Dinitrogen Pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition...

Dinitrogen Pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with a rate constatnt at 50 degrees Celcius 1.75 x 10^{-5 s-1}. Calculate the partial pressure of O2 produce from 1.00L of 0.500 M N2O5 solution at 50 degrees celcius over a period of 18.0 hours if the gas collected in a 10.0L container. (Assume that the products do not dissolve in chloroform.)

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Expert Solution

Activation energy plays no role in this problem. It would be of interest if you want to evaluate rates of reaction on different temperature levels.

From the given information you can calculate the change of reactant concentration with time. From this you get the change of concentration and subsequently the number of moles of reactant, which have reacted away. The amount of product formed is proportional to that. Then use ideal gas law to calculate the partial pressure.

How it works in detail:

We've got a first order reaction. i.e.
d[N?O?]/dt = - k?[N?O?]
Solving this differential equation with initial concentration [N?O?]? leads to the integrated rate law:
ln[N?O?] = - k?t + ln[N?O?]?

From this you can calculate the concentration of nitrogen pentoxide after 18h = 64800s
ln[N?O?] = - k?t + ln[N?O?]?
<=>
ln( [N?O?] / [N?O?]? ) = - k?t
<=>
[N?O?] = [N?O?]? ? e^(- k?t)
= 0.5M ? e^(-1.75

queen_honey_blossom answered 11 months ago

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