Q10 / 150 · MHT CET 2023 (11 May Shift 1) · Chemistry

For an elementary reaction $$ 2 \mathrm{~A}+\mathrm{B} \longrightarrow 3 \mathrm{C} $$ rate of appearance of $\mathrm{C}$ is $1.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$, the rate of disappearance of $\mathrm{A}$ is:

  • A
    $1.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  • B
    $2.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  • C
    $5.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
  • Correct answer
    D
    $8.66 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

Solution

$$ \begin{aligned} \text { Rate of reaction } & =-\frac{1}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\frac{1}{3} \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \\ \text { Rate of reaction } & =\frac{1}{3} \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \\ & =\frac{1}{3} \times 1.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =0.433 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned} $$ $\therefore \quad$ Rate of disappearance of $\mathrm{A}$ $$ \begin{aligned} & =2 \times 0.433 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =0.866 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =8.66 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned} $$

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

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