The half life of the following first order reaction is .given that tha initial pressure of c2h4o2 is 84 mm and total pressure at the end of 16 minutes is 110mm.
Answers
Answered by
0
In [A] = -kt + ln [A]o
[A]o = initial conc = 0.0086 M
t = 2.4 hours = 2.4 X 60 X 60 = 8640 s
k = 5.18 X 10^-4 s^-1
[A] = conc. after 2.4 hours
putting the values...
ln [A] = -5.18 X 10^-4 X 8640 + ln 0.0086
ln [A] = -4.476 - 4.756
ln [A] = -9.232
converting ln to log ...multiplying by 2.303
2.303 X log [A] = -9.232
log [A] = -9.232/2.303 = -4.009
taking antilog ....
[A] = 10^-4.009 = 9.795 X 10^-5 M
Similar questions