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A reaction of the form aA h Products is second order with a rate constant of 0.169 L/(mol āˆ™ s). If the ini- tial concentration of A is 0.159 mol/L, how many seconds would it take for the concentration of A to decrease to 5.50 Ɨ 10āˆ’3 mol/L?

Respuesta :

Answer:

1038.6324 seconds would it take.

Explanation:

The integrated rate law for second order is:

[tex]\frac {1}{[A]_t}=kt+\frac {1}{[A]_0}[/tex]

Given:

[tex][A]_t=5.50\times 10^{-3}\ M[/tex]

[tex][A]_0=0.159\ M[/tex]

K = 0.169 L/mol.s

So,

[tex]\frac {1}{5.50\times 10^{-3}}=0.169\times t+\frac {1}{0.159}[/tex]

Solving for t, we get:

t = 1038.6324 s

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