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CUET · CHEMISTRY · PYQ PAPER 2025

The decomposition of N\(_2\)\(\mathrm{O}_5\) in CCl\(_4\) at 318K has been studied by monitoring the concentration of N\(_2\)\(\mathrm{O}_5\) in the solution. Initially, the concentration of N\(_2\)\(\mathrm{O}_5\) is 2.33 mol/L and after 184 minutes, it is reduced to 2.08 mol/L. The reaction takes place according to the equation
2N\(_2\)\(\mathrm{O}_5\)(g) \(\rightarrow\) 4NO\(_2\)(g) + O\(_2\)(g)
The average rate of this reaction in different units will be:

  1. A \(4.07 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1},6.79 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \min ^{-1},1.13 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
  2. B \(2.907 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1},6.79 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \min ^{-1},1.13 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
  3. C \(12.07 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1},7.79 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \min ^{-1},5.13 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
  4. D \(8.1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1},13.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1},2.25 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
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Answer & Solution

Correct Answer

(A) \(4.07 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~h}^{-1},6.79 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \min ^{-1},1.13 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Step-by-step Solution

Detailed explanation

Average rate \( = - \frac{1}{2} \frac{\Delta[\mathrm{N}_2\mathrm{O}_5]}{\Delta t} \) Average rate \( = - \frac{1}{2} \frac{(2.08 - 2.33) \mathrm{~mol/L}}{184 \mathrm{~min}} \) Average rate…
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