We know, for first order kinetics,
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.
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