Question

This reaction was monitored as a function of time:

Answer 1

A) Since the plot 1/[AB] vs time gives straight line, the order of the reaction with respect to A is second order:
rate constant, K = slope = 5.5 x 10⁻² M⁻¹S⁻¹

b) Rate law : Rate = k[AB]²

c) half life period of the 2nd order is inversely proportional to the initial concentration of the reactants 
t 1/2 = $\frac{1}{K}$. $\frac{1}{A0}$
t 1/2 = $\frac{1}{(5.5 x 10^{-2}) (0.55M)} = 33 s$

d) k = 5.5 x 10⁻² M⁻¹s⁻¹
Initial concentration of AB, [A₀] = 0.250 M
concentration of AB after 75 s = [A]
k = $\frac{1}{t} [ \frac{1}{[A]} - \frac{1}{[Ao]} ]$
[A] = 0.123 M
Equation: AB → A + B
  concentration of AB after 75 s = 0.123 M
Amount of AB dissociated = 0.25 - 0.123 = 0.127 M
concentration of [A] produced = concentration of [B] produced = Amount of AB reacted = 0.127 M