Question

For the reaction $2\mathrm{A}+3\mathrm{B}+\left(3/2\right)\mathrm{C}\underset{}{\overset{}{\to }}3\mathrm{P}$, which statement is correct ?

• A. $\raisebox{1ex}{$d{n}_A$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\raisebox{1ex}{$d{n}_B$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\raisebox{1ex}{$d{n}_C$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$
• B. $\raisebox{1ex}{$d{n}_A$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{3}{2}\right)\raisebox{1ex}{$d{n}_B$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{3}{4}\right)\raisebox{1ex}{$d{n}_C$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$
• C. $\raisebox{1ex}{$d{n}_A$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{2}{3}\right)\raisebox{1ex}{$d{n}_B$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{4}{3}\right)\raisebox{1ex}{$d{n}_C$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$
• D. $\raisebox{1ex}{$d{n}_A$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{2}{3}\right)\raisebox{1ex}{$d{n}_B$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{3}{4}\right)\raisebox{1ex}{$d{n}_C$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$

Answer 1

The correct option is C

$\raisebox{1ex}{$d{n}_A$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{2}{3}\right)\raisebox{1ex}{$d{n}_B$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.=\left(\frac{4}{3}\right)\raisebox{1ex}{$d{n}_C$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$

The following way can give us the correct answer:

Step 1:

$2A+3B+\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)C\to P$

Step 2:

Rate of reaction:

$$\begin{gathered} ROR=\frac{1}{2}\left[\frac{-d\left[n_A\right]}{dt}\right]=\frac{1}{3}\left[\frac{-d\left[n_B\right]}{dt}\right]=\frac{2}{3}\left[\frac{-d\left[n_C\right]}{dt}\right]=\frac{1}{3}\left[\frac{+d\left[n_C\right]}{dt}\right] \\ \left[\frac{-d{n}_A}{dt}\right]=\frac{2}{3}\left[\frac{-d{n}_B}{dt}\right]=\frac{4}{3}\left[\frac{-d{n}_C}{dt}\right] \end{gathered}$$

Hence, option C is correct.