Rate law seconds and concentration

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The discussion focuses on a second-order reaction involving the decomposition of AB into A and B, with a rate constant of 0.2 L/mol∙S. The initial concentration of AB is 1.50 M. To determine how long it takes for the concentration of AB to reach one-third of its initial value, calculations show that it takes approximately 6.67 seconds. After 10 seconds, the concentration of A is calculated to be 0.375 M. Emphasis is placed on the importance of including units in calculations, specifically seconds for time and mol/L for concentration.
Mag
could you check this?

AB(g) → A(g) + B(g)

Rate = k[AB]^2
k = 0.2L/mol∙S

How long will it take for [AB] to reach 1/3 (one third) of its initial concentration of 1.50M? What is [AB] after 10.0 seconds?

2nd order overall reaction so:
part A
1/[A] - 1/[A]o = (k)(t)
1/0.5 - 1/1.50 = (0.2)(t)
6.667 = t

part B
1/[A] - 1/1.5 = (0.2)(10 s)
[A] = 0.375
 
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Correct, and correct !

Just remember to put in the units; time in seconds and concentration in mol/L or M (molar).
 
okie dokie. thanks.
 
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