MHT CET · Chemistry · Chemical Kinetics
Find the rate of formation of \(\mathrm{NO}_{2(\mathrm{~g})}\) in the following reaction.
\(2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\) \(\ {\left[\frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=0.02 \mathrm{~mol} \mathrm{dm}^{-3}\right]}\)
- A \(0.01 \mathrm{~mol} \mathrm{dm}^{-3}\)
- B \(0.02 \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(0.03 \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(0.04 \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(D) \(0.04 \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\text {Rate of reaction } =-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \)
\( =+\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\)
Rate of formation of
\(\mathrm{NO}_2 =\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \)
\( =-\frac{4}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \)
\( =-2 \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \)
\( =2 \times 0.02 \)
\( =0.04 \mathrm{~mol} \mathrm{dm}^{-3}\)
\( =+\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\)
Rate of formation of
\(\mathrm{NO}_2 =\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \)
\( =-\frac{4}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \)
\( =-2 \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \)
\( =2 \times 0.02 \)
\( =0.04 \mathrm{~mol} \mathrm{dm}^{-3}\)
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