Respuesta :
This is an incomplete question, here is a complete question.
Consider the second-order decomposition of nitroysl chloride:
[tex]2NOCl(g)\rightarrow 2NO(g)+Cl_2(g)[/tex]
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 = [tex]15.4atm^{-1}s^{-1}[/tex]
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:
[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]
where,
k = rate constant
t = time
[tex][A_t][/tex] = concentration at time 't'
[tex][A_o][/tex] = initial concentration
As we know that,
[tex]PV=nRT\text{ or }PV=CRT[/tex]
Thus, the expression of second order kinetics will be:
[tex]kt=RT\times \frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}[/tex]
[tex]\frac{k}{RT}t=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}[/tex]
As, [tex]k'=\frac{k}{RT}[/tex]
So, [tex]k't=\frac{1}{P_{(A_t)}}-\frac{1}{P_{(A_o)}}[/tex] ............(1)
Now put all the given values in the above expression 1, we get:
[tex](15.4atm^{-1}s^{-1})\times t=\frac{1}{0.0191atm}-\frac{1}{0.0737atm}[/tex]
[tex]t=2.52s[/tex]
Therefore, the time needed for NOCl is, 2.52 seconds.
