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 ............... \(mol\, L^{-1}\, min^{-1}\)
- A \(1.667\times10^{-2}\)
- B \(4.167\times10^{-3}\)
- C \(8.333\times10^{-3}\)
- D \(2.083\times10^{-3}\)
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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