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**Problem**: Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25oC. If a 1.6-L reaction vessel initially contains 750 torr of N2O5 at 25 oC, what partial pressure of O2 will be present in the vessel after 220 minutes?

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Answer:

152 torr

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We’re being asked to calculate partial pressure of O_{2} will be present in the vessel after 220 minutes.

The given equation is: **N _{2}O_{5}(g) → NO_{2}(g) + O_{2}(g)**

Balancing the equation is: **2 ****N _{2}O_{5}(g) → 4 NO_{2}(g) + O_{2}(g)**

The ** integrated rate law** for a first order reaction is as follows:

$\overline{){\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{t}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{+}}{\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{0}}}}$

where:

**[A] _{t}** = concentration at time, t

**k**= decay constant

**t**= time

**[A]**= initial concentration.

_{0}Also, recall that ** half-life** is the time needed for the amount of a reactant to decrease by 50% or one-half.

The half-life of a first-order reaction is given by:

$\overline{){{\mathbf{t}}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}{\mathbf{=}}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{k}}}$

We first need to calculate for the decay constant using the given half-life, **2.81 hr**:

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Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25^{o}C. If a 1.6-L reaction vessel initially contains 750 torr of N_{2}O_{5} at 25 ^{o}C, what partial pressure of O_{2} will be present in the vessel after 220 minutes?

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