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:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

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:

$kt=RT\times \frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}$

$\frac{k}{RT}t=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}$

As, $k'=\frac{k}{RT}$

So, $k't=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}$         ............(1)

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

$(15.4atm^{-1}s^{-1})\times t=\frac{1}{0.0191atm}-\frac{1}{0.0737atm}$

$t=2.52s$

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