Question

A reaction of the form aA h Products is second order with a rate constant of 0.169 L/(mol ∙ 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 × 10−3 mol/L?

Answer 1

Answer:

1038.6324 seconds would it take.

Explanation:

The integrated rate law for second order is:

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

Given:

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

$[A]_0=0.159\ M$

K = 0.169 L/mol.s

So,

$\frac {1}{5.50\times 10^{-3}}=0.169\times t+\frac {1}{0.159}$

Solving for t, we get:

t = 1038.6324 s