

2020 · 2 Sep · Shift 2 · Q13
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Chemical Kinetics and Nuclear Chemistry question

2020 · 2 Sep · Shift 2 · Q13

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The results given in the below table were obtained during kinetic studies of the following reaction 2A + B →\to→ C + D JEE Main 2020 (Online) 2nd September Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 128 English X and Y in the given table are respectively :
  1. A
    0.3, 0.4
  2. B
    0.4, 0.3
  3. C
    0.4, 0.4
  4. D
    0.3, 0.3
View written solutionFree

Correct answer: A

To determine XXX and YYY, we use the fact that for the reaction

2A+B→C+D2A + B \to C + D2A+B→C+D

the rate law is obtained from experimental kinetic data, while the stoichiometric rate relations connect disappearance and appearance rates.

However, the question refers to a table, and that table is not visible in the provided prompt. So the numerical values of XXX and YYY cannot be derived directly from first principles without the missing data.

Still, since this is a standard kinetics question, the usual interpretation is that XXX and YYY are based on stoichiometric rate relationships for

2A+B→C+D2A + B \to C + D2A+B→C+D

For this reaction,

\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt} = r$$ So if one species rate is given, the others are related in the ratio $$- rac{d[A]}{dt} : -\frac{d[B]}{dt} : \frac{d[C]}{dt} : \frac{d[D]}{dt} = 2:1:1:1$$ Among the options, the stored correct answer is **A: $0.3, 0.4$**. Since the required table is missing, I cannot independently recompute the values from the data, but I will compare with the stored answer. ### Conclusion Because the experimental table is absent, a strict derivation is not possible from the information shown. Accepting the provided official key, the answer is: $$X = 0.3, \quad Y = 0.4$$
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