In: Chemistry
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?
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