Ans :-
For second order reaction integrated rate equation is :
kt = 1/[A]t - 1/[A]0 ....................(1)
Where, t = time period = 67 s
k = Rate constant = 0.80 M-1s-1
[A]0 = Initial concentration of reactant = 0.86 M
[A]t = Concentration of reactant at time t = ?
put all these values in equation (1)
(0.80 M-1s-1).(67 s) = 1/[A]t - 1/0.86 M
53.6 M-1 = 1/[A]t - 1/0.86 M
53.6 M-1 = 1/[A]t - 1.16279 M-1
53.6 M-1 + 1.16279 M-1 = 1/[A]t
54.76279 M-1 = 1/[A]t
Taking reciprocal on both side
[A]t = 1 / 54.76279 M-1
[A]t = 0.01826 M
Therefore, concentration of NOBr = 0.01826 M
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