Question

rate constant of first order reaction is 0.0693 min inverse.calculate the percentage of the reactant remaining at the end of 60 min​

Answer 1

Given,

Rate constant of first order reaction, k = 0.0693 min⁻¹

For first order reaction,  

Rate constant, k = $\frac{2.303}{60}$log$\frac{[a₀]}{[a]}$, where [a₀] and [a] are the concentrations at initial value and after 60 mins respectively.

⇒log$\frac{[a₀]}{[a]}$ = $\frac{0.0693×60}{2.303}$

⇒$\frac{[a₀]}{[a]}$ = antilog(1.805)

                                        = 63.82 ≅ 64

So, percentage of the reactant remaining at end of 60 min

= $\frac{[a]}{[a₀]}$×100

= $\frac{1}{64}$×100

= 1.56%