Question

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.

Answer 1

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}