Question

At 450 K the rate constant is 15.4 atm-1s-1. How much time (in s) is needed for NOCl originally at a partial pressure of 56 torr to decay to 14.5 torr?

Answer 1

This is an incomplete question, here is a complete question.

Consider the second-order decomposition of nitroysl chloride:

`2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)`

At 450 K the rate constant is 15.4 atm⁻¹s⁻¹. How much time (in s) is needed for NOCl originally at a partial pressure of 53 torr to decay to 10.6 torr?

Answer : The time needed for NOCl is, 2.52 seconds.

Explanation : Given,

Rate constant =
`15.4atm^(-1)s^(-1)`

Initial partial pressure of NOCl = 56 torr = 0.0737 atm

final partial pressure of NOCl = 14.5 torr = 0.0191 atm

The expression used for second order kinetics is:

`kt=(1)/([A_t])-(1)/([A_o])`

where,

k = rate constant

t = time

`[A_t]` = concentration at time 't'

`[A_o]` = initial concentration

As we know that,

`PV=nRT\text{ or }PV=CRT`

Thus, the expression of second order kinetics will be:

`kt=RT* (1)/(P_((A_t)))-(1)/(P_((A_o)))`

`(k)/(RT)t=(1)/(P_((A_t)))-(1)/(P_((A_o)))`

As,
`k'=(k)/(RT)`

So,
`k't=(1)/(P_((A_t)))-(1)/(P_((A_o)))` ............(1)

Now put all the given values in the above expression 1, we get:

`(15.4atm^(-1)s^(-1))* t=(1)/(0.0191atm)-(1)/(0.0737atm)`

`t=2.52s`

Therefore, the time needed for NOCl is, 2.52 seconds.