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CH 14 #98

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

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CH 14 #98

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  1. 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. 2 N2O5 4 NO2 + O2 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

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