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The rate constant for this second‑order reaction is 0.990 M − 1 ⋅ s − 1 0.990 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.650 M 0.650 M to 0.380 M?

Respuesta :

Answer : The time taken would be, 1.10 s

Explanation :

The integrated rate law equation for second order reaction follows:

[tex]k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)[/tex]

where,

k = rate constant = [tex]0.990M^{-1}s^{-1}[/tex]

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.380 M

[tex][A]_o[/tex] = Initial concentration = 0.650 M

Now put all the given values in above equation, we get:

[tex]0.990=\frac{1}{t}\left (\frac{1}{(0.380)}-\frac{1}{(0.650)}\right)[/tex]

[tex]t=1.10s[/tex]

Hence, the time taken would be, 1.10 s

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