Q79 / 200 · NEET 2022 (Phase 2) · Chemistry
For a chemical reaction
$4 \mathrm{~A}+3 \mathrm{~B} \rightarrow 6 \mathrm{C}+9 \mathrm{D}$
rate of formation of $\mathrm{C}$ is $6 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and rate of disappearance of $\mathrm{A}$ is $4 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of reaction and amount of $B$ consumed in interval of 10 seconds, respectively will be:
- Correct answerA$1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
- B$10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
- C$1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
- D$10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
Solution
$$
\begin{aligned}
& \text { Explanation: } 4 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 6 \mathrm{C}+9 \mathrm{D} \\
& \qquad \begin{aligned}
\text { Rate of reaction } & =-\frac{1}{4} \frac{d[\mathrm{~A}]}{d t}=-\frac{1}{3} \frac{d[\mathrm{~B}]}{d t} \\
& =\frac{1}{6} \frac{d[\mathrm{C}]}{d t}=\frac{1}{9} \frac{d[\mathrm{D}]}{d t}
\end{aligned}
\end{aligned}
$$
Rate of reaction $=$ Rate of formation of $\mathrm{C}$.
$$
\begin{aligned}
\therefore \quad r & =\frac{1}{6} \frac{d[\mathrm{C}]}{d t}=\frac{1}{6} \times 6 \times 10^{-2} \\
& =1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
Also, rate of reaction $=$ Rate of disappearance of $B$
$$
\begin{aligned}
r & =-\frac{1}{3} \frac{d[\mathrm{~B}]}{d t}=\frac{d[\mathrm{~B}]}{d t}=3 \times r \\
& =3 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
Consumption of $B$ in 10 seconds $=3 \times 10^{-2} \times 10$
$$
=30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
$$
Hence, the correct answer is option: 'A'.