For the first order reaction2N2O5 (g) → 4 NO2(g) + O2 (g)

Engineering

Chemistry

Integrated Rate Law

Question

For the first order reaction

2N₂O₅ (g) → 4 NO₂(g) + O₂ (g)

the half-life of the reaction depends on the initial concentration of the reactant

the concentration of the reactant decreases exponentially with time.

the half-life of the reaction decreases with increasing temperature

the reaction proceeds to 99.6 % completion in eight half-life duration.

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$C_{A} = C_{A_{0}} e^{- kt} [A]$

$t_{\frac{1}{2}} = \frac{0 .693}{K} = \frac{0 .693}{A_{0} e^{- E_{a} / RT}} = \frac{0 .693}{A_{0}} e^{E_{a} / RT} [B]$

$\frac{0 .4}{100} = {(\frac{1}{2})}^{4} = \frac{4}{100} = n = \frac{\log (\frac{4}{100})}{\log (\frac{1}{2})} = 8 [D]$