MHT CET · Chemistry · Chemical Kinetics
For a reaction,
\(2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\)
\(\mathrm{N}_2 \mathrm{O}_5\) disappears at a rate of \(0.06 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}\). What is rate of \(\mathrm{NO}_{2(\mathrm{~g})}\) formation?
- A \(0.06 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}\)
- B \(0.12 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}\)
- C \(0 \cdot 18 \mathrm{moldm}^{-3} \cdot \mathrm{~s}^{-1}\)
- D \(0.24 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(0.12 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(2 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 4 \mathrm{NO}_2+\mathrm{O}_2\)
Overall rate of reaction can be expressed as:
\(-\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{~d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\)
Rate of formation of \(\mathrm{NO}_2\) :
\(\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=-\frac{4}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{4 \times 0.06}{2}\)\(=0.12 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
Overall rate of reaction can be expressed as:
\(-\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{~d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\)
Rate of formation of \(\mathrm{NO}_2\) :
\(\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=-\frac{4}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{4 \times 0.06}{2}\)\(=0.12 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
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