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

(D) $\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{4}{3} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}$

(A) $\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{3}{4} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}$

(B) $\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}$

(C) $\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{3}{2} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{3}{4} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Chemical Kinetics

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For the reaction

2 A + 3 B + \frac{3}{2} C \to 3 P

, which statement is correct?

(A) $\frac{dn _{A}}{dt} = \frac{2}{3} \frac{dn _{B}}{dt} = \frac{3}{4} \frac{dn _{C}}{dt}$
(B) $\frac{dn _{A}}{dt} = \frac{dn _{B}}{dt} = \frac{dn _{C}}{dt}$
(C) $\frac{dn _{A}}{dt} = \frac{3}{2} \frac{dn _{B}}{dt} = \frac{3}{4} \frac{dn _{C}}{dt}$
(D) $\frac{dn _{A}}{dt} = \frac{2}{3} \frac{dn _{B}}{dt} = \frac{4}{3} \frac{dn _{C}}{dt}$ ✓ Correct

Answer: (D)

Solution

For

2 A + 3 B + \frac{3}{2} C \to 3 P

the single rate of reaction ties to each species through its coefficient: rate

= - \frac{1}{2} \frac{d n _{A}}{d t} = - \frac{1}{3} \frac{d n _{B}}{d t} = - \frac{2}{3} \frac{d n _{C}}{d t}

. Reading off the relations between the consumption rates,

\frac{d n _{A}}{d t} = \frac{2}{3} \frac{d n _{B}}{d t}

and

\frac{d n _{A}}{d t} = \frac{4}{3} \frac{d n _{C}}{d t}

, so

\frac{d n _{A}}{d t} = \frac{2}{3} \frac{d n _{B}}{d t} = \frac{4}{3} \frac{d n _{C}}{d t}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Chemical Kinetics

Solution

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