1. 2 N2O5  4 NO2 + O2 RATE = k [N2O5] = 1.0 * 10-5 /S
CH 14 #98
2 N2O5  4 NO2 + O2
RATE = k [N2O5] = 1.0 * 10-5 /S
20 hr = 7.20 * 104 s, [N2O5] = 0.600
ln[A]t – ln[A]0 = -kt
ln[N2O5]t =-kt + ln[N2O5]0
ln[N2O5]t = -1.0 * 10-5 (7.2*104 s) + ln(0.600) =
ln[N2O5]t = e =0.292 M

2. Final concentration (0.292) – initial (0.600) = 0.308 mol
2 N2O5  4 NO2 + O2
0.600 mol
-2x = 0.308 4x x
0.292
mol
0.616
0.154
Final concentration (0.292) – initial (0.600) = mol
Use PV=nRT to convert moles of oxygen to P. it is atm