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

\(NO_2\) required for a reaction is produced by the decomposition of \(N_2O_5\) in \(CCl_4\) as per the equation  \(2N_2O_5 (g) \to 4NO_2 (g) + O_2 (g)\) The initial concentration of \(N_2O_5\) is \(3.00\, mol\, L^{-1}\) and it is \(2.75\, mol\, L^{-1}\) after \(30\, minutes\). The rate of formation of \(NO_2\) is ............... \(mol\, L^{-1}\, min^{-1}\)

  1. A \(1.667\times10^{-2}\)
  2. B \(4.167\times10^{-3}\)
  3. C \(8.333\times10^{-3}\)
  4. D \(2.083\times10^{-3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1.667\times10^{-2}\)

Step-by-step Solution

Detailed explanation

\(2N_2O_5 (g) \to 4NO_2 (g) + O_2 (g)\) \(t=0\) \(3.0\,M\) \(t=30\) \(2.75\,M\) \(\frac{{ - \,\Delta \,[{N_2}{O_5}]}}{{\Delta t}}\, = \,\frac{{0.25}}{{30}}\)…
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