In: Chemistry
We noted in an earlier practice exercise that at 25 celsius the decomposition position of N2O5(g) into NO2(g) follows first order kinetics with k= 3.4 x 10^-5 s^-1. How long will it take for a sample originally containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr?
For a first order reaction rate constant , k = ( 2.303 /t )x log (Po /P)
Where
Po = initial pressure = 2.0 atm = 2.0 x 760 torr Since 1 atm = 760 torr
= 1520 torr
P = partial pressure at time t = 380 torr
t = time = ?
k = rate constant = 3.4 x 10 -5 s-1
Plug the values we get k = ( 2.303 /t )x log (Po /P)
t = ( 2.303 /k )x log (Po /P)
= ( 2.303 / (3.4 x 10 -5 ))x log (1520/380)
= 40780 s
Therefore the time required is 4070 s