ch 14 #98
DESCRIPTION
CH 14 #98. 2 N 2 O 5 4 NO 2 + O 2 RATE = k [ N 2 O 5 ] = 1.0 * 10 -5 /S 20 hr = 7.20 * 10 4 s, [ N 2 O 5 ] = 0.600 ln[A] t – ln[A] 0 = -kt ln[A] t – ln[A] 0 = -kt ln[ N 2 O 5 ] t =-kt + ln[ N 2 O 5 ] 0 ln[ N 2 O 5 ] t = -1.0 * 10 -5 (7.2*10 4 s) + ln(0.600) = -1.231 - PowerPoint PPT PresentationTRANSCRIPT
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[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) = -1.231
ln[N2O5]t = e-1.231 =0.292 M
2 N2O5 4 NO2 + O2
0.600 mol 0 0
-2x = 0.308
4x x
0.292
mol
0.616
mol
0.154
mol
Final concentration (0.292) – initial (0.600) = 0.308
mol
Use PV=nRT to convert moles of oxygen to P. it is 0.402 atm