Q57 / 180 · KCET 2020 · Chemistry

For an elementary reaction $2 A+3 B \rightarrow 4 C+D$ the rate of appearance of $C$ at time ' $t$ ' is $2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}$. Rate of disappearance of $B$ at ' $t$ '. $t$ will be

  • A
    $\frac{4}{3}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}-1 \mathrm{~S}^{-1}$
  • Correct answer
    B
    $\frac{3}{4}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}^{-1} \mathrm{~S}^{-1}$
  • C
    $2\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}^{-1} \mathrm{~S}^{-1}$
  • D
    $\frac{1}{4}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}-1 \mathrm{~S}^{-1}$

Solution

$2 A+3 B \longrightarrow 4 C+D$ $\begin{aligned} \frac{d(C)}{d t} &=2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}(\text { at time } t) \\ -\frac{d(B)}{d t} &=? \\ -\frac{1}{3} \frac{d(B)}{\Delta t} &=+\frac{1}{4} \frac{d[C]}{\Delta t} \end{aligned}$ Rate of disappearance of $B=\frac{3}{4} \times$ rate of appearance of $C$ $=\frac{3}{4} \times 2.8 \times 10^{-3}$

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

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