Question

If a reaction is first order with a rate constant of 5.48*10^-2 sec^-1, how long is required for 3/4 of the initial concentration of reactant to be used up?

Answer 1

Given that K for first order reaction = 5.48*10^-2 sec^-1

The integrated first order rate law :

ln[A] = -k∙t + ln[A]₀
(k rate constant, [A]₀ initial concentration)

To arrange the above expression to calculate the time; t:
t = ln( [A]₀/ [A] ) / k

When 3/4 of initial reactant has consumed 1/4 is still there, i.e.
[A] = (1/4)[A]₀

Now put all the values and calculate the time:

t = ln( [A]₀/ [A] ) / k

t = ln( [A]₀/ (1/4)[A]₀ ) / 5.48*10^-2 sec^-1

t= ln( 4 ) / 5.48×10⁻² s⁻¹

t= 1.386 / 5.48×10⁻² s⁻¹

t = 25.29 s

Hence 25.29 s is required for 3/4 of the initial concentration of reactant to be used up.