Q6 / 150 · MHT CET 2025 (27 Apr Shift 2) · Chemistry

For the reaction, $A+3 B \rightarrow 2 C$ rate of consumption of $A$ is $1.4 \mathrm{~mol~} \mathrm{dm}^{-3}$ minute ${ }^{-1}$. Calculate rate of consumption of B .

  • Correct answer
    A
    $0.7 \mathrm{~mol~} \mathrm{dm}{ }^{-3}$ minute ${ }^{-1}$
  • B
    $4.2 \mathrm{~mol~} \mathrm{dm}^{-3}$ minute ${ }^{-1}$
  • C
    $2.1 \mathrm{~mol~} \mathrm{dm}{ }^{-3}$ minute ${ }^{-1}$
  • D
    $2.8 \mathrm{~mol~} \mathrm{dm}^{-3}$ minute $e^{-1}$

Solution

The stoichiometry of $A + 3B \to 2C$ relates the rates via $\text{Rate} = -\frac{d[A]}{dt} = -\frac{1}{3}\frac{d[B]}{dt}$.

Given $-\frac{d[A]}{dt} = 1.4\ \mathrm{mol\ dm}^{-3}\ \mathrm{min}^{-1}$, it follows that $-\frac{d[B]}{dt} = 3 \times 1.4 = 4.2\ \mathrm{mol\ dm}^{-3}\ \mathrm{min}^{-1}$.

Final answer: $\boxed{\text{B}}$

Hence, the correct answer is option: 'A'.

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