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# The reaction 2A → A2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M–1min–1. If the initial concentration of A was 2.75 M, what was the concentration of A (in M) after 180.0 min?

Mar 7, 2018

The concentration of A was 0.195 mol/L.

#### Explanation:

The chemical equation for your reaction is

${\text{2A" → "A}}_{2}$

Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.

The integrated rate law for a second-order reaction is

$\frac{1}{{\left[\text{A"]_t ) = 1/(["A}\right]}_{0}} + k t$

where

$\frac{1}{{\left[\text{A}\right]}_{t}} =$ the concentratom of $\text{A}$ at time $t$
$\frac{1}{{\left[\text{A}\right]}_{0}} =$ the concentratom of $\text{A}$ at time $t = 0$ (i.e., at the beginning of the reaction)
$k \textcolor{w h i t e}{m l l} =$ the rate constant for the reaction

In this problem,

["A"]_0 = "2.75 mol/L"
$k \textcolor{w h i t e}{m l} = \text{0.0265 L·mol"^"-1" "min"^"-} 1$
$t \textcolor{w h i t e}{m l l} = \text{180.0 min}$

1/(["A"]_t ) = 1/("2.75 mol/L") + "0.0265 L·mol"^"-1" color(red)(cancel(color(black)("min"^"-"1))) × 180.0 color(red)(cancel(color(black)("min")))

$= \text{0.3636 L·mol"^"-1" + "4.770 L·mol"^"-1" = "5.134 L·mol"^"-1}$

["A"]_t = 1/"5.314 L·mol"^"-1" = "0.195 mol/L"

The concentration of A was 0.195 mol/L.