Question

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate constant for this reaction?

Answer 1

`k=1.20x10^(-3) s^(-1)`

Step-by-step explanation:

For a first order reaction the rate law is:

`v=(-d[A])/([A])=k[A]`

Integranting both sides of the equation we get:

`\int\limits^a_b {(d[A])/([A])} \, dx =-k\int\limits^t_0 {} \, dt`

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

`ln([A])/([A]_(0) ) =-kt`

`[A]=[A]_(0)e^(-kt)`

`ln[A]=ln[A]_(0) -kt`

`k=(ln[A]_(0)-ln[A])/(t)`

therefore k is

`k=(ln1-ln0,485)/(600)=1,20x10^(-3)`