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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

\(\mathrm{N}_{2} \mathrm{O}_{5(9)} \rightarrow 2 \mathrm{NO}_{2(9)}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}\) In the above first order reaction the initial concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(2.40 \times 10^{-2}\, \mathrm{~mol} \,\mathrm{~L}^{-1}\) at \(318 \,K.\) The concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(1\, hour\) was \(1.60 \times 10^{-2}\, \mathrm{~mol} \,\mathrm{~L}^{-1}\), The rate constant of the reaction at \(318\, \mathrm{~K}\) is \(.....\,\times 10^{-3} \mathrm{~min}^{-1}\). (Nearest integer) [Given: \(\log 3=0.477, \log 5=0.699\) ]

  1. A \(5\)
  2. B \(6\)
  3. C \(7\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(7\)

Step-by-step Solution

Detailed explanation

\(k=\frac{2.303}{t} \log \frac{\left[A_{0}\right]}{[A]}\) \(k=\frac{2.303}{t} \log \frac{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}\)…
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