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 $2\mathrm{A} + 3\mathrm{B} + \frac{3}{2}\mathrm{C} \rightarrow 3\mathrm{P}$ , which statement is correct?

Options

a
$\frac{dn_A}{dt} = \frac{3}{2}\frac{dn_B}{dt} = \frac{3}{4}\frac{dn_C}{dt}$
b
$\frac{dn_A}{dt} = \frac{dn_B}{dt} = \frac{dn_C}{dt}$
c
$\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}$
✓
d
$\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{3}{4}\frac{dn_C}{dt}$

Correct Answerc

$\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}$

Detailed Solution

The rate of reaction is defined as: $r = -\frac{1}{2}\frac{dn_A}{dt} = -\frac{1}{3}\frac{dn_B}{dt} = -\frac{1}{3/2}\frac{dn_C}{dt} = \frac{1}{3}\frac{dn_P}{dt}$

So: $\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt}$ and $\frac{dn_A}{dt} = \frac{2}{3/2}\frac{dn_C}{dt} = \frac{4}{3}\frac{dn_C}{dt}$

Answer: Option (3)

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