Question

In: Chemistry

Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction...

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 ∘C.

If a 1.7-L reaction vessel initially contains 760 torr of N2O5 at 25 ∘C, what partial pressure of O2 is present in the vessel after 205 minutes?

Solutions

Expert Solution

half- life = 2.81 h

rate constant = k = 0.693 / t1/2 = 0.693 / 2.81

                    k = 0.247 h-1

initial pressure = 760 torr

t = 205 min = 205 / 60 = 3.42 hour

k = 1/t * ln (Po / Pt)

0.247 = 1/3.42 * ln (760 / Pt)

Pt = 326.6 torr

remaining pressure of N2O5 after 205 min

2N2O5 -----------------------> 4 NO2 + O2

760                                       0             0 ----------------> initial

760 -2x                                4x            x ---------------------> after 205 min

760 -2x = 326.6

x = 216.7

partial pressure of O2 = x = 216.7 torr


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