KCET · Chemistry · Chemical Kinetics
For the reaction \(2 \mathrm{~N}_2 \mathrm{O}_{5_{(8)}} \rightarrow 4 \mathrm{NO}_{2_{(\mathrm{g})}}+\mathrm{O}_{2_{(8)}}\) initial concentration of \(\mathrm{N}_2 \mathrm{O}_5\) is \(2.0 \mathrm{molL}^{-1}\) and after 300 \(\min\), it is reduced to \(1.4 \mathrm{molL}^{-1}\). The rate of production of \(\mathrm{NO}_2\left(\right.\) in \(\mathrm{molL}^{-1} \mathrm{~min}^{-1}\) ) is
- A \(2.5 \times 10^{-4}\)
- B \(4 \times 10^{-4}\)
- C \(2.5 \times 10^{-3}\)
- D \(4 \times 10^{-3}\)
Answer & Solution
Correct Answer
(D) \(4 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
Rate of formation of \(\mathrm{NO}_2=2 \times\) rate of disappearance of \(\mathrm{N}_2 \mathrm{O}_5\)
\(\begin{aligned} & =2 \times\left[\frac{-(1.4-2)}{300}\right] \\ & =4 \times 10^{-3}\end{aligned}\)
\(\begin{aligned} & =2 \times\left[\frac{-(1.4-2)}{300}\right] \\ & =4 \times 10^{-3}\end{aligned}\)
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