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In: Chemistry

We noted in an earlier practice exercise that at 25 celsius the decomposition position of N2O5(g)...

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?

Solutions

Expert Solution

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

queen_honey_blossom answered 2 months ago
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