For a first order reaction 2N2O5⟶k4NO2+O2, the half-life period is given by

For a first order reaction $2{\mathrm{N}}_{2}{\mathrm{O}}_{5}\stackrel{k}{⟶}4{\mathrm{NO}}_{2}+{\mathrm{O}}_{2}$, the half-life period is given by

1. A

2. B

$0.693/\mathrm{k}$

3. C

4. D

$0.693/2\mathrm{k}$

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Solution:

For $2{\mathrm{N}}_{2}{\mathrm{O}}_{5}\to 4{\mathrm{NO}}_{2}+{\mathrm{O}}_{2}$, we have

or

On integrating, we get

$\mathrm{ln}\frac{{\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]}_{0}}{{\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]}_{t}}=2\mathrm{kt}.$   For a half-life  ${\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]}_{t}=\left(1/2\right){\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]}_{0}$ Hence ${t}_{1/2}=\frac{\mathrm{ln}2}{2k}=\frac{0.693}{2k}$

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