Given a second order process: 2 AB → A 2 + B 2 with k = 0.0908 M-1min-1; and the initial concentration of AB = 0.941 M, calculate the concentration of A2 after exactly 20 min. Enter the result with 3 sig. figs. exponential notation and no units.
Answered Essay: Given a second order process: 2 AB → A 2 + B 2 with k = 0.0908 M-1min-1; and the initial
Expert Answer
Integrated form of second order equation is
1/[A]t = kt + 1/[A]0
where,
[ A ]t = concentration of reactant at time t
[ A ]0 = concentration of reactant at initial,0.941M
t = time , 20 min
k = rate constant , 0.0908M^-1 s^-1
Therefore,
1/[A]t = (0.0908M^-1 min^-1×20min) + 1/0.941M
=1.816M^-1 + 1.0627M^-1
=2.878M^-1
[A]t = 0.3475M
Concentration of AB at 20min = 0.3475M
Concentraion of AB consumed = 0.9410M – 0.3475M= 0.5935M
concentraion of A2 formed = concenration of AB consumed/2
= 0.5935M/2
= 0.297M
So , the answer is 0.297M
