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Chemical Kinetics and Nuclear Chemistry question

2020 · 3 Sep · Shift 2 · Q3
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Chemical Kinetics and Nuclear Chemistry question

2020 · 3 Sep · Shift 2 · Q3

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For the reaction 2A + 3B + 32{3 \over 2}23​ C →\to→ 3P, which statement is correct ?
  1. A
    dnAdt=dnBdt=dnCdt{{d{n_A}} \over {dt}} = {{d{n_B}} \over {dt}} = {{d{n_C}} \over {dt}}dtdnA​​=dtdnB​​=dtdnC​​
  2. B
    dnAdt=23dnBdt=34dnCdt{{d{n_A}} \over {dt}} = {2 \over 3}{{d{n_B}} \over {dt}} = {3 \over 4}{{d{n_C}} \over {dt}}dtdnA​​=32​dtdnB​​=43​dtdnC​​
  3. C
    dnAdt=32dnBdt=34dnCdt{{d{n_A}} \over {dt}} = {3 \over 2}{{d{n_B}} \over {dt}} = {3 \over 4}{{d{n_C}} \over {dt}}dtdnA​​=23​dtdnB​​=43​dtdnC​​
  4. D
    dnAdt=23dnBdt=43dnCdt{{d{n_A}} \over {dt}} = {2 \over 3}{{d{n_B}} \over {dt}} = {4 \over 3}{{d{n_C}} \over {dt}}dtdnA​​=32​dtdnB​​=34​dtdnC​​
View written solutionFree

Correct answer: D

  1. For the reaction 2A+3B+32C→3P2A+3B+\frac{3}{2}C \to 3P2A+3B+23​C→3P the stoichiometric rate relation is −12dnAdt=−13dnBdt=−1(3/2)dnCdt=13dnPdt-\frac{1}{2}\frac{dn_A}{dt}=-\frac{1}{3}\frac{dn_B}{dt}=-\frac{1}{(3/2)}\frac{dn_C}{dt}=\frac{1}{3}\frac{dn_P}{dt}−21​dtdnA​​=−31​dtdnB​​=−(3/2)1​dtdnC​​=31​dtdnP​​

  2. Ignoring the common negative sign for reactants, the magnitudes of disappearance rates are proportional to their stoichiometric coefficients: dnAdt:dnBdt:dnCdt=2:3:32\frac{dn_A}{dt}:\frac{dn_B}{dt}:\frac{dn_C}{dt}=2:3:\frac{3}{2}dtdnA​​:dtdnB​​:dtdnC​​=2:3:23​

  3. Now check option D: dnAdt=23dnBdt=43dnCdt\frac{dn_A}{dt}=\frac{2}{3}\frac{dn_B}{dt}=\frac{4}{3}\frac{dn_C}{dt}dtdnA​​=32​dtdnB​​=34​dtdnC​​ From this,

  • if dnAdt=23dnBdt\frac{dn_A}{dt}=\frac{2}{3}\frac{dn_B}{dt}dtdnA​​=32​dtdnB​​, then dnAdt:dnBdt=2:3\frac{dn_A}{dt}:\frac{dn_B}{dt}=2:3dtdnA​​:dtdnB​​=2:3
  • if dnAdt=43dnCdt\frac{dn_A}{dt}=\frac{4}{3}\frac{dn_C}{dt}dtdnA​​=34​dtdnC​​, then dnAdt:dnCdt=4:3\frac{dn_A}{dt}:\frac{dn_C}{dt}=4:3dtdnA​​:dtdnC​​=4:3 But 2:32=4:32:\frac{3}{2}=4:32:23​=4:3 So option D is consistent.
  1. Check others briefly:
  • A says all are equal, false.
  • B gives A:C=3:4A:C=3:4A:C=3:4, false.
  • C gives A:B=3:2A:B=3:2A:B=3:2, false.

Therefore, the correct option is: D\boxed{D}D​

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