The thermal decom

We know, for first order kinetics,

k =

where, k = rate constant.

t = time taken.

a =initial concentration.

a-x = concentration left after t amount of time.

Now, k=2.4 x 10^-3 s^-1 as given in the question.

Now, we have to find out the time taken for the decomposition of three-fourths of the compound.

Decomposed part(x) = 0.75a

2.4 x 10^-3=

∴ t =

= = 577.6sec. [If, log x = a, then log () = -a]

So, it will it take 577.6 sec for three- fourths of initial quantity of HCO₂H to decompose.