A plot of #ln[A]# vs. #t# for the reaction #A -> B + C# gives a straight line, with slope #-4.6 xx 10^(-3) "s"^(-1)#. Calculate the rate constant?

1 Answer
Truong-Son N.
Sep 28, 2017

Since a plot of #ln[A]# vs. #t# for the reaction

#A -> B + C#

gives a straight line, it is a first-order reaction, whose integrated rate law is

#overbrace(ln[A])^(y) = overbrace(-k)^(m)overbrace(t)^(x) + overbrace(ln[A]_0)^(b)#,

where:

  • #k# is the rate constant in the appropriate units at the temperature of the reaction.
  • #[A]# is the molar concentration of #A#.
  • #[A]_0# is the initial molar concentration of #A#.

You can see that the integrated rate law is of the form

#y = mx + b#,

so the slope is #-k#. Therefore, the rate constant is just:

#color(blue)(k) = -"slope"#

#= color(blue)(4.6 xx 10^(-3) "s"^(-1))#

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