JEE Main · 2020 · Shift-IImediumCK-005

For the reaction 2A + 3B + 32C 3P, which statement is correct?

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For the reaction 2A+3B+32C3P2\mathrm{A} + 3\mathrm{B} + \frac{3}{2}\mathrm{C} \rightarrow 3\mathrm{P}, which statement is correct?

Options
  1. a

    dnAdt=32dnBdt=34dnCdt\frac{dn_A}{dt} = \frac{3}{2}\frac{dn_B}{dt} = \frac{3}{4}\frac{dn_C}{dt}

  2. b

    dnAdt=dnBdt=dnCdt\frac{dn_A}{dt} = \frac{dn_B}{dt} = \frac{dn_C}{dt}

  3. c

    dnAdt=23dnBdt=43dnCdt\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}

  4. d

    dnAdt=23dnBdt=34dnCdt\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{3}{4}\frac{dn_C}{dt}

Correct Answerc

dnAdt=23dnBdt=43dnCdt\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}

Detailed Solution

The rate of reaction is defined as: r=12dnAdt=13dnBdt=13/2dnCdt=13dnPdtr = -\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}

So: dnAdt=23dnBdt\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} and dnAdt=23/2dnCdt=43dnCdt\frac{dn_A}{dt} = \frac{2}{3/2}\frac{dn_C}{dt} = \frac{4}{3}\frac{dn_C}{dt}

Answer: Option (3)

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