Question

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?

Answer 1

Given that;

half-life t1/2 = 2.81 h = 168.6 min and time = 205 min

rate constant k = 0.693 / t1/2

                          = 0.693 / 168.6

                          = 0.00411 min-1

2 N2O5 (g) ----------------------> 4 NO2 (g) + O2 (g)

760 - 2x       4x    x

k = 1/t ln (Po / Pt)

0.00411 = 1 / 205 ln (760 / 760 - 2x)

0.00411= 0.00488 ln (760 / 760 - 2x)

0.842 = ln(760 / 760 - 2x)

760 / 760 - 2x= e^0.842

760 / 760 - 2x= 2.32

760= 17632- 4.64 x

4.64 x= 1003.2

X = 1003.2/4.64

= 216.2 torr

partial pressure of O2 is = 216 torr