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Question: Answered & Verified by Expert
In the reversible reaction,
$2 \mathrm{NO}_{2} \frac{\mathrm{k}_{1}}{\rightleftharpoons \mathrm{k}_{2}} \mathrm{~N}_{2} \mathrm{O}_{4}$
the rate of disappearance of $\mathrm{NO}_{2}$ is equal to
ChemistryChemical KineticsVITEEEVITEEE 2011
Options:
  • A $\frac{2 \mathrm{k}_{1}}{\mathrm{k}_{2}}\left[\mathrm{NO}_{2}\right]^{2}$
  • B $2 \mathrm{k}_{1}\left[\mathrm{NO}_{2}\right]^{2}-2 \mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]$
  • C $2 \mathrm{k}_{1}\left[\mathrm{NO}_{2}\right]^{2}-\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]$
  • D $\left(2 \mathrm{k}_{1}-\mathrm{k}_{2}\right)\left[\mathrm{NO}_{2}\right]$
Solution:
2298 Upvotes Verified Answer
The correct answer is: $2 \mathrm{k}_{1}\left[\mathrm{NO}_{2}\right]^{2}-\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]$
$$
\begin{array}{l}
2 \mathrm{NO}_{2} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}} \mathrm{N}_{2} \mathrm{O}_{4} \\
\text { Rate }=-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=k_{1}\left[\mathrm{NO}_{2}\right]^{2}-k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]
\end{array}
$$
$\therefore$ Rate of disappearance of $\mathrm{NO}_{2}$ i.e.,
$$
-\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=2 k_{1}\left[\mathrm{NO}_{2}\right]^{2}-k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]
$$

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