This reaction was monitored as a function of time:
1 answer:
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 =
. 
t 1/2 =
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]} ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bt%7D%20%5B%20%5Cfrac%7B1%7D%7B%5BA%5D%7D%20-%20%20%5Cfrac%7B1%7D%7B%5BAo%5D%7D%20%5D)
[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
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 =
t 1/2 =
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 =
[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
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