Question

at 300 K the reaction below obeys the rate law rate=k[NOCl]² where k= 2.8x10^-5 M^-1* S^-1

2NOCl ------> 2NO + Cl₂

Suppose 1.0 mole of NOCl is introduced into a 2.0- liter container at 300 k. How much NOCl will remain after 30 minutes?

the answer is 0.95 mol but how do you get this?

Answer 1

from the unit of rate constant, it is evident that the reaction is of 2nd order

Now, initial concentration of NOCl = moles of NOCl/volume of solution in litres = 1/2 = 0.5 M

Now, the governing equation for 2nd order reaction is :-

rate constant(k)*time(t) = {1/[A]} - {1/[A₀]} ; where [A₀] & [A] are the initial concentration and concentration of NOCl after time 't' respectively

Thus, (2.8*10^-5)*30*60 = (1/[A]) - (1/0.5)

or, 1/[A] = 2.0504

or, concentration after 30 min = [A] = 0.4877 M

thus, moles of NOCl left = molarity*volume in litres = 0.4877*2 = 0.9754 moles