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?

1 Answer
Mar 7, 2018

Answer:

The concentration of A was 0.195 mol/L.

Explanation:

The chemical equation for your reaction is

#"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

#1/(["A"]_t ) = 1/(["A"]_0) + kt#

where

#1/(["A"]_t ) = # the concentratom of #"A"# at time #t#
# 1/(["A"]_0) =# the concentratom of #"A"# at time #t = 0# (i.e., at the beginning of the reaction)
#kcolor(white)(mll) =# the rate constant for the reaction

In this problem,

#["A"]_0 = "2.75 mol/L"#
#k color(white)(ml)= "0.0265 L·mol"^"-1" "min"^"-"1#
#tcolor(white)(mll) = "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")))#

#= "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.