g What is the half-life for a particular reaction if the rate law is rate = (1291 M⁻¹*min⁻¹)[A]² and the initial concentration of A is 0.250 M?

In this case, for the calculation of the half-life for the mentioned reaction we first must realize that considering the units of the rate constant and the form of the rate law, it is a second-order reaction, therefore, the half-life expression is:

[tex]t_{1/2}=\frac{1}{k[A]_0}[/tex]

Therefore, we obtain:

[tex]t_{1/2}=\frac{1}{1291\frac{1}{M*min}*0.250M}\\\\t_{1/2}=3.10x10^{-3}min=0.186s[/tex]

Regards.

Eduard22sly Eduard22sly · Answer 2 · 2021-12-09T15:08:12+01:00

The half-life for the reaction is 0.003 mins

From the question given above, the following data were obtained:

Rate law = (1291 M⁻¹min⁻¹)[A]²

Initial concentration of A = 0.250 M

Half-life (t½) =?

From the rate law given, we can see that the reaction is indicating a 2nd order reaction.

Thus, the half-life can be obtained as follow:

t½ = 1 / K[A₀]

t½ = 1 / (1291 × 0.25)

t½ = 0.003 mins

Therefore, the half-life for the reaction is 0.003 mins

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