AP EAMCET · Chemistry · Chemical Kinetics
Consider a gas phase reaction which occurs in a closed vessel
\(2 \mathrm{~A} \rightarrow 4 \mathrm{~B}+\mathrm{C}\)
The concentration of \(\mathrm{B}\) is found to be increased by \(5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\) in 10 seconds.
The rate of disappearance of \(\mathrm{A}\) (in \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\) ) is
- A \(4.75 \times 10^{-4}\)
- B \(7.5 \times 10^{-4}\)
- C \(1.25 \times 10^{-4}\)
- D \(2.5 \times 10^{-4}\)
Answer & Solution
Correct Answer
(D) \(2.5 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{1}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{2} \times \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \\ & =\frac{1}{2} \times…
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