Rate law seconds and concentration

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SUMMARY

The discussion focuses on a second-order reaction involving the decomposition of AB(g) into A(g) and B(g), with a rate constant (k) of 0.2 L/mol∙S. The calculations demonstrate that it takes 6.667 seconds for the concentration of AB to decrease to one-third of its initial concentration of 1.50 M. Additionally, after 10 seconds, the concentration of AB is calculated to be 0.375 M. Proper unit notation is emphasized, specifically using seconds for time and mol/L for concentration.

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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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